Download Chapter 5A - Polynomial Functions

©
: Pre-Calculus - Chapter 5A
Chapter 5A - Polynomial Functions
Definition of Polynomial Functions
A polynomial function is any function px of the form
px  p n x n  p n−1 x n−1    p 2 x 2  p 1 x  p 0 ,
where all of the exponents are non-negative integers.
The coefficients of the various powers of x, that is, the p i ’s are assumed to be known real
numbers, and p n ≠ 0. The degree of this polynomial function is n. If n is even the
polynomial is said to be of even degree and if n is odd, the polynomial is said to be of odd
degree.
The coefficient p 0 is called the constant term and the non-zero p n is called the leading
coefficient.
By a polynomial of degree 0, we mean a non-zero constant function. The zero polynomial
function is sometimes said to have degree equal to − .
In the table below we list a few polynomials and their degrees.
polynomial
degree constant term leading coefficient
px  5
0
5
5
px  3x − 8
2
−8
3
px  −6x  x − 5x  9
5
9
−6
px  2x 126
126
0
2
2
5
4
We have studied two examples of polynomial functions already. Linear functions (fx  mx  b,
m ≠ 0) are polynomial functions of degree 1, and quadratic functions (fx  ax 2  bx  c, a ≠ 0) are
polynomial functions of degree 2.
Below are some examples of polynomials of degree 3, 4, and 5.
Example 1:
Let px  5 − 6x  13x 3 . What is the degree of px? What are the leading coefficient
and constant terms?
Solution:
The degree of px  5 − 6x − 13x 3 is 3. The largest exponent. The leading coefficient
is −13 and the constant term is 5.
Example 2:
Let px  3x 4 − 3x 3  5x. What is the degree of px? What are the leading coefficient
and constant terms?
Solution:
The degree of px  3x 4 − 3x 3  5x is 4. The leading coefficient is 3, and the constant
term, the coefficient of x 0 , is 0.
Example 3:
If fx  16 − 3x 2  27x 4 − 5x 7  8x 13 , what are the degree, leading coefficient, and
constant term equal to?
Solution:
The degree of fx  16 − 3x 2  27x 4 − 5x 7  8x 13 is 13. The leading coefficient is 8,
and the constant term is 16.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5A
Behavior of Polynomial Functions
In this page we will discuss how the values px of polynomial functions behave for large values of x,
both positive and negative. It turns out that the behavior of polynomial functions for large values of x
is determined by the degree of the polynomial. Polynomials of odd degree behave one way, think of the
graph of a linear function, and polynomials of even degree behave in a different way, picture the graph
of a quadratic function.
Example 1:
Solution:
Plot the third degree polynomial x 3 .
y
x
Notice that the function values of this polynomial as x goes to  also go to , and as x goes to −  the
function values also go to − .
Example 2:
Solution:
Plot the third degree polynomial −x 3 .
y
x
Notice here that as x goes to  the function values go to − . The reverse of the situation with x 3 . What
is true, for both examples, is that as x gets large so to do the function values, and if one end of the graph
goes up, the other end goes down.
Example 3:
Solution:
Plot the fourth degree polynomial fx  x 4 .
y
x
Notice both ends of the graph get large in the same direction.
Example 4:
Solution:
Plot the fourth degree polynomial fx  −x 4 .
y
x
Notice that both ends of the function get large in the same direction.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5A
What do you expect the graph of x 5 to look like?
Question:
Answer:
The graph of x 5 is the graph of a polynomial of odd degree, so I would expect it (for
large values of x) to look like the graph of x 3 or the graph of x. In the sense that if one end goes to plus
infinity, then the other end goes to minus infinity.
What do you expect the graph of x 16 to look like?
Question:
Answer:
The graph of x 16 is the graph of a polynomial of even degree, so I would expect it (for
large values of x) to look like the graph of x 4 or the graph of x 2 . In the sense that if one end goes to plus
infinity, then the other end goes to plus infinity also.
We summarize our findings:
Theroem:
Let px  p 0  p 1 x    p n x n be a polynomial of odd degree.
If p n  0 , then as x goes to positive infinity px goes to positive infinity,
and as x goes to negative infinity px goes to negative infinity
If p n  0 , then as x goes to positive infinity px goes to negative infinity,
and as x goes to negative infinity px goes to positive infinity
Proof:
The idea of the proof is fairly simple, we factor out the highest power of x in px,
and notice how each factor must behave for large values of x.
px  p 0  p 1 x    p n x n
p1
p
p
 x n n0  n−1
   n−1
x  pn
x
x
Now if we examine each of the terms in the second factor we see that as x gets large either
positively or negatively every one of the quotients must get smaller and smaller. That is every
p
term which of the form n−ii goes to zero as long as the exponent n − i is positive. So, for large
x
x the second factor gets closer and closer to p n , and for large x,
px ≈ p n x n . Remember, n is an odd integer.
If x goes to positve infinity, x n goes to positive infinity. Thus, if p n  0, px will go to positive
infinity. If x goes to negatve infinity, then x n goes to negative infinity too, and if p n  0, then
px will go to negative infinity. Remember that n is odd, so the sign of x n is the same as the
sign of x. If p n  0, the situation is reversed.
The way to remember this theorem is to remember what the graph of x 3 looks like compared to the
graph of −x 3 . The leading coefficient of x 3 is positive and the leading coefficient of −x 3 is negative.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5A
Example 5:
Show that as x goes to positive infinity the odd degree polynomial
px  3 − x 2  5x 4 − 6x 9 goes to − .
Solution:
The theorem states that if the leading coefficient of an odd degree polynomial is
negative, then as x goes to infinity the values of the polynomial go to negative infinity. The following
calculation which parrots the above proof reinforces this.
px  3 − x 2  5x 4 − 6x 9
 x 9 39 − 17  55 − 6 .
x
x
x
Now as x goes to , the second factor goes to −6, which means that px, for large values of x, looks
like −6x 9 . Thus, px must go to −  as x goes to .
Example 6:
Show that as x goes to negative infinity the odd degree polynomial
px  3 − x 2  5x 4 − 6x 9 goes to .
Solution:
The theorem states that if the leading coefficient of an odd degree polynomial is negative,
then as x goes to negative infinity the polynomial values go to positive infinity. The following
calculation which parrots the above proof reinforces this.
px  3 − x 2  5x 4 − 6x 9
 x 9 39 − 17  55 − 6 .
x
x
x
As x goes to −  the second factor gets close to −6 just as in the previous example. Here though, as x
goes to −  the term x 9 also goes to − . Thus, the product of −6 and x 9 will go to positive infinity.
Example 7:
What will the graph of px  5 − x  3x 4 − 8x 7 look like for large values of x?
Solution:
px is an odd degree polynomial and its leading coefficient, −8, is negative. Thus as x
goes to infinity, px will go to − , and as x goes to − , px will go to positive infinity.
Theorem:
Let px  p 0  p 1 x    p n x n be a polynomial of even degree.
If p n  0 , then as x goes to positive infinity px goes to positive infinity,
and as x goes to negative infinity px goes to positive infinity
If p n  0 , then as x goes to positive infinity px goes to negative infinity,
and as x goes to negative infinity px goes to negative infinity
Proof:
The idea of the proof is exactly the same as in the case where px is a polynomial of
odd degree.
px  p 0  p 1 x    p n x n
p1
p
p
 x n n0  n−1
   n−1
x  pn
x
x
Now if we examine each of the terms in the second factor we see that as x gets large either
positively or negatively every one of the quotients must get smaller and smaller. That is every
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5A
p
term which of the form n−ii goes to zero as long as the exponent n − i is positive. Thus, for
x
large x the second factor gets closer and closer to p n . Thus, for large x,
px ≈ p n x n .
So if x goes to plus infinity x n goes to plus plus infinity. Thus, if p n  0, px will go to plus
infinity. If x goes to minus infinity, then x n goes to plus infinity, and if p n  0, then px will
go to plus infinity. Remember that n is even, so the sign of x n is always positive. If p n  0, the
situation is reversed.
The way to remember this theorem is to remember what the graph of x 4 looks like compared to the
graph of −x 4 . The leading coefficient of x 4 is positive and the leading coefficient of −x 4 is negative.
Example 8:
Describe what happens to the values of px  −5  5x − 8x 3 − x 4  8x 10 as x goes to
plus and minus infinity.
Solution:
The theorem states that for even degree polynomials if the leading coefficient is
positive then the values of px must go to positive infinity for large values of x whether positive or
negative. The following calculation parrots the above proof and is given here to reinforce these ideas.
px  −5  5x − 8x 3 − x 4  8x 10
 x 10 − 510  59 − 87 − 16  8 .
x
x
x
x
As x goes to infinity the second factor gets close to 8. Thus, px looks like 8x 10 , and as x gets large,
positively or negatively, x 10 goes to positive infinity and so will 8x 10 .
Example 9:
What will the graph of px  9x 3 − 15x 16 look like for large values of x?
Solution:
This is an even degree polynomial whose leading coefficient (−15) is negative. Thus,
from the theorem, as x goes to positive or negative infinity px goes to negative infinity.
Let’s see what happens if we factor out x 16 .
px  9x 3 − 15x 16  x 16 913 − 15 .
x
9
As x gets large the term 13 gets smaller and smaller, so the second factor gets close to −15. Thus, for
x
large x, ps looks like −15x 16 , and as x gets large, either positively or negatively, −15x 16 goes to − .
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5A
Exercises for Chapter 5A - Polynomial Functions
1.
What is the constant term in the polynomial px  3x 4 − 5x 2  13 − 2x  x 5 ?
2.
What is the leading coefficient of −2x  3x 2 − 8 ?
3.
If px is a polynomial of degree 2, with leading coefficient −3, constant term equal to 8, and
p1  5, what is px?
4.
Which of the following expressions is a polynomial?
4
a. x 2 − 3 b. −x 2  x c. x − 6x  2 d. 5x e. 5 x
5 − 83
5.
What are the leading coefficient and constant term of px  4 − 5x  11x 2  7x 9 ?
6.
What are the degrees of the following polynomials?
a. 9x 3  5x 4 − 16x  2 b. 9x − 12 c. 22x 45 − 17x 23 − 1 d. 3x − 5  67x 5  x 4
7.
Describe the behavior of x 3 for large values of x
8.
If px is a polynomial of degree 5 and values of px go to  as x goes to − , what happens
to px for large positive values of x?
9.
Can a polynomial px have the following properties: px is a polynomial whose leading
coefficient is −2 and px goes to plus infinity as x gets large in either direction?
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5A
Answers to Exercises for Chapter 5A - Polynomial
Functions
1.
13
2.
3
3.
We know that px  −3x 2  bx  8. We also know that 5  p1  −3  b  8  b  5 .
This implies that b  0, and hence, px  −3x 2  8 .
4.
Items a, c, and e are polynomials.
5.
The leading coefficient is 7 and the constant term is 4.
6.
a. 4
7.
For large values of positive x the values of the polynomial x 3 get very large positively. For
large negative values of x the polynomial x 3 also gets large in a negative sense.
8.
The values of px go to −  as x goes to  .
9.
No, since px behaves the same as x gets large in either direction, the degree of px must
be even. But, if the leading coefficient of this polynomial is negative, then the polynomial’s
values must go to minus infinity, not plus infinity.
©
b. 1
: Pre-Calculus
c. 45
d. 5
©
: Pre-Calculus - Chapter 5B
Chapter 5B - Rational Functions
Definition of a Rational Function
In this page we define a rational function and look at some graphs of rational functions.

