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Chapter 9
Exponential and
Logarithmic
Functions
§ 9.1
The Algebra of Functions;
Composite Functions
Operations on Functions
It is possible to add, subtract, multiply, and
divide functions.
The results of these operations will also be
functions (assuming we don’t divide by zero).
Martin-Gay, Intermediate Algebra, 5ed
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Operations on Functions
Algebra of Functions
Let f and g be functions. New functions from f and g
are defined as follows:
Sum
(f + g)(x) = f(x) + g(x)
Difference (f – g)(x) = f(x) – g(x)
Product
(f · g)(x) = f(x) · g(x)
Quotient
f
f ( x)
 ( x) 
, g ( x)  0
g ( x)
g
Martin-Gay, Intermediate Algebra, 5ed
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Operations on Functions
Example:
If f(x) = 4x + 3 and g(x) = x2, then find each of the
following
•
(f + g)(x)
4x + 3 + x2 = x2 + 4x + 3
•
(f – g)(x)
4x + 3 – x2 = -x2 + 4x + 3
•
(f · g)(x)
(4x + 3)x2 = 4x3 + 3x2
Martin-Gay, Intermediate Algebra, 5ed
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Operations on Functions
Example:
f
If f(x) = 4x + 3 and g(x) = x2, then find  (x)
g
4x  3 4 3
  2
2
x
x x
x0
Martin-Gay, Intermediate Algebra, 5ed
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Function Composition
We can also combine functions through a function
composition.
A function composition uses the output from the
first function as the input to the second function.
Composition of a Function
The composition of function f and g is
( f  g )( x)  f ( g ( x))
This means the value of x is first substituted into
the function g. Then the value that results from the
function g is input into the function f.
Martin-Gay, Intermediate Algebra, 5ed
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Function Composition
Notice, that with function composition, we
actually activate the functions from right to left in
the notation. The function named on the right
side of the composition notation is the one we
substitute the value for the variable into first.
Martin-Gay, Intermediate Algebra, 5ed
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Function Composition
Example:
If f(x) = 4x + 3 and g(x) = x2, then find
( f  g )( x)
f(g(x)) = 4(x2) + 3 = 4x2 + 3
( g  f )( x)
g(f(x)) = (4x + 3)2 = 16x2 + 24x + 9
Notice the results are different with a different order.
Martin-Gay, Intermediate Algebra, 5ed
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Function Composition
Example:
If H(x) = x3 + 3, name two functions whose
composition will result in H(x).
Note: There may be more than one way to select
the two functions. Answers are not necessarily
unique.
Let f(x) = x + 3, and g(x) = x3
( f  g )( x)  f ( g ( x))  x3  3
Martin-Gay, Intermediate Algebra, 5ed
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§ 9.3
Exponential Functions
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Exponential Expressions
We have previously worked with exponential
expressions, where the exponent was a
rational number
The expression bx can actually be defined for
all real numbers, x, including irrational
numbers.
However, the proof of this would have to wait
until a higher level math course.
Martin-Gay, Intermediate Algebra, 5ed
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Exponential Functions
Exponential Functions
A function of the form
f(x) = bx
is called an exponential function if b > 0, b is
not 1, and x is a real number.
Martin-Gay, Intermediate Algebra, 5ed
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Exponential Functions
We can graph exponential functions of the form
f(x) = 3x, g(x) = 5x or h(x) = (½)x by substituting in
values for x, and finding the corresponding
function values to get ordered pairs.
We would find all graphs satisfy the following
properties:
• 1-to-1 function
• y-intercept (0, 1)
• no x-intercept
• domain is (-, )
• range is (0, )
Martin-Gay, Intermediate Algebra, 5ed
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Graphs of Exponential Functions
We would find a
pattern in the graphs
of all the exponential
functions of the type
bx, where b > 1.
y
x
Martin-Gay, Intermediate Algebra, 5ed
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Graphs of Exponential Functions
y
We would find a
pattern in the graphs
of all the exponential
functions of the type
bx, where 0 < b < 1.
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Graphs of Exponential Functions
y
We would find a
pattern in the graphs
of all the exponential
functions of the type
bx-h, where b > 1.
(h, 1)
x
The graph has the
same shape as the
graph for bx, except
it is shifted to the
right h units.
Martin-Gay, Intermediate Algebra, 5ed
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Graphs of Exponential Functions
y
We would find a
pattern in the graphs
of all the exponential
functions of the type
bx+h, where b > 1.
(-h, 1)
x
The graph has the
same shape as the
graph for bx, except
it is shifted to the
left h units.
Martin-Gay, Intermediate Algebra, 5ed
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Uniqueness of
x
b
Uniqueness of bx
Let b > 0 and b  1. Then bx = by is equivalent
to x = y.
Example:
Solve 6x = 36
6x = 6 2
x=2
Martin-Gay, Intermediate Algebra, 5ed
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Solving Exponential Functions
Example:
Solve 92x+1 = 81
92x+1 = 92
2x + 1 = 2
2x = 1
x=½
Martin-Gay, Intermediate Algebra, 5ed
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Solving Exponential Functions
Example:
1
 32 x
Solve
27
3-3 = 32x
–3 = 2x
3
x
2
Martin-Gay, Intermediate Algebra, 5ed
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Solving Exponential Functions
Example:
Solve 43x-6 = 322x
(22)3x-6 = (25)2x
(22)3x-6 = 210x
26x-12 = 210x
6x – 12 = 10x
–12 = 4x
x = –3
Martin-Gay, Intermediate Algebra, 5ed
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Applications of Exponential Functions
Many applications use exponential functions of
various types.
Compound interest formulas are exponential functions
used to determine the amount of money accumulated
or borrowed.
Exponential functions with negative exponents can be
used to describe situations of decay, while those with
positive exponents can be used to describe situations
of growth.
Martin-Gay, Intermediate Algebra, 5ed
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Applications of Exponential Functions
Example:
Find the total amount invested in a savings account if $5000
was invested and earned 6% compounded monthly for 18
years. Round your answer to two decimal places.
The formula that is used for calculating compound interest is
 r
A  P 1  
 n
nt
where P is the initial principal invested, r is the interest rate, n
is the number of times interest is compounded each year, t is
the time of the investment (in years) and A is the amount of
money in the account.
Continued.
Martin-Gay, Intermediate Algebra, 5ed
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Applications of Exponential Functions
Example continued:
1218
nt
 r
 0.06 
A  P1    50001 

