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Chapter Seven
Logarithmic Functions
The logarithm of a number is the power to which a base must be raised to
obtain that number. Using a base of 3, the logarithm of 9, written log 3 9, is
2, because 9 = 32. Using a base of 5, the logarithm of 125 (log 5 125) is 3,
since 125 = 53. A logarithmic function is a function that assigns to each
x-value the power to which the base must be raised to produce that x-value.
A logarithmic function is the inverse of an exponential function. A function
such as f(x) = 2 x has a graph that is always increasing from left to right and
therefore (page 10) has an inverse function. This inverse determines the
exponent to use to obtain the value you wish for 2x.
For example, if you would like to know what x-value would make 2x = 8, you
must solve the problem that is the inverse of computing 2x. This particular
problem is fairly easy, since you know that 23 = 8.
The exponential problem is to compute f(3), which is 8 since 2 2 2 = 8.
The inverse problem is to find the power of 2 that produces the function value
8, 2? = 8, and the answer to this question is 3. The number 3 used in this way
is called the logarithm of 8, base 2 (log2 8). The logarithm is the exponent
to which you must raise the base to obtain the desired number.
Graphs of 2x and log 2 x, showing the
inverse relationship between them.
Some examples:
log4 16 = 2, since 42 = 16
log9 3 =
log5
log8 4 =
, since
=
=3
= –2, since
=
=
, since
Find: log2 32, log27 3, log4
log49
.
= 22 = 4
=
Ans.: 5,
log16
=
, since
=
=
=
=
, –3,
,
.
, log4 8,
120 CHAPTER 7: LOGARITHMIC FUNCTIONS
Properties of Logarithms
Since logarithms are exponents of the base being used, logarithms obey the
same rules that exponents do. You saw in the preceding chapter that when
you multiply together numbers with the same bases the exponent of the
product (with the same base) is the sum of the exponents of the factors. The
same rule written in terms of logarithms (using the same base for each
logarithm) is
(1) log (a b) = log a + log b.
If log q 2 = .356, logq 3 = .565, and
logq 5 = .827, find:
a. logq 10
b. logq 6
c. logq 45
log 15 = log 5 + log 3
The logarithm of a product of two numbers is equal to the
logarithms of the two factors.
sum of the
You also know that when you divide one power of some base by another,
your quotient is the same base raised to the difference in the exponents of the
numerator and denominator. This rule written in terms of logarithms is
(2) log
Ans.:
a. 1.183 b. .921 c. 1.957
Find:
d. logq
e. logq
f. logq
= log a – log b.
log 6 = log 12 – log 2
The logarithm of a quotient of two numbers is equal to the difference in the
logarithms of the two numbers.
Ans.:
d. .209 e. –.471 f. –.356
Third, if you take a number raised to a power and raise that to another power,
for example
single power,
, or
, you can write the complete calculation as a
. This last power is the product of the two others. In terms
of logarithms this is
(3) log
Find:
= n log a.
log
=
g. logq 9 h. logq 8 i. logq
log x
The logarithm of a power of a number is equal to the product of the power
with the logarithm the number.
This third property can be particularly useful for computations like finding the
logarithm of the square or cube root of a number, like
log2
= log2
=
log2 8 =
3 = 1.5.
Ans.:
g. 1.130 h. 1.068 i. .4135
What number could q be?
Ans.: A number between 6 and 8.
CHAPTER 7: LOGARITHMIC FUNCTIONS
These rules are useful in simplifying some complicated algebraic expressions
and you should get used to using logarithmic computations. They make
dealing with logarithmic expressions easier than non-logarithmic expressions
in some cases.
For example,
log
can be simplified to 4 log (x – 2) +
log (x + 5) – 5 log (x – 3).
Properties of Logarithms:
(1) log (a b) = log a + log b.
(2) log (a ÷ b) = log a – log b.
(3) log an
= n log a.
You apply logarithms most frequently in problems involving exponential
expressions. The way to solve for a variable in an exponent is to take
logarithms with the right choice of base.
To solve the equation 4 x = 16, you must find the power of 4 that equals
16, or find log4 16. Taking logs of both sides of the equation, you want
to use a base that is easy to calculate for both sides of the equation. The
natural choice here is base 4.
log4
= log4 16
x log4 4 = log4 42
x 1=2
and the solution of the original equation 4x = 16 is x = 2.
Notice the use of property (3) for logarithms in the second line.