Definition:
A rational function is the quotient of two polynomials. That is, if rx is a
px
.
rational function, then there are two polynomials px and qx such that rx 
qx
− 5 . The quotient of px and
If px  3x − 5 and qx  x 2 − 2x, then rx  3x
x 2 − 2x
− 5 is shown below. Notice the behavior of rx for x
qx, is a rational function of x. A plot of 3x
x 2 − 2x
close to 0 or 2. We say that the function has vertical asymptotes at x  0 and at x  2.
Example 1:
y
x
x=2
px
. Since
qx
polynomials are defined for all x, the only real numbers which are not in the domain of any rational
function are those values of x for which the denominator is zero. Thus, the domain of any rational
px
is the set of x for which qx ≠ 0.
function
qx
The domain of a rational function is the set of real numbers x for which one can compute
If we look at the rational function in the above example, its denominator equals x 2 − 2x  xx − 2.
Thus, the numbers x  0 and x  2 (The values for which the denominator equals zero.) are not in the
− 5 , and all other values of x are in the domain.
domain of the function 3x
x 2 − 2x
The x-intercepts of a rational function rx are those real numbers x for which rx  0. Note that a
fraction can equal zero only if its numerator is zero. Thus, the x-intercepts are those values of x for
which the numerator px  0.
The y-intercept is that number y 0 such that r0  y 0 .
Notice that there may be many x-intercepts, but there is at most one y-intercept.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
Question:
When will there be no y-intercept?
Answer:
When 0 is not in the domain of the rational function.
− 5 , x  0 is not in the domain of this rational function, hence
In the previous example rx  3x
x 2 − 2x
there is no y-intercept. The x-intercept is x  5/3. Since that is the only value of x for which the
numerator of rx is zero, there is only one x-intercept.
Example 2:
2
Find the domain and intercepts of the rational function rx  x −3 4x − 2 .
x −1
Solution:
To determine the domain we need to find those real numbers x for which x 3 − 1  0.
The only solution to this equation is x  1. Thus, the domain consists of all real numbers except x  1.
The y-intercept is r0  −2  2. The x-intercepts are those values of x for which x 2 − 4x − 2  0.
−1
The solutions are x  2  6 and x  2 − 6 . A plot of this rational function is shown below.
y
x=1
x
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
Horizontal Asymptotes
There are times when we need to understand how a rational function behaves for very large (positive
and negative) values of x. Before defining what a horizontal asymptote is we’ll look at the graphs of
several rational functions.
Example 1:
Let rx  1x . Take note that the domain of this function is all non-zero x and it has
no intercepts. We have plotted 1x below. Notice that for large values of x, either positive or negative,
the function values get close to zero. We describe this phenomenon by saying that the line y  0 is a
horizontal asymptote.
y
x
2
Let rx  2x 2− x  5 . Since the denominator is never zero, the domain is all real
x 1
numbers. The y-intercept is the point 0, 5, and the x-intercepts are the solutions of the equation
2x 2 − x  5  0, which has no real solutions. (Use the quadratic formula.). Thus, there are no
x-intercepts. An examination of a plot of this rational function shows that the line y  2 is a horizontal
asymptote.
Example 2:
y
y=2
x
Notice that the graph crosses the horizontal asymptote. There is no truth to the idea that the graph of a
function cannot cross a horizontal asymptote.
In order to understand the concept of a horizontal asymptote we need to understand the idea of the
limiting behavior of a function rx as x tends to plus or minus infinity. In the table below some values
2
of rx  2x 2− x  5 are shown for large values of positive and negative x.
x 1
x −2 −20
−200
−2000 2
20
200
2000
rx 3
2. 0574 2. 0051 2. 0005 2. 2 1. 9576 1. 9951 1. 9995
This computation indicates that for large values of x, positive or negative, the function values of rx
are getting close to 2.
The following notation is used to indicate this.
lim
rx  2 and lim rx  2.
x→
x→−
These are read as ”The limit, as x goes to infinity, of rx equals 2” and ”The limit, as x goes to
negative infinity, of rx equals 2” respectively.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
If you go on to take calculus, you will see this notation a lot. For now, just think of it as a shorthand to
describe what happens to the values of a function as x gets large in either a positive or negative sense.
The horizontal line y  a is called a horizontal asymptote of the function rx if
rx  a, or if lim rx  a.
lim
x→
x→−
Determine the limit as x goes to infinity of 1x .
Solution:
We first make a table of values of 1x for large values of x.
x 1 10 100 1, 000 10, 000 100, 000 10 n
1 1 . 1 . 01 . 001 . 0001 . 00001 10 −n
x
It seems clear that as x gets larger and larger the values of 1x get closer and closer to zero. Thus, we say
1  0.
the limit of 1x as x goes to infinity is zero, and we write lim
x→ x
Example 3:
Determine the domain, intercepts and horizontal asymptotes of rx  2x − 1 .
x1
Solution:
The domain consists of all x for which the denominator x  1 is not zero. Thus, the
domain equals all x ≠ −1. The y-intercept equals r0  −1, and the x-intercept equals those x for
which 2x − 1  0. That is, x  1 is the x-intercept. To find horizontal asymptotes we need to
2
determine if the values of rx approach a limiting value as x goes to plus or minus infinity. We
construct a table of values to see if such is the case.
Example 4:
x
10
100
1000
10, 000
10 6 10 10
rx 1. 727 3 1. 970 3 1. 997 1. 999 7 2. 0 2. 0
These values indicate that lim
rx  2. So we will have the line y  2 as a horizontal asymptote. A plot
x→
of 2x − 1 is shown on the next page.
x1
y
y=2
x
x= 1
Notice that rx approaches 2 as x goes to negative infinity as well.
Any function of the form 1n where n is any positive integer goes to zero as x goes to
x
plus or minus infinity. That is, each of these functions has the line y  0 as an horizontal asymptote.
To convince ourselves of this we plot several such functions for x  0, and then for x  0. The graphs
of 13 , 14 , and 15 for x  0 are plotted below.
x
x x
Example 5:
1 is in red
x3
1 is in black
x4
1 is in green
x5
y
(1, 1)
x
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
Notice that
if 0  x  1, then 13  14  15 and
x
x
x
if 1  x, then 15  14  13 .
x
x
x
It seems clear from the graphs that for large values of x each of these rational functions gets close to 0. In fact
y  0 is a horizontal asymptote for each of these functions. The graphs of 13 , 14 , and 15 for x  0 are plotted
x
x
x
below.
1 is in red
x3
1 is in black
x4
1 is in green
x5
y
( 1, 1)
x
( 1, 1)
For values of x  0, these functions behave differently than they do for x  0. For example, if the exponent is
even ( 14 in blue) then the graph is symmetric with respect to the y-axis, while if the exponent is odd ( 13 in red,
x
x
and 15 in green) the graph is odd with respect to the origin. However, regardless of the oddness or evenness of
x
the exponent, each of these functions has the line y  0 as a horizontal asymptote.
3
2
Let rx  −x 3  x 2 21x − 45 . Determine the horizontal asymptotes of rx.
x − 3x − 6x  8
Solution:
To determine if there is a horizontal asymptote we compute the limit as x →  of rx. The trick is to
divide numerator and denominator by the highest power of x which occurs in rx.
3
2
rx  −x 3  x 2 21x − 45 the highest power of x is 3
x − 3x − 6x  8
so we divide numerator and
−1  1x  212 − 453
x
x
denominator by x 3 .