12 
 n

 50001  0.005
216
 50001.005
216
 $14683.83
Martin-Gay, Intermediate Algebra, 5ed
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Applications of Exponential Functions
Example:
An accidental spill of 100 grams of radioactive material in a
local stream has led to the presence of radioactive debris
decaying at a rate of 5% each day. Find how much debris still
remains after 30 days.
The formula that would be used for this problem is
y  A(2.7)  rt
where A is the amount of radioactive material to start, r is the
rate of decay, t is the number of days and y is the amount of
radioactive material after the time period.
Continued.
Martin-Gay, Intermediate Algebra, 5ed
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Applications of Exponential Functions
Example continued:
y  A(2.7)  rt  100(2.7) 0.0530
 100(2.7) 1.5
(exact answer)
 22.54 grams
Martin-Gay, Intermediate Algebra, 5ed
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9.1
#'s 1-23 odd
9.3 #'s 1, 5, 17, 19,
21, 23, 29, 31, 33
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§ 9.4
Logarithmic Functions
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Graph of a Exponential Function
If we graph an exponential function where the
base > 1, we get an increasing function, as
y
shown below.
x
Martin-Gay, Intermediate Algebra, 5ed
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Graph of a Logarithmic Function
We can graph the inverse of the function, as shown below.
y
x
This inverse function is referred to as a logarithmic function.
Martin-Gay, Intermediate Algebra, 5ed
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Logarithmic Functions
Logarithmic Definition
If b > 0 and b ≠ 1, then
y = logb x means x = by
for every x > 0 and every real number y.
Martin-Gay, Intermediate Algebra, 5ed
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Writing Exponential Functions
Example:
Write each of the following as an exponential equation.
a) log4 16 = 2
4² = 16
b)
log8 ⅛ = –1
8–1 = ⅛
c)
log3 3 = ½
31/ 2  3
Martin-Gay, Intermediate Algebra, 5ed
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Writing Logarithmic Functions
Example:
Write each of the following as a logarithmic equation.
a) 54 = 625
log5 625 = 4
b) 2–3 = ⅛
log2 ⅛ = –3
c) 41/3 =
3
4
log4 3 4 = ⅓
Martin-Gay, Intermediate Algebra, 5ed
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Values of Logarithmic Expressions
Example:
Find the value of each of the following logarithmic
expressions.
a) log2 32
Since 25 = 32, then log2 32 = 5
b)
1
log5 25
Since
5–2
=
1
25
, then
1
log5 25
= –2
c) log4 2
Since 4½ = 2, then log4 2= ½
Martin-Gay, Intermediate Algebra, 5ed
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Solving Logarithmic Equations
Example:
Solve log3 1 = x for x.
First we rewrite the equation as an exponential
equation.
3x = 1
Since 30 = 1, then x = 0.
Martin-Gay, Intermediate Algebra, 5ed
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Solving Logarithmic Equations
Example:
Solve logx 81 = 4 for x.
First we rewrite the equation as an exponential
equation.
x4 = 81
Since 34 = 81, then x = 3.
Martin-Gay, Intermediate Algebra, 5ed
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Solving Logarithmic Equations
Example:
Solve log6 x = 2 for x.
First we rewrite the equation as an exponential
equation.
62 = x
Since 62 = 36, then x = 36.
Martin-Gay, Intermediate Algebra, 5ed
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Properties of Logarithms
Properties of Logarithms
1) logb 1 = 0
2) logb bx = x
3) b logb x = x
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Properties of Logarithms
Example:
Simplify each of the following expressions
1) log4 46
From Property 2, log4 46 = 6.
2) 7 log7 –3
From Property 3, 7
log7 –3 = –3.
Martin-Gay, Intermediate Algebra, 5ed
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Logarithmic Functions
Logarithmic Function
If x is a positive real number, b is a constant positive
real number, and b is not 1, then a logarithmic
function is a function that can be defined by
f(x) = logb x
The domain of f is the set of positive real numbers, and
the range of f is the set of real numbers.
A logarithmic function is an inverse function of an
exponential function.