To solve 52x = 125, you would apply logarithms (base 5) to both sides of
the equation
log5 52x = log5 125
2x log5 5 = log5 53 = 3 log5 5
2x = 3
x=
121
122 CHAPTER 7: LOGARITHMIC FUNCTIONS
To solve the equation 8x = 16, you can find a base that both sides of the
equation can use by realizing that both 8 and 16 are powers of 2 (23 and
24 respectively). Then take the logarithms, base 2, of each side of the
equation:
log2 8x = log2 16
log2 23x = log2 24
3x log2 2 = log2 24
3x = 4
Solve for x:
a.
= 27
b.
=
c.
= 36
x=
To solve
= 27, take logarithms base 3 of both sides since you know
that 27 is a power of 3.
Ans.: a.
log3
b.
c. 3
= log3 27
( 1 – 4x) log3 3 = log3 33
1 – 4x = 3
–4x = 2
x=
You will ordinarily restrict yourself to using the same numbers for bases of
logarithms as you do for your exponential functions, that is, numbers greater
than 1. You may use numbers between 0 and 1 as bases for logarithms but
Expect to stick to
you will usually not need to do so since
= –logb x.
Do
you understand why
= –log b x? Go through the
steps to show this is true.
bases that are greater than 1.
As with the exponential functions, the most important base for most
applications is e, which appears in any problem involving natural growth.
This is also the base used, almost exclusively, in calculus. Since this base is
so important, it has its own notation. Instead of writing loge x for logarithms
with base e, write ln x, read "natural log of x", to indicate that you are using
base e. This is the most common notation used, but be careful because some
calculus books simply write log x to indicate a logarithm of base e, since this
is the base the reader usually uses in that subject. Many elementary books
write log x with no base specified when working with base 10, since base 10
logarithms were at one time frequently used for calculations. Your calculator
has both a ln key for logarithms base e and a log key for base 10
logarithms (which you will probably not use much, since base 10 logs are of
no particular importance). To find the logarithm, just enter your number after
pressing the logarithm key you want, and then the ENTER key.
Using your calculator, find both log10
and ln of:
a. 3 b. 8 c. 30 d. 100
Ans.:
a. .4771 and 1.0986
b. .9031 and 2.0794
c. 1.4771 and 3.4012
d. 2 and 4.6052
CHAPTER 7: LOGARITHMIC FUNCTIONS
123
Changing Bases
Since your calculator only has base 10 and base e logarithms built-in, you
may have to change bases in order to do a calculation in another base. To do
this, follow the following steps:
(1) a =
This is true because when you raise b to the exponent (base b) that gives you
a, you will get a! Make sure you understand that statement. Log b a is the
correct exponent of b to produce a. So when you do raise b to that power,
you get a.
6=
=
12 =
Find:
a. log5 13
c. log11 28
b. log4 7
d. log9 16
Ans.:
a. 1.5937
c. 1.3896
b. 1.4037
d. 1.2619
=
7=
(2) c = ax =
=
with x = loga c
also, logb c = x logb a = loga c logb a , and
(3) loga c =
For example, log3 7 =
=
= 1.7712
Take another look at the equation 8x = 16. You can now solve this by
first taking logs, base 8, of both sides:
log8
= log8 16
x=
=
To find how long it will take a population to double in size when the
annual increase is 7%, you solve the equation
Using natural logs:
ln
= 2.
How long will it take money earning
interest at a rate of 5% per year to
double in value?
= ln 2
x ln 1.07 = ln 2
x=
= 10.244 years
The population will double in 10 years and approximately 3 months.
Ans.:
x = 14.2 years.
= 2, solution is
124 CHAPTER 7: LOGARITHMIC FUNCTIONS
Note that to change bases using your calculator you may use either
the base 10 key or the natural log key. For example, to find log7 5,
either use base 10 logs:
log7 5 =
= (from your calculator)
= .8271
or natural logs:
log7 5 =
= (from your calculator)
= .8271
Logarithmic Functions
Since a logarithm gives the exponent of the base (positive and usually larger
than 1) necessary to obtain a given number, you cannot ask for the logarithm
of zero or of a negative number. Raising positive numbers to powers always
produces positive results. No power of 3 is ever equal to 0; no power of 5 is
ever negative. So there is no possible answer to the question, "what is log 3
0?" or "what is log5 –2?". The domains of logarithmic functions must always
exclude values that ask for the logarithm of zero or of negative numbers.
Domains of logarithmic functions are always limited to positive input values.
The first thing to do when graphing a logarithmic function is to remove from
consideration any x-values that do not belong in the domain. This is the same
thing you did with the rational functions, but rational functions usually require
the elimination of a few discrete values, while logarithmic functions
commonly need to have entire intervals of values removed.