1 − 3x − 62  83
x
x
Okay, we’ve rewritten rx. How does this help? The next observation is that any expression of the form cn
x
where c is any constant and n is any positive number must go to zero as the absolute value of x gets large. Thus,
rx approaches −1 as x goes to plus infinity. Thus, the line y  −1 is a horizontal asymptote. A plot of rx is
1
shown below.
Example 6:
y
x
y= 1
This trick, dividing both numerator and denominator by the highest power of x should be used everytime.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
px
qx
we need to be able to determine if the values of rx are approaching some number as x goes to positive
or negative infinity. Fortunately, there is an easy way to determine what this limiting behavior is. This
behavior is determined solely by the degrees of the polynomials px and qx.
As we saw in the previous examples, to find horizontal asymptotes of a rational function rx 
Theorem:
px
be a rational function of x, where
qx
Let rx 
px  p m x m  p m−1 x m−1    p 0 and
qx  q n x n  q n−1 x n−1    q 0 .
m is the degree of px and n the degree of qx, then
a. If m  n (degree of denominator larger than degree of numerator) lim rx  0.
x→
p
b. If m  n, (numerator and denominator have the same degree) then lim rx  qmn . The ratio
x→
of the coefficients of the largest powers of x.
c. If m  n, (degree of denominator smaller than degree of numerator) then rx has no limit as
x tends to plus or minus infinity. In fact, the absolute value of rx becomes arbitrarily large.
Example 7:
Let rx 
2x − 5 . Calculate the limit of rx as x goes to infinity.
x2  x − 6
Solution:
Since the degree of the numerator is 1, and this is less than 2 which is the degree of the
denominator, we know that lim rx  0
x→
Example 8:
3
− 7x  15 . Calculate the limit of rx as x goes to infinity.
Let rx  −12x
3
3x  x 2  x − 6
Solution:
Since the degree of the numerator is 3, and this equals the degree of the denominator,
we know that lim rx equals the ratio of the coefficients of the highest powers of x. Thus,
x→
lim rx  −12  −4.
x→
3
Example 9:
Let rx 
2x 3 − 5 . Calculate the limit of rx as x goes to infinity.
x x−6
2
Solution:
Since the degree of the numerator is 3, and this is larger than the degree of the
denominator, the limit does not exist.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
Vertical Asymptotes
In this page vertical asymptotes are defined and several examples are shown.
If we look at a plot of the function 1x , we notice that as x gets close to zero the function values get
larger and larger. In fact, the plot of this function in an interval about 0 looks like a vertical line going
straight up (for x  0) and down (for x  0). We describe this phenomenon by saying that 1x has a
vertical asymptote at x  0. Notice that the value of x for which this function has a vertical asymptote,
x  0, is a point at which the denominator equals zero. This is the secret to locating vertical asymptotes.
That is, look for those values of x for which the denominator equals 0. A little care has to be taken as
we shall see in one of the examples below.
Plot the function fx 
Example 1:
1 , and locate its vertical asymptotes.
x2
Solution:
Before plotting this function we notice that it is not defined at x  −2, as that is where
the denominator equals zero, and we suspect that there is a vertical asymptote at x  −2. A plot of the
graph of fx is shown below. Note that as x gets close to −2, the function values get large in either a
positive or a negative sense.
A plot of
1
x+2
-2
1 . Construct a table of values of fx for x close to −2.
x2
Let fx 
Example 2:
Solution:
x
−2. 1
−2. 01
−2. 001
−2. 0001
−1. 9999
−1. 999 −1. 99 −1. 9
fx −10. 0 −100. 0 −1, 000. 0 −10, 000. 0 10, 000. 0 1, 000. 0 100. 0 10. 0
Notice that the closer x gets to −2 the absolute value of fx becomes larger. As x gets close to −2 from
above, i.e., x is larger than −2, the numbers fx get arbitrarily large ( lim  fx  ).
x→−2
As x gets close to −2 from below, i.e., x is smaller than −2, the numbers fx get arbitrarily large in a
negative sense ( lim − fx  −).
x→−2
The above calculations indicate that the function
©
: Pre-Calculus
1 has a vertical asymptote at x  −2.
x2
©
: Pre-Calculus - Chapter 5B
Plot the function fx  x2 − 1 , and locate its vertical asymptotes.
x −1
Solution:
The denominator of this rational function is zero at x  1 and x  −1. From the
previous examples we suspect that there are vertical asymptotes at both of these values. However, we
are wrong, there is no vertical asymptote at x  1. The numerator is also zero at x  1, which cancels
the zero in the denominator. A plot of this function appears below. Notice there is only one place where
the function values get larger and larger; that is, at x  −1.
Example 3:
A plot of
x-1
x 2 -1
-1
The open circle at the point
(1,1/2) indicates that this
point is not on the graph of
the function. Remember that
x =1 is not in the domain.
We know how to spot a vertical asymptote from a graph (wherever the plotted values get arbitrary
large). We next give an analytical definition of a vertical asymptote.
Definition:
We say a function y  fx has a vertical asymptote at x  a if the numbers
|fx| get arbitrarily large as x approaches the value a from the right or the left. In the language
of calculus we say that the limit as x approaches a from above or below is infinite and write
lim fx   or x→a−
lim fx   respectively.
x→a
The fact that the function values must become infinite is why x  1 is not a vertical asymptote in the
last example. To explore this further read the examples below.
In summary, to find a vertical asymptote for a rational function


px
do the following:
qx
px
in lowest terms. That is, cancel all common factors.
qx
px
If
is in lowest terms, this rational function will have a vertical asysmptote at all values of
qx
px
will be the locations of
x for which qx  0. That is, all values of x not in the domain of
qx
vertical asymptotes.
Put
x
, and locate its vertical asymptotes.
x  3x − 4
Solution:
We notice that the denominator of this function is zero at the points x  −3 and at
x  4, which leads us to believe that there are two vertical asymptotes. One at each of the values −3 and
4.
Example 4:
Plot the function fx 
A plot of
-3
©
: Pre-Calculus
1
(x+3)(x-4)
4
©
: Pre-Calculus - Chapter 5B
Let fx  x2 − 1 . Show by constructing tables of values that this function has a
x −1
vertical asymptote at x  −1 and does not have a vertical asymptote at x  1.
Solution:
We look at values of x close to −1 first.
Example 5:
x
−1. 1
−1. 01
−1. 001
−1. 0001
−. 9
−. 99
−. 999
−. 9999
fx −10. 0 −100. 0 −1, 000. 0 −10, 000. 0 10. 0 100. 0 1, 000. 0 10, 000. 0
We notice that the function values are getting arbitrarily large as x gets close to −1, and conclude that
x  −1 is a vertical asymptote.
The following table shows the values of fx for x close to 1.
x
1. 1
1. 01
1. 001
1. 0001
0. 9
0. 99
0. 999
0. 9999
fx . 476 19 . 497 51 . 499 75 . 499 98 . 526 32 . 502 51 . 500 25 . 500 03
In this case the function values are getting closer and closer to 1 as x approaches 1. In fact, if we
2
rewrite this function as fx  x2 − 1  1 , we see that as x approaches 1, x  1 approaches 2, and
x1
x −1
1
its reciprocal, which is fx, approaches .
2
We now examine what happens to vertical and horizontal asymptotes of the function 1x as it undergoes
various transformations. As a reminder the domain of 1x is all real numbers except x  0, its range is
all real numbers y ≠ 0, and it has a horizontal asymptote at y  0, and a vertical asymptote at x  0.
Example 6:
Translate fx  1x two units to the right. Determine the domain, range, and all
horizontal and vertical asymptotes. Plot the function.
Solution:
The function rx  1 is the translate of 1x two units to the right. To find the
x−2
horizontal asymptote we note that as x approaches   the function values of rx approach 0. We also
see that rx is in lowest terms, and that its denominator equals 0 when x  2. Thus, x  2 is a vertical
asymptote. We summarize these remarks in the table below.
A plot of
Domain
all x ≠ 2
Range
all y ≠ 0
Horizontal Asymptote
y0
Vertical asymptote
x2
1 is shown below.
x−2
y
2
©
: Pre-Calculus
x
©
: Pre-Calculus - Chapter 5B
Example 7:
Translate fx  1x three units down. Determine the range, domain, and all
asymptotes.
Solution:
The function rx  1x − 3 is the downward translation of 1x by three units. The
domain, range, and asymptotes are tabulated below.
Domain
all x ≠ 0
Range
all y ≠ −3
Vertical Asymptote
x0
Horizontal Asymptote
y  −3
A plot of rx follows
y
x
y = -3
The table below lists the domain, range and asymptotes of 1x under general horizontal and vertical
transformations.
Domain
Range
Vertical Asymptote Horizontal Asymptote
1
xa
y0
x − a all x ≠ a all y ≠ 0
1 − a all x ≠ 0 all y ≠ −a
x0
y  −a
x
1 . Determine the domain, range, and asymptotes of this function.
x−5
Solution:
The domain consists of all x ≠ 5. We note that if x  5, then fx  0 and if x  5, then
fx  0, and that we have a vertical asymptote at x  5. Moreover from the sign of fx we expect that
as x approaches 5 through values of x  5, that fx will approach − , and as x approaches 5 through
values of x  5, that fx will approach .
Example 8:
Let fx 
A table of function values for x close to 5 is displayed below.
x
4. 9
4. 99
4. 999
5. 001
5. 01
5. 1
fx −10. 0 −100. 0 −1000. 0 1000. 0 100. 0 10. 0
This function has a vertical asymptote at x  5 and a horizontal asymptote at y  0. Its domain is all
x ≠ 5 and its range is all y ≠ 0. A graph of 1 is shown below.
x−5
y
5
x
Notice that the graph of fx is the graph of 1x shifted to the right by 5 units.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
Example 9:
Let fx 
1  1. What are the range, domain, and asymptotes of this function?
x−3
Solution:
 The only real number x which cannot be evaluated by f is x  3. For if we try to compute f3
we first compute 3 − 3, this equals 0, but then we have to take the reciprical of 0, which is not
possible. Thus, the domain is all real numbers except 3.
 The range equals all y such that for some x ≠ 3 we have
y  1 1
x−3
y−1  1
x−3
1  x−3
y−1
x  1 3
y−1
The only value of y for which the above computation is not possible is y  1. Thus, the range of
fx is all y ≠ 1.
 The vertical asymptote is x  3 and the horizontal asymptote is y  1.
A plot of fx is shown below:
y
y=1
x=3
x
Notice that the graph of fx is obtained from the graph of 1x by shifting to the right 3 units and
up 1 unit.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
Exercises for Chapter 5B - Rational Functions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
©
Which of the following functions is rational?
3
2
c. x  x x − 3 d. x 2 − 3x  5
a. 2x, b. 4 x − 1
2
x  2x − 5
x
−
1
Is fx 
a rational function.
x
Find the domain of each of the following functions.
x−1
a.
x2 − 1
3
b. 2x −3 5x  8
x −8
5
−1
x
c.
x2  x − 2
Find the range of the following functions.
1
a.
x−2
1 −1
b.
x2
c. 3 − 2
x5
In problems 5 through 9 find the x and y-intercepts of the rational function:
2x − 6
x2  1
x2 − 4
2x
x
2x  1
3
x1
x 2 − 3x − 4
2x
Find the limit as x →  of 2x x− 1
2
Find the limit as x →  of 2x x− 1
2
Find the horizontal asymptote of rx  2x 2 − 6 and plot the function
−x  2
3
2
Find the limit as x →  of x − x 4 5x − 18 , and plot the function for large values of x
19x  15
2
x
−
2x