Martin-Gay, Intermediate Algebra, 5ed
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Graphs of Logarithmic Functions
To graph a logarithmic function, we first
write the equation in exponential notation.
Then we find ordered pairs that satisfy the
equation and plot their corresponding
points.
Martin-Gay, Intermediate Algebra, 5ed
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Graphs of Logarithmic Functions
Example:
Graph y = log2 x.
y
Write the exponential form 2y = x.
Then choose y-values and find
corresponding x-values.
x
y
4
2
2
1
1
0
½
-1
¼
-2
x
Martin-Gay, Intermediate Algebra, 5ed
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Graphs of Logarithmic Functions
Example:
Graph y = log½ x.
y
Write the exponential form
(½)y = x.
Then choose y-values and find
corresponding x-values.
x
y
¼
2
½
1
1
0
2
-1
4
-2
x
Martin-Gay, Intermediate Algebra, 5ed
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Logarithmic Functions
In general, from these two previous examples, we would
find that for the logarithmic function
f(x) = logb x, b > 0, b ≠ 1,
The function
• is a 1-to-1 function.
• has a domain of (0, ∞).
• has a range (–∞, ∞).
• has an x-intercept of (1, 0).
• has no y-intercept.
Martin-Gay, Intermediate Algebra, 5ed
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§ 10.1
The Parabola and
the Circle
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Conic Sections
Conic sections derive their name because each
conic section is the intersection of a right circular
cone and a plane.
Circle
Ellipse
Parabola
Martin-Gay, Intermediate Algebra, 5ed
Hyperbola
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The Parabola
Just as y = a(x – h)2 + k is the equation of a parabola that
opens upward or downward, x = a(y – k)2 + h is the
equation of a parabola that opens to the right or to the left.
y = a(x – h)2 + k
y
a>0
x = a(y – k)2 + h
y
y
y
(h, k)
(h, k)
y=k
(h, k)
a<0
(h, k)
x
x=h
y=k
x
x
a>0
a<0
x
x=h
Martin-Gay, Intermediate Algebra, 5ed
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The Parabola
Example:
Graph the parabola x = (y – 4)2 + 1.
• a > 0, so the parabola opens to the right.
• The vertex of the parabola is (1, 4).
• The axis of symmetry is y = 4.
Martin-Gay, Intermediate Algebra, 5ed
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The Parabola
Example continued:
y
The table shows ordered
pairs of the solutions of
x = (y – 4)2 + 1.
x
1
y
4
2
3
2
17
17
5
0
8
y=4
2
x
2
Martin-Gay, Intermediate Algebra, 5ed
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The Parabola
Example:
Graph the parabola y = x2 + 12x + 25.
• Complete the square on x to write the equation in
standard form.
y – 25 = x2 + 12x
Subtract 25 from both sides.
• The coefficient of x is 12. The square of half of
12 is 62 = 36.
y – 25 + 36 = x2 + 12x + 36
Add 36 to both sides.
Martin-Gay, Intermediate Algebra, 5ed
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The Parabola
Example continued:
y + 11 = (x + 6)2
Simplify the left side
and factor the right side.
y = (x + 6)2 – 11
Subtract 11 from both sides.
• a > 0, so the parabola opens upward.
• The vertex of the parabola is (– 6, – 11).
• The axis of symmetry is x = – 6.
Martin-Gay, Intermediate Algebra, 5ed
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The Parabola
Example continued:
y
y = x2 + 12x + 25
3
3
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The Distance Formula
Distance Formula
The distance d between any two points (x1, y1) and
(x2, y2) is given by
d  (x2  x1)2  ( y2  y1)2.
y
(x2, y2)
d
b = y 2 – y1
x
(x1, y1)
a = x 2 – x1
Martin-Gay, Intermediate Algebra, 5ed
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The Distance Formula
Example:
Find the distance between (– 6, – 6) and (– 5, – 2).
d  (x2  x1)2  ( y2  y1)2
 [5  (6)]2  [2  (6)]2
 (5  6) 2  (2  6) 2
 (1) 2  (4) 2
 1  16
 17  4.123
Martin-Gay, Intermediate Algebra, 5ed
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The Midpoint
• The midpoint of a line segment is the point located
exactly halfway between the two endpoints of the
line segment.
Midpoint Formula
The midpoint of the line segment whose endpoints
are (x1, y1) and (x2, y2) is the point with the
coordinates
 x1  x2 , y1  y2  .
 2