Next, you usually find a starting point (but often you cannot use x = 0, so
there may be no intercept). If there is no intercept, you may want to make use
of the fact that log 1 = 0, no matter what the base is (since a non-zero number
raised to the zero power is always equal to 1). When you choose a value
causing you to take the logarithm of 1, you get zero for the logarithm.
Show the end-behavior. To do this, keep in mind that obtaining large
numbers requires you to use large exponents of your base, and obtaining
numbers just above 1 requires small positive exponents. To obtain numbers
less than 1 (but greater than 0), negative exponents of the base are necessary.
In the following examples, the bases are not specified because, like the
exponential functions, the graphs of logarithm functions with different bases do
not differ much from each other although they pass through different points.
The most common base to use for logarithms, as for exponential functions is e.
What values must be excluded from
the domains of:
a. f(x) = log (3 – x)
b. f(x) = log (x2 – 1)
c. f(x) = log (x4 + 2)
Ans.:
a.
b.
c.
x = 3 all values of x larger
than 3
x = 1, x = –1, and all
values between them.
No values are excluded.
CHAPTER 7: LOGARITHMIC FUNCTIONS
For a first example, take f1(x) = log x (or ln x).
You must limit the domain of this function to the positive numbers, since
you cannot take logarithms of zero or of negative numbers.
The graph only appears on the right side of the y-axis and there is no
intercept. You may use any positive number for x to obtain a starting
point, but you may find x = 1 easiest, since you know that log 1 = 0 for
any base. This gives you the starting point (1, 0) which lies on the x-axis.
You may notice that this point is the inverse of the starting point used for
an exponential function
, (0, 1).
As you use larger values for x, the logarithms are the exponents necessary
to obtain these x-values, and therefore must be increasing. However, you
do not have to change an exponent much to obtain a significant change in
the value obtained, so the logarithms do not increase at a very rapid rate.
On the right side the function always continues to rise, but it rises at a
very slow rate.
For x-values less than x = 1, remember that the exponents you need to use
to get such numbers are negative, and become extremely negative as you
get closer to x = 0. It takes a very negative power to get a number close
to zero.
The graph does not appear until you get to the right of the y-axis, where
it starts with very negative values, then rises rapidly to go through the
point (1, 0), and continues to rise to the right, but slowly.
For the next example, let f2(x) = log (1 + x).
The easiest way to graph f2(x) is to realize that its graph is identical to that
of f1(x), but translated one unit to the right. Once you see this, you can
easily complete the graph. If you do not notice the translation, you can
still get the graph with the usual procedure.
Since you cannot take the logarithm of zero or of negative numbers, you
must remove x = –1 and all values less than –1 from the domain. You
may only use x-values larger than –1.
Since zero is a possible x-value, you can find the intercept
(0, log 1) = (0, 0). The graph passes through the origin.
Graph of f1(x) = log x.
ln 2 = 0.693
ln 20 = 2.996
ln 200 = 5.298
ln 2000 = 7.601
ln 20,000 = 9.903
ln 2 = 0.693
ln .2 = –1.609
ln .02 = –3.912
ln .002 = –6.215
ln .0002 = –8.517
125
126 CHAPTER 7: LOGARITHMIC FUNCTIONS
As you increase the value of x, you take the logarithm of larger numbers.
This gives larger function values, since larger numbers require you to
raise your base to higher powers. Since only small increases in exponents
can lead to substantial changes in the numbers, this increase is slow. The
graph rises at a slow rate after passing through the origin.
Using values of x less than zero, down to just above x = –1, you produce
the logarithms of numbers less than 1, which are negative, and more
negative the closer you are to x = –1. The graph decreases sharply to the
left of the origin.
Graph of f2(x) = log (1 + x).
Next, try f3(x) = ln (3 – x).
One way to do this is by moving the graph of ln x to the left 3 units - by
replacing x with x – 3, then reflecting across the y-axis - by changing the
sign of (x – 3) to –(x – 3) = (3 – x). In other words, take
f3(x) = f1(–[x – 3]). The graph shows a curve that looks like that of f1(x)
but moved over 3 units and increasing in the opposite direction.
If you don't look at the function this way, you obtain the graph using the
usual procedure. First, realize that you cannot set x equal to 3, nor to
values greater than 3, since these values would make 3 – x 0. You can
use x = 0, and the intercept is (0, ln 3). Since ln 3 is not a simple number
to calculate (its value is 1.0986 according to your calculator) nor to locate
on the graph, you can find an easier starting point by using x = 2 and get
f3(2) = ln (3 – 2) = ln 1 = 0. This starting point is (2, 0).