7
Does the function
have a horizontal asymptote?
x−1
Find a rational function which has the line y  6 as a horizontal asymptote.
Find a rational function which has 2 as an x-intercept and the line y  −1 as a horizontal
asymptote.
Find a rational function which has −3 as an x-intercept, 7 as a y-intercept, and has the line
y  1 as a horizontal asymptote.
For each of the following functions locate their vertical asymptotes, and then plot the
functions.
x
a.
x−1
x2 − 2
b.
2
x − 3x  2
x−1
c.
x 2  2x − 3
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
2x 2 − 1
x2 − x − 6
2
Locate all vertical and horizontal asymptotes of y  x − 1
2x  3
2
Find all intercepts and asymptotes of the function y  x 2− 4
−3x  27
2
3x
−
1
Find all asymptotes of the function y  3
x −1
4
x
− x 3 − 4x  7
Find all asymptotes of the function y 
5x 2 − 6
Find a rational function which has y  1 as a horizontal asymptote and the lines x  3 and
x  5 as vertical asymptotes
Find a rational function which has the line y  5 as a horizontal asymptote and the lines
2
x  4, x  5, and x  −2 as vertical asymptotes
2
2
Determine the horizontal and vertical asymptotes for fx  ax 2 − ab2
cd − cx
19. Locate all vertical and horizontal asymptotes of y 
20.
21.
22.
23.
24.
25.
26.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
Answers to Exercises for Chapter 5B - Rational Functions
1.
2.
3.
The first two and the last function are rational, the third one is not. 2 x is not a polynomial in
x.
No, the denomiator x is not a polynomial in x.
a. All x such that x 2 − 1 ≠ 0. That is, x ≠  1.
b. All x such that x 3 − 8 is not zero. Thus, all x ≠ 2.
c. All x such that x 2  x − 2  x − 1x  2 ≠ 0. Thus, all x not equal to 1 or −2.
4. Find the range of the following functions.
a. All y ≠ 0.
b. Range is all y  −1. For if y  12 − 1  y  1  12  x 2  1 This is
y1
x
x
possible if y ≠ −1.  x   1
This is possible if y  1  0. Thus, there is
y1
an x in the domain of fx as long as y  −1. Another line of reasoning is as
follows. The range of the function 12 is all positive numbers. Hence, the range
x
of 12 − 1 must be all numbers larger than −1.
x
c. y  3 − 2  y − 3  − 2  x  5  2  x  2 − 5. Thus, the
x5
x5
3−y
3−y
range of fx  3 − 2 is all y ≠ 3.
x5
5. The x-intercept is x  3, and the y-intercept is y  −6.
2
6. The x-intercepts are x  2. There is no y-intercept, since 0 is not in the domain of x − 4 .
2x
7. x  0 is the x-intercept, and y  0 is the y-intercept.
8. There is no x-intercept, and the y-intercept is 3.
9. There are two x-intercepts, x  −1 and x  4. There is no y-intercept since 0 is not in the
domain of the function.
10. 2
11. If we carry out the division we rewrite the function as 2x − 1x . As x goes to infinity the 1x
term goes to zero and the 2x term becomes infinite. We say in this case that the limit does
not exist.
2x 2 − 6  −2, the line y  −2 is a horizontal asymptote.
12. Since lim
x→ −x 2  2
y
x
3
y= 2
13. The denominator has degree larger than the numerator. Thus, the function goes to zero as x
goes to infinity.
y
x
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
14. No the values of the function become unbounded as x goes to plus or minus infinity. Note
that the degree of the numerator is larger than the degree of the denominator.
15. The are an infinite number of such rational functions. Three of them are y  6,
3
2
and y  6x −3 5x − x − 8
y  12x − 8 ,
2x  5
x  2x − 4
x is one such rational function. There are of course many rational functions which fit
16. 2 −
x
the bill.
7x  3
17. One solution is y
. One way to find such functions is to look for a solution of the
7x  3
ax

b
form y 
. Then the criteria that y must satisfy impose conditions on the coefficients
cx  d
criteria
condition
−3 an x-intercept
−3a  b  0
b 7
7 the y-intercept
d
a 1
1 the horizontal symptote
c
of the rational function, and they are:
These conditions are a system of three equations in four unknowns. The system is
−3a  b  0
b − 7d
 0
a−c
 0
One solution to this system is d  3, c  7, b  21, and a  7.
18.
a.
x1
y
y=1
x
x =1
b.
The denominator factors into x 2 − 3x  2  x − 1x − 2. Since the zeros in the
denominator are not cancelled by corresponding zeros in the numerator, we have
the lines x  1 and x  2 as vertical asymptotes.
y
x=1
x
x=2
c.
x−1
x−1