2


Martin-Gay, Intermediate Algebra, 5ed
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The Midpoint
Example:
Find the midpoint of the line segment that joins points
P(0, 8) and Q(4, – 6).
x1  x2 y1  y2 

midpoint  
,

2
2


0  4 8  (6) 


,

2
2


4 2
  , 
2 2
  2,1
Martin-Gay, Intermediate Algebra, 5ed
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The Cirlce
A circle is the set of all points in a plane that are the same
distance from a fixed point called the center. The
distance is called the radius.
Circle
The graph of (x – h)2 + (y – k)2 = r2 is a circle with center
(h, k) and radius r.
y
r
(h, k)
x
Martin-Gay, Intermediate Algebra, 5ed
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The Circle
Example:
Graph (x – 3)2 + y2 = 9.
The equation can be written as (x – 3)2 + (y – 0)2 = 32.
h = 3, k = 0, and r = 3.
y
r=3
(3, 0)
Martin-Gay, Intermediate Algebra, 5ed
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The Circle
Example:
Find the equation of the circle with center (– 7, 6)
and radius 2.
• h = – 7, k = 6, and r = 2.
• (x – h)2 + (y – k)2 = r2.
The equation can be written as [x – (– 7)2] + (y – 6)2 = 22.
(x + 7)2 + (y – 6)2 = 4.
Simplify.
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