If you next use input values greater than your starting x-value, the
function values you get are lower than your original function value since
3 – x becomes smaller when x is larger. If the input values are near x = 3,
3 – x is near 0, and the logarithm of 3 – x gets very negative. The graph
decreases to the right of the starting point, very rapidly near x = 3.
Now use x-values less than your starting value. Since you subtract your
x-value from 3 (remember to change the sign for negative numbers), the
further to the left your x-value is, the larger 3 – x is. Larger values of 3
– x yield larger values of log (3 – x). The values of the function increase
slowly as you move toward the left.
You can see that this graph does look like a reflection of f 1(x) that was
first translated 3 units to the right.
Graph of f3(x) = ln (3 – x).
CHAPTER 7: LOGARITHMIC FUNCTIONS
For your next example, look at f4(x) = ln (x2 – 1).
Now there are two x-values which make x2 – 1 equal to zero, both x = 1
and x = –1. You cannot use either of these values, nor any values
between x = –1 and x = 1, which make x2 – 1 negative. The graph has
two separate pieces, on the left for x-values less than x = –1, and on the
right for values greater than x = 1. You may notice that these two pieces
are symmetric, since this function is an even function - changing the sign
of an input value does not change the value you obtain for the function.
Noting this before you start saves you half the work.
You can use any valid starting x-value to obtain symmetric starting points
for each branch. To make use of the fact that ln 1 = 0, you might try
using x =
(and –
on the other side) because then, x2 is 2, x2 – 1
= 1 and ln (x2 – 1) = 0. So, you can use (
, 0) and (–
, 0) for starting
points.
For x-values that are very large (or very negative) x2 – 1 is very large and
the logarithms are also large - but remember that logarithms get large
slowly.
When the x-values are near 1 (or –1), x2 – 1 is near 0, and the logarithm
is extremely negative in value.
The graph steadily rises to the extreme left and falls as you move to the
right, through your starting point, then rapidly drops to very negative
values as you near x = –1. There is no graph at all until you move beyond
x = 1, when the graph again appears but has very negative values. The
graph then climbs through your starting point and keeps rising, although
at a slower rate as you move further to the right.
Graph of f4(x) = log (x2 – 1).
For a different looking graph try f5(x) = ln (1 – x2).
Again, both x = –1 and x = 1 must be eliminated, as well as x-values to
2
the left of x = –1 and x-values to the right of x = 1, since (1 – x
) is
negative for such x-values.
The entire domain lies between x = –1 and x = 1. The graph is again
symmetric, since this function is even. You square every input value
before using it further.
127
128 CHAPTER 7: LOGARITHMIC FUNCTIONS
You can obtain an intercept, since 0 is a valid x-value, and the function
value is ln (1 – 0) = ln 1 = 0. The intercept is at the origin.
Also notice that 1 – x2 cannot ever be higher than 1. Therefore, all values
for the function (except at the origin) are negative. If your x-values are
close to x = –1 and x = 1, then x2 – 1 is small, and the logarithm is very
negative.
The graph is a curve rising from extremely negative values for x-values
just above x = –1 to a high point at the origin, then down through negative
values, getting lower as you approach x = 1. There is no graph outside
these values.
Graph of f5(x) = log (1 – x2).
For a final example, graph f6(x) = log (x2 + 4).
Since x2 + 4 is positive, no matter what x-value is selected, there are no
values that you need to remove from the domain.
The intercept is at (0, log 4). The value of log 4 depends on the base used
for the logarithm, but you can be sure that it is a positive number since log
4 is surely larger than log 1.
Squaring extremely negative values (and adding 4), produces large
numbers and relatively large logarithms. The same is true for large
positive input values. Note that this graph is also symmetric since the
function is even.
The lowest value is at the intercept, and the graph is be a curve that drops
down from high function values on the left, through the intercept, and
then rises to high values on the right. However, the curve does not quite
look like a parabola, since the ends do not rise as sharply as in a parabola.
Graph of f6(x) = log (x2 + 4).
CHAPTER 7: LOGARITHMIC FUNCTIONS
Applications
The most useful application of logarithms is in the solution of exponential
problems because a logarithm is the inverse of an exponential function.
Logarithms are used to undo the results of raising to powers.
Besides solving exponential equations like the examples done earlier in this
chapter, another problem in which you could apply logarithms is given in
exercise #9e, on page 113. To graph the function f(x) =
, you need
to find any x-values that make the denominator equal to zero. The answer
(p. 117, or on your calculator) showed that a negative x-value was eliminated
from the domain. To find this value, you have to solve the following
equation.