 1 if x ≠ 1. Thus, only the line x  −3 is a
x3
x  3x − 1
x 2  2x − 3
vertical asymptote
y
x
x=
3
19. Since the degrees of the numerator and denominator are the same, there is a horizontal
asymptote and it equals the ratio of the coefficients of the largest powers of x. In this case
that ratio equals 2. Thus, y  2 is a horizontal symptote. The denominator factors into
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5B
x 2 − x − 6  x − 3x  2. Thus the vertical lines x  3 and x  −2 are vertical asymptotes.
20. Since the degree of the numerator is larger than the degree of the denominator, there is no
horizontal asymptote. There is a vertical asymptote at x  −3 .
2
21. x-intercepts: x  2. y-intercept: y  −4 . vertical asymptotes at x  3. horizontal
27
asymptote y  −1 . The function is plotted below.
3
y
x= 3
x=3
x
y = 1/3
22. horizontal asymptote: y  0
23. horizontal asymptote: none
x2
x − 3x − 5
5x 3
25. y 
2x − 4x − 5x  2
a
26. horizontal asymptote: y  −c
does not equal |d| )
vertical asymptote: x  1
vertical asymptote: x  
6
5
24. y 
©
: Pre-Calculus
vertical asymptotes: x  −d, x  d (Note: As long as |b|
©
: Pre-Calculus - Chapter 5C
Chapter 5C - Exponential Functions
Review of Exponents
In this chapter we will define and discuss the properties of exponential functions. These are functions of
the form a x where a is a positive real number, and x is any real number. Before discussing this function,
we’ll quickly review the laws of exponents, and then show how a x is defined for irrational numbers. A
reminder of terminology: a is called the base and x is called the power or exponent.
Below we quickly review exponentiation, and the laws of exponents. For a more thorough review see
the previous chapter on Exponents. The following lists those x for which we can compute a x , and how
to do so. Remember a is any positive real number.
1.
2.
3.
4.
5.
If x  0, then a x is defined to be 1.
If x is a positive integer, then a x is computed by multiplying a by itself x times. Thus,
a 5  aaaaa
−x
5
. Thus, a −5  1
If x is a negative integer, then a x  1
a
a .
If x is the reciprocal of an integer, e.g., x  1 , then a x  b where b 1/x  a. Note: if x  1/5,
5
then 1/x  5. Thus, a 1/5  b, if and only if b 5  a.
That is, a x is the x th root of a. There is of course a computational problem here. It may not
be easy to compute the x th root of a.
x
m 1/n
If x is a rational number, i.e., x  m
n where m and n ≠ 0 are integers, then a  a  .
The table below lists the algebraic properties of exponents:
1 a m  a n  a mn
m
2 a n  a m−n
a
3 a −n  1n
a
0
4 a  1 for a ≠ 0
5 ab n  a n b n
a n  an
6
b
bn
n
7 a m   a mn
If you don’t already have these rules memorized, stop right
now, and memorize them.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5C
What does 16 −3/4 equal?
Question:
Answer:
16 −3/4  2 4  −3/4
 2 −3
 8 −1
 1
8
x
The problem we now face is what does a mean if x is an irrational number? For example what do 2  ,
or 3 2 equal? See the next page for a discussion of this.
In this page the computation of a x is discussed for the case when x is irrational. We need one property
of rational numbers before we can compute a x , and this property is:
Given any irrational number x, there is a rational number m
n which is as close to x as we
want. Another way to express this, is to say that any irrational number can be approximated
as closely as we desire with a rational number.
This is the key to calculating (approximating) a x , we find a rational number m
n which is
m/n
m/n
very close to x, and then compute a . This number, a , can then be shown to be close to
something. This something is called a x .
The plot below shows 2 x for various rational values of x between −1 and 1, they are the red circles. The
blue curve is the plot of 2 x on the interval −1 to 1.
y
2
(0.5, 1.414)
1
( 0.5 , 0.707)
1
0.5
0
0.5
1
x
There are only a few red dots and lots of blue. This is the general situation. There are a lot more
irrational numbers then there are rational numbers, but the amazing thing is that any irrational number
can be approximated with a rational number.
In the following example we use a calculator to compute 2 x for a sequence of rational numbers x which
are getting close to the irrational number 2 . These numbers 2 x will be getting close to 2 2 .
Example 1:
Compute 2 x for a sequence of x’s getting close to 2 .
Solution:
The rational numbers we will use to approximate 2 are 1. 4, 1. 41, 1. 414, 1. 414 2,
1. 414 21, and finally 1. 414 213. The table below list these values of x and below them the
corresponding values of 2 x .
x
2
The value of 2
2
x
1. 4
1. 41
1. 414
1. 4142 1. 41421 1. 414213
2. 63902 2. 65737 2. 66475 2. 66512 2. 66514
2. 66514
as computed by our calculator to 7 decimal places is
2 2  2. 665144 1
Just imagine how long it would have taken to compute 2 1.414 with out the aid of our calculator, and even
then we are only within 2 decimal place accuracy of 2 2 .
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5C
We list one more time the laws of exponents. This time with the remark that the powers are now
allowed to be any real number.
1.
2.
3.
4.
5.
6.
7.
©
For any positive real number a, and any real numbers x and y, the following properties hold:
a x a y  a xy
a x  a x−y
ay
a −x  1x
a
a 0  1 for a ≠ 0
ab x  a x b x
a x  ax
b
bx
x y
xy
a   a
: Pre-Calculus
©
: Pre-Calculus - Chapter 5C
Exponential Functions
In this page we discuss the behavior of the exponential functions y  a x . It turns out that there are three
different modes of behavior: one if a satisfies 0  a  1, a different mode for a’s which satisfy 1  a,
the third type of behavior occurs when a equals 1.
We formally define what we mean by the exponential function with base a.
Definition:
The function fx  a x , with 0  a and x any real number, is called the
exponential function with base a.
x
1
2
In the table below we list values of
x
fx  1
gx  2 x
2
and 2 x for various values of x, and then plot both functions.
x −3 −2 −1 0
1/2 x
2x
1
2
3
1
4
1
8
4
8
8
4
2 1
1
2
1
8
1
4
1
2
2
1
There are several items to notice from this table of data.
0
1. 2 0  1
 1. This is of course true for any positive a. We always have a 0  1.
2 −x
for any x. Note: that this is also true for any positive a.
2. 2 x  1
2
These facts are true for any base a but deserve to be repeated for what they will imply about the graphs
x
of the two functions y  a x and y  1
a . The plot below demonstrates this more clearly than words
can.
y
2
x
y=1
x
x
1
and 2 x .
2
are the reflections of each other through the y-axis.
Plots of
The plots of 2 x and
1
2
x
Example 1:
If the point 2, 36 is on the graph of a x , what must a equal?
Solution:
The statement that 2, 36 is on the graph of a x means that a 2  36. Thus,
a  36  6.
Question:
Answer:
©
If 3 x  81, what must x equal?
4.
3 4  3 2  2  9 2  81
: Pre-Calculus
©
: Pre-Calculus - Chapter 5C
1 x on the same graph.
3
A table of values is first constructed. The two plots follow the table.
Plot the functions y  3 x and y 
Example 2:
Solution:
−2 −1 0
1 1 1
9 3
x
3x
1
3
x
9
3
1
1
2
3
3
9
27
1
3
1
9
1
27
y
3
x
y=1
x
Notice that as with 2 x and
x
1
2
the plots are the reflections of each other through the y-axis.
Example 3:
Using technology fill in the following table of values. Then plot these values and
connect the dots to get an approximation to the graph of 4 x .
x
−2 −1. 5 −1 −0. 5 0. 5 1 1. 5 2
4x
Solution:
x
4
x
−2
−1. 5 −1 −0. 5 0. 5 1 1. 5
. 0 625 . 125
1
4
2
0. 5 2. 0 4 8. 0 16
16
8
4
2
0.5
1
1.5
2
We continue our discussion of the behavior of the exponential functions y  a x . We saw in the last
x
behave differently. Below we characterize this behavior in terms of the
page that y  2 x and y  1
2
x
size of the base of a .
©
: Pre-Calculus
©
1.
: Pre-Calculus - Chapter 5C
If a  1, then the graph of a x looks like the graph of 2 x . That is,
y
2
x
x
Note: see property 5. below.
2.
If 0  a  1, then the graph of a x looks like the graph of
1
2
x
. That is,
y
1
2
x
x
3.
Note: see property 6. below.
If a  1, then the graph of a x is the horizontal line y  1.
y
1
x
Below is a list of the properties of these exponential functions.
1.
2.
3.
4.
5.
6.
The domain of fx  a x is all real numbers.
The range of fx  a x , if a ≠ 1, is all positive real numbers. If a  1, then the range is the
set consisting of the number 1 only. Note: a x can never equal 0.
If a  1, then fx  a x goes to infinity as x goes to infinity, and fx  a x goes to zero as x
goes to minus infinity.
If 0  a  1, then fx  a x goes to zero as x goes to infinity, and fx  a x goes to infinity
as x goes to minus infinity.
If 1  a and x  y, then a x  a y . fx  a x is an increasing function if a  1.
If 0  a  1 and x  y, then a x  a y . fx  a x is a decreasing function if 0  a  1.
Memorize these properties.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5C
6
5
Which number is bigger 1
or 1 ?
3
3
Solution:
Since a  1 is less than 1, property 6. above tells us that the larger the exponent is
3
5
6
is larger than 1 .
the smaller the value of the exponential function. Thus, the number 1
3
3
Example 4:
Example 5:
What does the plot of 2  2 x look like compared to the plot of 2 x ?
Solution:
There are two ways to answer this. One is to say that the plot of 2  2 x is just the plot of 2 x
magnified along the vertical axis by a factor of 2. A perhaps better description is to use the following
law of exponents
2  2 x  2 1 2 x  2 x1 .
This tells us that the graph of 2  2 x is the translate by one unit along the x-axis of the graph of 2 x . Both
graphs are shown below.
y
2
x+1
2
x
x
We now compare the graphs and related properties of exponential functions with different bases.
Below are plots of y  1. 5 x , y  2 x , y  2. 5 x , y  3 x and y  3. 5 x for −1 ≤ x ≤ 1.
a = 3.5
3.5
a=3
3
a = 2.5
2.5
a=2
2
a = 1.5
1.5
1
0.5
-1
-0.5
0
0.5
y = a x for 1 < a
x
1
The next plot compares the graphs of y  a x for different values of a less than 1. In particular a  0. 25,
0. 5, and 0. 75 for −1 ≤ x ≤ 1.
0.25
x
4
3.5
3
0.5
x
2.5
0.75
x
1.5
2
1
0.5
-1
-0.5
0
0.5
x
1
y = ax for 0 < a < 1
From these graphs we summarize how the values of a x compare to each other.
If a  b and x  0, then a x  b x .
If a  b and x  0, then a x  b x .
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5C
Example 6:
Which number is larger 2 − 2 or 3 − 2 ?
Solution:
With the above rules in mind we observe that 2  3, and that the exponents are the
same and negative. Thus, 2 − 2 is bigger than 3 − 2 . To reassure ourselves we use a calculator to
compute approximations to these numbers.
2 − 2 ≈ . 375 214 23
3−
2
 . 211 469 94 .
Example 7:
Let fx  2 x . Plot the vertical shift of the graph of fx which is given by fx  3.
Solution:
This vertical shift displaces the graph of 2 x three units upwards.
y
x
y=2 +3
4
y=3
1
x
0
The domain of 2 x is all real numbers and its range is all positive real numbers.
The domain of 2 x  3 is also all real numbers, however its domain is all numbers greater than 3.
Example 8:
Let fx  2 x . Plot the horizontal translate fx  1  2 x1 of fx.
Solution:
This horizontal shift displaces the graph of fx  2 x one unit to the left.
y
y=2
x+1
2
1
0
x
The domain and range of 2 x1 are the same as the domain and range of 2 x . Domain of 2 x1 is all real
numbers, and the range of 2 x1 is all positive real numbers.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5C
The Natural Exponential Function
For every positive number a there is a corresponding exponential function. From all of these many
exponential functions there is one which is called the natural exponential function. Its base is denoted
by the letter e to distinguish it from all of the other bases. In the following we discuss the number e, and
why e x is the most commonly used exponential function.
For reasons discussed later the number e is calculated by computing 1  1
n
values of n. In the language of calculus e is defined to be the limit of 1  1
n
infinity.
n
for large
n
as n goes to
n
In the table below the values of 1  1n
are shown for various n.
n
10
100
1000
10, 000
n
1 1
2. 5937425 2. 7048138 2. 7169239 2. 7181459
n
It seems as though the value of this number e is approximately 2. 71. A better approximation
to e is
e ≈ 2. 71828182845905 .
It can be shown that the number e is not only an irrational number, but is also a transcendental
number. A number c is said to be algebraic if there is a polynomial px with integer coefficients such
that pc  0. If the number c is not algebraic, it is said to be transcendental.
Clearly, if only pencil and paper were available to compute e x , the computation would take a long time.
Fortunately calculators can rapidly and accurately compute these numbers.
The number e lies between 2 and 3. Thus, e x lies between 2 x and 3 x . The plots of these three
exponential functions are shown below.
y
3x
ex
2x
x
Example 1:
Compute e x for x  −3, −2, −1, 0, 1, 2, and 3.
Solution:
e −3  4. 97871  10 −2 , e −2  0. 135335,
e 1  2. 71828, e 2  7. 38906, e 3  20. 0855
Question:
e −1  . 367879, e 0  1. 0
Place the following numbers in increasing order. 2. 79 5 , 2. 78 5 , and e 5 .
Answer:
e 5 , 2. 78 5 , 2. 79 5 . The number e is approximately 2. 71. Thus, e  2. 78  2. 79 and raising
these numbers to the fifth power preserves the inequalities.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5C
n
n
1 1
, where nlim
1 1
represents the unique
We saw in the previous page that e  nlim
n
n
→
→
n
number which the terms 1  1n
get closer and closer to as n gets larger and larger.
We will use this formula to get a similar formula for e x .
n
is close to e. Therefore the number
For large n the number 1  1
n
x
e . We now manipulate this latter expression.
n x
nx
ex ≈
1 1
 1 1
let y  nx
n
n
 1  xy
y
 1  nx
n
1 1
n
n
x
should be close to
now replace y with n
.
Note that if x  0, then y goes to  if and
if n goes to  .
This gives us the formula
e x  nlim
1  nx
→
n
.
Notice that if x  1, we have the original formula for e.
The argument used to derive this formula is not a proof, there are too many missing steps. Think of it as
an heuristic reason for believing that the formula is true. If you want to see a real proof of this formula,
calculus is where to go.
Example 2:
Solution:
n
Use the formula e x  nlim
1  nx
to find an approximation to the number e 3.1 .
→
n
for n  100, 1000, and 10, 000.
We use our calculator to compute 1  3.n1
The value of e 3.1
Example 3:
Solution:
values.
n
100
1000
10, 000
n
1  3.n1
21. 1771 22. 0918 22. 1873
computed by our calculator is
e 3.1  22. 197951 .
n
Use the formula e x  nlim
1  nx
to approximate e −1 .
→
The table below contains values of the expression 1  −1
n
n
1  −1
n
1000
n
10, 000
100, 000
n
for n equal to various
1, 000, 000
. 36769542 . 36786105 . 3678776 . 36787926
We should feel comfortable using the value 0. 36787926 obtained with n  1, 000, 000. The actual
value of e −1 to 8 places is
e −1 ≈ 0. 36787944
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5C
Exercises for Chapter 5C - Exponential Functions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
3
Simplify 242
6
Simplify the expression 4 2 8 −3 15 2 3 −4
Simplify 2 4 2 −3  5
If a x  16, what is a x/2 ?
If a 3  27, what is a ?
If a 2  5 and b 4  81, what does ab 8 equal?
If a 3  7 and b −3  5, what does a 2 b 3 equal ?
Evaluate the function fx  2 x at 10 different values of x and plot the ordered pairs x, fx.
Evaluate the function fx  0. 9 x at 10 different values of x and plot the ordered pairs
x, fx
x
Evaluate the function fx  1
and fx  1 at 10 different values of x. Then plot both
3
sets of data on the same graph. That is, if you’ve calculated f2 and f2  1 plot the
ordered pairs 2, f2 and 2, f2  1.
Graph the two functions fx  4 x and gx  4. 5 x on the same plot. Before doing so with
your graphing calculator sketch the two graphs using pencil and paper paying particular
attention to which one lies below the other.
Which gets bigger faster 2 x or x 2 as x goes to infinity ? Use graphs to answer this.
Plot on the same graph the two functions fx  1. 5 x and gx  1. 5 x  2.
Which gets bigger faster 2 x or x 3 as x goes to infinity ? Compute the values of both functions
for several values of x to try and decide the answer to this question.
Which gets bigger faster 2 x or x 10 as x goes to infinity ? Compute the values of both
functions for several values of x to try and decide the answer to this question.
Note: it can be shown that no matter what n equals, the function 2 x will eventually get larger
than x n .
The plots of the three functions 3 −x , 2 x−1 , and 4 x are shown below. Which plot belongs to
which function ?
a.
|
2
|
1
10
|
5
|
|
y
15
|
0
|
1
|
2
x
|
2
c.
©
: Pre-Calculus
|
1
4
|
2
|
6
|
b.
|
0
: Pre-Calculus - Chapter 5C
|
4
4
|
2
|
6
|
©
|
2
|
0
17. The graph of a function of the form ka x is shown below. From the graph determine k and a
(2, 45)
5
18. If e x  3 and e y  7, what does e xy equal ?
n
1  nx
to approximate e 3 by taking n  50, 000. Then compare
19. Use the formula e x  nlim
→
this approximate value with what your calculator says e 3 equals.
20. If e x  5, what does e 2x equal ?
21. The number e lies between 2 and 3. Between what two numbers must e 5 lie ?
22. The natural number e’s decimal approximation to 5 places is e ≈ 2. 718 28. If we set a  2.
718 28, then a  e, and if x  0 we have a x  e x . What does e 5 − a 5 equal ?
23. The number e satisfies 2. 7  e  2. 8. Using this inequality what can you say about the
number e −1 ?
24. Solve for x: 8 2x−1  16 x5
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5C
Answers to Exercises for Chapter 5C - Exponential
Functions
1.
2.
3.
4.
5.
6.
7.
3
24 3  4  6  4 3 6 3  4 3 6  384
62
62
62
2
2
2 2 2
2
4 2 3  5 2
4 2 8 −3 15 2 3 −4  4 3154 
 3 3 4 4 5 3  35 2  25
3 4
288
8 3
2 3 4
2 3 4
2  4 3
4 −3 5
4−3 5
1 5
5
2 2   2   2   2  32
a x/2  a x  1/2  16 1/2  4
a  a 3  1/3  27 1/3  3
2
2
 5 2 81  5 4 81 2  4, 100 , 625. One could also
ab 8  a 4 b 4  2  a 2  2 b 4
8
realize that a  5 1/2 and b  81 1/4 . Thus, ab 8  5 1/2 81 1/4   5 4 81 2 .
2
a 2 b 3  a 3  2 b −3  −1  7 2 5 −1  7  49
5
5
8.
y
( 1,
(2, 4)
)
x
9.
y
( 1, 1.111)
(1, 0.9)
x
10.
y
(0, 1)
(0, 1/3)
x
11. Since 4  4. 5, the graph of 4 x will lie below the graph of 4. 5 x for x  0 and above it for
x  0.
4.5x
2
4x
1
0.6
-0.4
-0.2
0
0.2
0.4
x
12. 2 gets bigger faster than x 2 as is shown in the plot below.
2x
60
40
x
20
0
©
2
4
: Pre-Calculus
6
2
©
: Pre-Calculus - Chapter 5C
13.
y
3
1
x
x
14.
2
x
2
4
8
16
4. 0 16. 0 256. 0 65, 536. 0
x 3 8. 0 64. 0 512. 0
4, 096. 0
From this table it appears that 2 x gets bigger faster than x 3 even though x 3 is bigger for
smaller values of x.
15.
x
2
4
8
16
2x
4
16
256
65, 536. 0
32
64
128
4. 29497  10 9 1. 84467  10 19 3. 40282  10 38
x 10 1024 1, 048, 576 1. 07374  10 9 1. 09951  10 12 1. 12590  10 15 1. 152 92  10 18 1. 18059  10 21
From this table it appears that 2 x gets bigger faster than x 10 even though x 10 is bigger for
smaller values of x.
16. a. is 4 x ,
b. is 3 −x ,
and c. is 2 x−1
17. Since the graph passes through the point 0, 5 we must have k  5. The other data point
2, 45 then gives us the equation
45  5a 2 
9  a2 
3  a.
18. e xy  e x e y  3  7  21
n
19. From the formula e x  nlim
1  nx , we have e 3  nlim
1 3
n
→
→
n
expression 1  3
at
n

50,
000,
we
have
n
e 3 ≈ 20. 083 729
Using our calculator we have
e 3 ≈ 20. 085 537 .
20. e 2x  e x  2  5 2  25
21. e 5 must satisfy the inequality
2 5  e 5  3 5 or
32  e 5  243
©
: Pre-Calculus
n
. Evaluating the
©
: Pre-Calculus - Chapter 5C
22.
e 5 − 2. 718 28 5  148. 413 159 102 577 − 148. 412 659 950 842
 . 000 499 151 735
23. From 2. 7  e  2. 8 we have
1  1  1
e
2. 8
2. 7
1
. 357 142 8  e . 370 370 3
Remember too, that 1e  e −1 .
24.
8 2x−1  16 x5
2 3  2x−1  2 4  x5
2 6x−3  2 4x20
Thus, 6x − 3  4x  20
2x  23
x  23/2
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
Chapter 5D - Logarithmic Functions
Definition of Logarithms
Logarithms were the invention of John Napier, a Scotsman, who invented them in the early 1600’s.
Before the advent of computers these functions provided a means of easily performing many tedius
computations. In fact, if you look at any mathematics text book more than 10 years old more than likely
a table of logarithms will be in the back of the book. However, while it is true that logarithms are no
longer needed for computations they still serve as a valuable tool in mathematics and science. In
particular they are useful in solving equations which contain exponential functions.
Logarithms are difficult for students to master. More than likely you will need to read this material
several times before it starts to make sense. If it is still hazy after several readings, keep going.
Eventually you will understand them, and wonder why it was such a big deal.
We start by defining what is meant by the logarithm of a number to a particular base, and then look at
several examples.
Definition of a logarithm to a base a
Let a be any positive number not equal to 1. The logarithm of x to the base a is y if and only
if
ay  x
The number y is denoted by
y  log a x
Another way to say the same thing is to say that y is the log of x to the base a, y  log a x if when we
raise a to the y th power we get x.
Note that the logarithm to a base a is the inverse function of a x . We have chosen to define log functions
directly rather than merely stating that they are particular inverse functions.
The symbolism log a x is read as ”the log of x to the base a”, or ”the log to the base a of x”.
Logarithms of various numbers to various bases are listed below. Be sure you understand each one of
these.
The logarithm of 10 to the base 10 equals 1: 10 1  10, log 10 10  1.
The logarithm of 10 to the base 100 equals 0. 5: 100 1/2  10, log 100 10  1 .
2
2
The logarithm of 10 to the base 10 equals 2:
10
 10, log 10 10  2.
The logarithm of 100 to the base 10 equals 2: 10 2  100, log 10 100  2.
The logarithm of 1000 to the base 10 equals 3: 10 3  1000, log 10 1000  3.
The logarithm of 25 to the base 5 equals 2: 5 2  25, log 5 25  2.
The logarithm of 64 to the base 2 equals 6: 2 6  64, log 2 64  6.
The logarithm of 64 to the base 4 equals 3: 4 3  64, log 4 64  3.
The logarithm of 64 to the base 8 equals 2: 8 2  64, log 8 64  2.
©
: Pre-Calculus
©
Question:
Answer:
Question:
Answer:
Example 1:
Solution:
: Pre-Calculus - Chapter 5D
If 2. 3 4  x, what is a in log a x  4?
2. 3
If the logarithm of x to the base 3 is 4, then x must equal?
81. a y  3 4  81.
If log 8 x  4, what is x?
The statement log 8 x  4 has the same meaning as
x  8 4  4096 .
Example 2:
Solution:
If log a 9  2, what is a?
log a 9  2 is equivalent to the statement
a 2  9 or
a   9 and since a  0 we have
a
9  3.
Example 3:
You will later learn how to use your calculator to compute log 3 6. It turns out that
log 3 6 ≈ 1. 6309. When a piece of technology is used to perform some calculation always ask yourself
if the answer is reasonable. So, is 1. 6309 reasonable? Well 3 1  3 which is less than 6, so 1 should be
less than the log of 6 to the base 3. The number 2 is larger than 1. 6309, so 3 2 which is 9 should be
larger than 6 which it is. Thus, by these simple tests 1. 6309 seems reasonable.
We now look at some plots of several log functions paying attention to their domains and ranges.
Example 4:
Plot log 10 x.
y
(10, 1)
10
x
There are several things to note:
1. log 10 x is not graphed for x ≤ 0. That is, the domain of log 10 x is all positive real numbers.
We’ll see why later.
2. The range of values of log 10 x is all real numbers.
3. log 10 1  0
4. The graph of log a x looks like the graph of log 10 x as long as a  1.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
In the next example we look at a plot of a log function where the base a is less than 1.
Example 5:
Plot log 1/2 x
y
2
x
( 2, 1)
This plot certainly looks different than the preceding one. However, there are certain similarities.
1. log 1/2 x is not graphed for x ≤ 0. That is, the domain of log 1/2 x is all positive real numbers.
2. The range of values of log 1/2 x is all real numbers.
3. log 1/2 1  0.
4. The graph of log a x looks like the graph of log 1/2 x as long as 0  a  1.
To understand why the domain of any log function is only the positive real numbers, we look at the
definition. That is, log a x  y if and only if a y  x. However, we’ve seen that numbers of the form a y
are never less than or equal to zero. That is, a y  x  0, and the x’s make up the domain of the log
function.
The fact that the range of log a x is all real numbers comes from the fact that in the second equation
log a x  y