=0
10 =
=
taking natural logs on both sides, you obtain
ln
= –.5x
=x
or
x = –1.0216
which does appear to be the value eliminated in the correct graph.
Logarithmic functions are also used in models for functions that describe
quantities that increase, but at slow rates (slower than polynomials).
For example, you can model the amount of knowledge a person
accumulates in a lifetime by a function such as K(x) = log (2x + 5), where
x represents how long her life has been.
The values you must eliminate from your domain are the values x
which would be meaningless in the model anyway.
–2.5,
The intercept, which is (0, log 5) represents the knowledge she had
already gained by the time she was born.
As her life proceeds, her knowledge first increases rapidly, then at a
slower rate, but it does always increase. The graph is shown at the right.
Graph of the knowledge model.
129
130 CHAPTER 7: LOGARITHMIC FUNCTIONS
Exercises
1. Find the following logarithms:
a. log10 1000
b. log4
c. log27 9
d. log25
e. log16 4
2. Given log 2 = .416, log 3 = .659, and log 7 = 1.167, find:
a. log 6
b. log 4
c. log 42
d. log
e.
log
f.
g. log
h. log
i.
j.
log
log
log 36
3. Simplify:
a. log
b. log
c. log
4. Solve the following equations:
a. 3 + 2 x −1 = 11
b. 175 − 2 ⋅ 32x + 3 = 13
c. 5 x +1 = 9
d. (1.23 ) = 5
x
e. log6 (x + 1) = 3
5. Find:
a. log4 7
b. log11 5
6. Graph:
a.
b.
c.
d.
f(x) = log (x + 4)
f(x) = log (2x – 6)
f(x) = log (9 – x2)
f(x) = ln (e + 3x2)
e. f(x) = log
f. f(x) = log (1 – x3)
g. f(x) = log (x3 – 27)
h. f(x) = log
( )
i. f (x) = ln e x
2
j. f(x) = log (log x)
c. log5 e
d. log8 10
e. log7.4 12
CHAPTER 7: LOGARITHMIC FUNCTIONS
7. The amount of a radioactive substance present in grams is given by the function A(t) = 100
131
, where
t is given in centuries. Sketch a graph of this function. How much of the substance was present at first?
How long will it take the amount to decay to only 70 grams? When will there only be one-half the original
amount left?
8. A person's strength decays from the time he is about 25. One model for this decay gives the percentage
of his strength at age 25 that he retains at age x.
Suppose the model is given by
S(x) = 100 – 24 ln (x – 24). Graph this model. How much strength will this person retain at age 50?
When does the model indicate he will have lost all his strength?
9. The value of one model of automobile when it is n years old is given by V(n) = 12,000
. Graph this
function. What did the car originally cost? What will it be worth in 5 years? When should the owner sell
the car if she hopes to be able to get $5000 for it?
132 CHAPTER 7: LOGARITHMIC FUNCTIONS
Answers
1. a. 3
b. –3
c.
d.
e.
2. a. 1.075
f. .208
b. .832
g. –.092
c. 2.242
h. .486
d. .508
i. 2.150
e. –.416
j. 1.2535
3. a. log (x – 2) – log (5x + 3)
b.
c. 3 log (4x + 1) + 2 log (7x – 3) – 5 log (x + 2)
4. a.
= 11 – 3 = 8
log2
= log2 8 = 3
x–1=3
x=4
b.
162 = 2
81 =
log3 81 = log3
4 = 2x + 3
x=
c.
log5
= log5 9
x + 1 = log5 9
x=
d.
x ln (1.23) = ln 5
x=
e.
– 1 = .365
= 7.7745
x+1=
x=
– 1 = 215
CHAPTER 7: LOGARITHMIC FUNCTIONS
5. a.
d.
= 1.4036
= 1.1073
b.
= .6712
e.
= 1.2415
c.
= .6213
6. a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
133
134 CHAPTER 7: LOGARITHMIC FUNCTIONS
7. At first there were 100 grams present. After 29.7 centuries there will be 70 grams left. Half the original
amount (50 grams) will be present after 57.8 centuries.
Graph of radioactive decay model.
8. At age 50 he will still have 22% of his strength. It will all be lost when he is about 88 years old.
Graph of the strength model.
9. The car originally cost $12,000. After 5 years the value is approximately $7085. It will be worth $5000 after
8.3 years.
Graph of car depreciation model.