a  x
y
the y’s can be any real number.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
Properties of Logarithms
We list below some of the algebraic properties of logarithms. These properties should be memorized.
1.
2.
3.
4.
5.
6.
7.
8.
The domain of log a x is all positive real numbers and its range is all real numbers.
a log a x  x
log a a x   x
log a x  log a y  x  y
log a 1  0
log a xy  log a x  log a y
log a x y   y log a x
log a xy  log a x − log a y
Proofs of the above properties:
1. This follows from the fact that the log a x is the inverse function of a x , and the fact that the
domain of a x is all real numbers, while its range is all positive real numbers.
2. This follows directly from the definition. That is, log a x  y where y is a number such that
a y  x. Thus, x  a y  a log a x .
3. This property also follows directly from the definition of log a x. That is, log a x  y if and
only if a y  x. Thus, log a a x   y if and only if a y  a x . That is, x  y  log a a x .
4. If x  y, then log a x  log a y, for if not, then the fact that a x is a one-to-one function implies
that x ≠ y. Conversely if log a x  log a y, then we have x  a log a x  a log a y  y .
5. Since a 0  1 no matter what a equals, we have log a 1  0 for any base a.
6. Let z  log a xy, then we must have a z  xy. However, we also have
a log a xlog a y  a log a x a log a y  xy. Thus, a z  a log a xlog a y from which we have
log a xy  z  log a x  log a y.
7. Let z  log a x y . Then a z  x y . Now compute a y log a x . a y log a x  a log a x  y  x y  x y . Thus,
we have a z  a y log a x . From this we deduce that log a x y   z  y log a x .
8. log a xy  log a xy −1   log a x  log a y −1   log a x − log a y
Example 1:
Let fx  log 2 x − 1. What is the domain and range of fx? Plot this function.
Solution:
For any log function its argument must be greater than 0. Thus, domain of
fx  x : x − 1  0  x : x  1. The range of fx all real numbers.
x=1
2
4
6
x
0
log2(x - 1)
-5
-10
©
: Pre-Calculus
8
10
12
14
©
Example 2:
Solution:
: Pre-Calculus - Chapter 5D
Simplify the expression 2 2 log 4 5 .
2 2 log 4 5  2 2  log 4 5
 4 log 4 5
 5.
Example 3:
Solution:
Simplify log 3 16  log 3 5 − log 3 8.
log 3 16  log 3 5 − log 3 8  log 3 16  5 − log 3 8
 log 3 16  5
8
 log 3 2  5
 log 3 10
Example 4:
Solution:
Solve the equation 2  3 log 5 x  15.
2  3 log 5 x  15
3 log 5 x  13
log 5 x  13
3
x  5 13/3
≈ 1068. 73
Example 5:
Solve the equation 4 x−6  13.
Solution:
The easiest way to solve an equation in which the unknown is part of an exponent is to
take the log of both sides of the equation.
take the log 4 of both sides
4 x−6  13
x − 6  log 4 4 x−6   log 4 13 solve for x
x  6  log 4 13
As we mentioned previously logarithms are the inverse functions of the exponential functions. On the
next page we examine this relationship more closely for the two functions log 2 x and 2 x . Since log 2 x is
the inverse function of 2 x we have
log 2 x  y if and only if
x  2y .
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
The following table demonstrates this relationship with specific numbers:
20  1
2 −1  1
2
23  8
2 −4  1
16
6
2  64

log 2 1  0
log 2 1  −1
2
log 2 8  3
log 2 1  −4
16
log 2 64  6




There is nothing special about base 2 in this relationship. For every base a it is true that
log a x  y if and only if
x  ay
There is another way to write these relationships, and it is
log a a x   x
a log a x  x
Question:
Answer:
If a 3  4. 56, what is log a 4. 56?
log a 4. 56  log a a 3   3
The graph below contains a plot of 2 x and log 2 x.
(2,4)
y=2
x
(1,2)
(4,2)
(0,1)
(2,1)
(1,0)
y=x
y = log 2 x
Notice that each plot is the reflection of the other about the line y  x.
Question:
Answer:
©
If log 2 15  y, what does 2 y equal ?
15. If log 2 15  y, then 2 y  2 log 2 15  15 .
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
Natural Logarithm
The natural log function is that log function which has base e. This function is so commonly used that instead of
writing log e x we write ln x. Hence, ln represents log e . To repeat ourselves ln x is the inverse function of e x , and the
usual relationships between a function and its inverse hold.
lne x   x
e ln x  x .
Example 1:
Solution:
Compute the natural log of 5, 0. 45, 17, and 1.
ln 5 ≈ 1. 609 44
ln 0. 45 ≈ − 0. 798 508
ln 17 ≈ 2. 833 21
ln 1  0
Example 2:
Plot e x and ln x on the same graph.
y
x
Question:
Answer:
In the above plot which of the two curves is the graph of ln x?
ln x is the graph drawn in red.
Question:
Answer:
In the above plot what are the coordinates of the point where the red curve crosses the x-axis ?
1, 0
Question:
Answer:
In the above plot what are the coordinates of the point where the blue curve crosses the y-axis ?
0, 1
It is possible to write every log function in terms of the natural log function. We show how to do so here.
Suppose y  log a x. Then x  a y . The next step is to express a in terms of e. We have a  e ln a . Thus,
x  ay
 e ln a 
y
 e y ln a .
Taking the natural log of both sides of this equation, we derive
ln x  y ln a
y  ln x or
ln a
log a x  ln x .
ln a
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
There are two formulas that should be remembered, and we list them once more.
a x  e x ln a
log a x  ln x
ln a
This formula will be useful in many of the exercises at the end of this chapter. Be sure you understand its
derivation.
Example 3:
Solution:
Write log 8 15 using the natural logarithm.
log 8 15  ln 15
ln 8
2.
708 05
≈
2. 079 44
≈ 1. 302 3
Example 4:
In an earlier chapter we said that a function has exponential growth if it has the form fx  a kx .
Write this function in terms of the natural exponential function.
Solution:
fx  a kx
 e ln a 
kx
 e k ln ax .
This last example shows us that any function which has exponential growth can be written in terms of the natural
base e ≈ 2. 718 28.
Example 5:
Solution:
Solve the equation e 5x−6  18.
The way to solve this equation is to take the natural log of both sides.
e 5x−6  18
5x − 6  ln 18
5x  ln 18  6
x  ln 18  6
5
x ≈ 1. 778 07
Example 6:
and k.
Solution:
Suppose that fx  ae kx for some value of k and a. Suppose that f1  2 and f3  1. Find a
The values of fx when x  1 and x  3 lead to the following equations.
2  f1  ae k
1  f3  ae 3k
We have a system of two equations in two unknowns. One way to solve a system which involves exponential
functions is to first take the natural log of both sides. Doing so we get
ln 2  lnae k   ln a  lne k   ln a  k
ln 1  lnae 3k   ln a  lne 3k   ln a  3k
Since ln 1  0, the second equation implies that ln a  −3k. Substituting this into the first equation we have
ln 2  −3k  k  −2k
k  ln 2 .
−2
3
ln
2
Hence, ln a  −3k  −3

ln 2. Thus, a  2 3/2 . Substituting these values into the formula for fx, we
2
−2
have
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
ln 2 x
fx  2 3/2 e −2
 2 3/2 e ln 2 
−x/2
 2 3/2 2 −x/2
3−x
2 2
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
Exercises for Chapter 5D - Logarithmic Functions
1.
Who invented logarithms?
2.
What is the domain of log 2 x  1?
3.
What are the domain and range of log 3 2x − 1?
4.
What are the domain and range of log 5 1  x − 1 ?
5.
If x  log a 5, what does a x equal ?
6.
Does log 4 2 equal log 2 4 ?
7.
Compute the following logarithms: log 10 15, log 8 15, log 6 15, and log 4 15. Do you see a
relationship between the numbers ?
8.
Compute the following logarithms: log 16 10, log 4 10, and log 2 10. Do you see a relationship
between these numbers ?
9.
Solve log 5 25  x  4.
10. Solve the equation 5 2x−1  6. 5.
11. Which number is larger log 9 3 or log 3 9 ?
12. If x  log 4 23, what does 2 x equal ?
13. Express the equation log 15 x − 2  5 in exponential form.
14. Solve 4 x2  15.
15. If log 4 6  log 2 x, what does x equal?
16. If x  0, what does log x x equal ?
17. Express the equation 15 x
2 −3
 9 in logarithmic form, then solve for x.
18. Which number is larger log 10 100 or log 9 100?
19. Which number is larger log 10 1 of log 9 1 ?
2
2
20. 5 is larger than 4, thus, for x  1 we expect log 4 x to be larger than log 5 x. It takes more 4’s to
equal x than it does 5’s. Plot these two functions for x  1 and graphically verify this
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
conjecture.
21. Solve log 4 x − log 4 x − 1  5.
22. Simplify log 10 x 2   log 10 x 3  − log 10 5x.
23. Simplify log 3 2x − 7 log 3 x  5.
24. Simplify 1 log 3 16 − log 3 24.
2
25. Simplify log a x − log a 1x .
26. Simplify log a x 3 − log a 3x  log a 1x .
27. Solve the equation 2 log 5 x − log 5 x  1. 3.
28. Solve the equation log 3 2x  log 3 x 4  − 5  0.
29. If log 4 x  log 4 43, what must x equal ?
30. Write log 9 15 in terms of the natural logarithm.
31. Compute the numbers: ln 64, ln 32, ln 16, and ln 8.
32. Solve the expression x ln 32  ln 2.
33. What is the domain of lnx  4?
34. Write log a 2 x in terms of log a .
35. If log 10 x  3. 5, then log 15 x equals?
36. Which is the larger number log 5 x or ln x ?
37. Find a relationship between log a x and log a 2 x.
38. Find the domain of y 
e xa
ln2x − b
39. Solve for x: x lna b   lna
40. Given log b 2  a, log b 3  c, and log b 5  f. Find log b 1800.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
Answers to Exercises for Chapter 5D - Logarithmic
Functions
1.
2.
John Napier, a Scotsman.
In the function log 2 x  1 the expression x  1 must be positive, since the domain log 2 x is
all positive real numbers. Thus, we must have
x1  0
x  −1 .
3.
4.
If x is in the domain of log 3 2x − 1, then we must have
2x − 1  0
2x  1
x 1
2
1
So the domain of log 3 2x − 1 is x  . Morever as x varies over the interval 1/2, ,
2
2x − 1 varies over the interval 0, . This means that the range of log 3 2x − 1 is the same
as the range of log 3 x, i.e., all real numbers.
There are two problems in determining the domain of log 5 1  x − 1 . One we need to
ensure that
1 x−1  0
and that x − 1 ≥ 0 so that we can take its square root. Since square roots are always
non-negative adding 1 to x − 1 will certainly make 1  x − 1 positive. Thus, the domain
of log 5 1  x − 1 is
5.
6.
7.
x ≥ 1.
To determine the range of log 5 1  x − 1 we observer that as x varies over the interval
1, , the expression 1  x − 1 varies over the interval 1, , and thus the values of
log 5 1  x − 1 will vary over the interval 0, .
5
log 2 4  2
No.
log 4 2  12
log 10 15 ≈ 1. 176 09
log 8 15 ≈ 1. 302 3
log 6 15 ≈ 1. 511 39
log 4 15 ≈ 1. 953 45
As far as a relationship is concerned the smaller the base the larger the value of the
logarithm.
8.
log 16 10 ≈ . 830 482
log 4 10 ≈ 1. 660 96
log 2 10 ≈ 3. 321 93
Notice that as we take the square root of the base the logarithm doubles.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
9.
log 5 25  x  4
2x  4
x2
10.
5 2x−1  6. 3
2x − 1  log 5 6. 3
2x  1  log 5 6. 3
1  log 5 6. 3
x
2
≈ 1. 071 8
log 9 3  1
log 3 9  2
2
log
23
 4 1/2  4  4 log 4 23  1/2  23 1/2 ≈ 4. 795 83
11. log 3 9 is larger.
12. 2 x  2 log 4 23
13. x − 2  15 5
14. 4 x2  15 implies
x  2  log 4 15
x  log 4 15 − 2
x ≈ 1. 953 45 − 2
x ≈ −4. 655 47  10 −2
15. log 4 6  log 2 x implies
x  2 log 2 x
 2 log 4 6
 4 1/2 
log 4 6
 4 log 4 6  1/2
 6 1/2 .
16. 1
2
17. To express 15 x −3  9 as a logarithmic equation take the log to base 15 of both sides.
2
15 x −3  9
x 2 − 3  log 15 9
≈ 0 . 811 368
This leads to the equation
x 2  3. 811368
x   3. 811368
≈  1. 952 27
18. log 9 100 is larger, for the reason that it will take a higher power of 9 to equal 100 than for 10.
In fact, these numbers are approximately
log 10 100  2. 0
log 9 100 ≈ 2. 095 9
19. log 10 1
2
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
20.
y
log 4( 25 ) = 2.3215
log 5( 25 ) = 2
25
21. log 4 x − log 4 x − 1  5 implies log 4
x
x−1
x
x  4 5 This leads to the equation
x−1
x  x − 14 5
5
x  x  45 − 45
x1 − 4 5   −4 5
−4 5
1 − 45
≈ 1. 000 98
x
22.
log 10 x 2   log 10 x 3  − log 10 5x
 2 log 10 x  3 log 10 x − log 10 x − log 10 5
 4 log 10 x − log 10 5
 log 10 x 4  − log 10 5
4
 log 10 x .
5
23.
log 3 2x − 7 log 3 x  5  log 3 2x − log 3 x  5 7
2x
 log 3
x  5 7
24.
1 log 16 − log 24  log 16 1/2 − log 24
3
3
3
3
2
 log 3 4
24
 log 3 1
6
 − log 3 6
25.
log a x − log a 1x  log a x − log a x −1
 log a x  log a x
 2 log a x
 log a x 2
26.
log a x 3 − log a 3x  log a 1x
 3 log a x − log a 3  log a x − log a x
 log a x − log a 3
 log a x
3
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
26.
log a x 3 − log a 3x  log a 1x
 3 log a x − log a 3  log a x − log a x
 log a x − log a 3
 log a x
3
27.
2 log 5 x − log 5 x  1. 3
2 log 5 x − 1 log 5 x  1. 3
2
3 log x  1. 3
5
2
log 5 x  2. 6
3
2.6/3
x5
≈ 4. 034 35
28.
log 3 2x  log 3 x 4  − 5  0
log 3 2  log 3 x  4 log 3 x  5
5 log 3 x  5 − log 3 2
5 − log 3 2
log 3 x 
5
log 3 2
 1−
5
Thus,
x3
1−
log 3 2
5
 3  3 −log 3 2/5
 33 log 3 2  −1/5
 31/5
2
≈ 2. 611 65
29. 43
30. log 9 15  ln 15
ln 9
31.
ln 64 ≈ 4. 158 88
ln 32 ≈ 3. 465 74
ln 16 ≈ 2. 772 59
ln 8 ≈ 2. 079 44
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
32.
x ln 32  ln 2
x  ln 2
ln 32
 ln 25
ln2 
 ln 2
5 ln 2
 1
5
33. If x is in the domain of lnx  4, then we must have
x4  0
x  −4 .
34.
log a 2 x  ln x2
ln a
 ln x
2 ln a
 1 log a x
2
35.
log 15 x  ln x
ln 15
ln 10log 10 x

ln 15
ln 103. 5

ln 15
8.
059
07
≈
2. 708 05
≈ 2. 975 97
36. ln x is log e x. If x  1, since e is less than 5, ln x will be larger than log 5 x. If 0  x  1, ln x
will be smaller than log 5 x.
37.
log a 2 x  ln x2
ln a
 ln x
2 ln a
 1 ln x
2 ln a
 1 log a x .
2
38. Here we must check two things. We must make sure we are only taking the log of positive
numbers and we must make sure we are not dividing by zero:
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 5D
2x − b  0
2x  b
x b
2
and
ln2x − b ̸ 0
2x − b ̸ 1
2x ̸ 1  b
x ̸ 1  b
2
Thus, the domain is
b, 1b
2
2

1  b ,
2
39.
x lna b   lna
xb lna  lna
lna
x
b lna
x  1 b ̸ 0
b
40. We want to write 1800 as the product of powers of 2, 3, and 5. One way to do this is as
follows:
1800  2598
 5 2 3 2 2 3 
Thus,
log b 1800  log b 5 2 3 2 2 3 
 log b 5 2   log b 3 2   log b 2 3 
 2 log b 5  2 log b 3  3 log b 2
 2f  2c  3a
©
: Pre-Calculus