Transcript Exponential and logarithmic functions

SECTION 4.1
Exponential Functions
5
Define an exponential function.
Graph exponential functions.
Use transformations on exponential functions.
Define simple interest.
Develop a compound interest formula.
6
Understand the number e.
1
2
3
4
EXPONENTIAL FUNCTION
A function f of the form
f x   a x ,
a  0 and a  1,
is called an exponential function with base
a. Its domain is (–∞, ∞).
EXAMPLE 1
Evaluating Exponential Functions
a. Let f x   3 . Find f 4 .
x2
b. Let g x   2 10 x. Find g 2 .
x
 1
 3
c. Let h x     . Find h    .
 9
 2
d. Let F(x) = 4x. Find F(3.2).
EXAMPLE 1
Evaluating Exponential Functions
Solution
a. f  4   3
4 2
3 9
2
1
1
b. g  2   2  10  2  2  2 
 0.02
10
100
2
 3 1
c. h      
 2 9

3
2
 9
3

1 2

3
2
 9  27
d. F(3.2) = 43.2 ≈ 84.44850629
RULES OF EXPONENTS
Let a, b, x, and y be real numbers with a > 0
and b > 0. Then
a a  a
x
y
x y
a
,
x
a
x y
a ,
y
a
 ab 
x

y
 a xy ,
a  1,
0
x
x
a b ,
x
x
a
x
1 1
 x   .
a
a
EXAMPLE 2
Graphing an Exponential Function with
Base a > 1 – Exponential Growth
Graph the exponential function f x   3 .
x
Solution
Make a table of values.
Plot the points and draw a smooth curve.
EXAMPLE 2
Graphing an Exponential Function with
Base
a>1
Solution continued
This graph is typical for exponential functions
when a > 1.
EXAMPLE 3
Graphing an Exponential Function with
Base 0 < a < 1 – Exponential Decay
x
1
Sketch the graph of y    .
2
Solution
Make a table of values.
Plot the points and draw a smooth curve.
EXAMPLE 3
Graphing an Exponential Function with
Base 0 < a < 1
Solution continued
As x increases
in the positive
direction, y
decreases
towards 0.
PROPERTIES OF EXPONENTIAL FUNCTIONS
Let f (x) = ax, a > 0, a ≠ 1.
1. The domain of f (x) = ax is (–∞, ∞).
2. The range of f (x) = ax is (0, ∞); the entire graph lies
above the x-axis.
3. For a > 1, Exponential Growth
(i)
f is an increasing function, so the graph rises to
the right.
(ii)
as x → ∞, y → ∞.
(iii)
as x → –∞, y → 0.
4. For 0 < a < 1, - Exponential Decay
(i)
f is a decreasing function, so the graph falls to the
right.
(ii) as x → – ∞, y → ∞.
(iii)
5.
6.
7.
as x → ∞, y → 0.
The graph of f (x) = ax has no x-intercepts, so it never
crosses the x-axis. No value of x will cause f (x) = ax to
equal 0.
The graph of is a smooth and continuous curve, and it
passes through the points
The x-axis is a horizontal asymptote for every
exponential function of the form
f (x) = ax.
TRANSFORMATIONS ON EXPONENTIAL
FUNCTION f (x) = ax
Transformation Equation
Horizontal
y = ax+b
Shift
Effect on Equation
Shift the graph of
y = ax, |b| units
(i) left if b > 0.
(ii) right if b < 0.
y = ax + b
Shift the graph of
y = ax, |b| units
(i) up if b > 0.
(ii) down if b < 0.
Vertical
Shift
TRANSFORMATIONS ON EXPONENTIAL
FUNCTION f (x) = ax
Transformation Equation
Stretching or
Compressing
(Vertically)
y = cax
Effect on Equation
Multiply the y
coordinates by c.
The graph of y = ax
is vertically
(i) stretched if c > 1.
(ii) compressed if
0 < c < 1.
TRANSFORMATIONS ON EXPONENTIAL
FUNCTION f (x) = ax
Transformation Equation
Reflection
Effect on Equation
y = –ax
Reflect the graph
of y = ax in the
x-axis.
y = a–x
Reflect the graph
of y = ax in the
y-axis.
EXAMPLE 6
Sketching Graphs
Use transformations to sketch the graph of each
function.
a. f  x   3x  4
b. f  x   3x1
c. f  x   3x
d. f  x   3x  2
State the domain and range of each function and
the horizontal asymptote of its graph.
EXAMPLE 6
Sketching Graphs
Solution
a. f  x   3x  4
Domain: (–∞, ∞)
Range: (–4, ∞)
Horizontal
Asymptote: y = –4
EXAMPLE 6
Sketching Graphs
Solution continued
b. f  x   3x1
Domain: (–∞, ∞)
Range: (0, ∞)
Horizontal
Asymptote: y = 0
EXAMPLE 6
Sketching Graphs
Solution continued
c. f  x   3
x
Domain: (–∞, ∞)
Range: (–∞, 0)
Horizontal
Asymptote: y = 0
EXAMPLE 6
Sketching Graphs
Solution continued
d. f  x   3  2
x
Domain: (–∞, ∞)
Range: (–∞, 2)
Horizontal
Asymptote: y = 2
General Exponential Growth/Decay Model
Rate of decay (r < 0),Growth (r > 0)
Amount after t time
periods
Original amount
Number of time
periods
COMPOUND INTEREST – Growth
Compound interest is the interest paid on both the principal and
the accrued (previously earned) interest. It is an application of
exponential growth.
Interest that is compounded annually is paid once a year. For
interest compounded annually, the amount A in the account after
t years is given by
Rate of decay (r < 0),Growth (r > 0)
Amount after t time
periods
Number of time
periods
Original amount
EXAMPLE 2
Calculating Compound Interest
Juanita deposits $8000 in a bank at the interest
rate of 6% compounded annually for five
years.
a. How much money will she have in her
account after five years?
b. How much interest will she receive?
EXAMPLE 2
Calculating Compound Interest
Solution
a. Here P = $8000, r = 0.06, and t = 5.
b. Interest = A  P
= $10,705.80  $8000 = $2705.80.
COMPOUND INTEREST FORMULA


r
A  P 1 
n

nt
A = amount after t years
P = principal
r = annual interest rate (expressed as a decimal)
n = number of times interest is compounded
each year
t = number of years
EXAMPLE 3
Using Different Compounding Periods to
Compare Future Values
If $100 is deposited in a bank that pays 5%
annual interest, find the future value A after one
year if the interest is compounded
(i)
(ii)
(iii)
(iv)
(v)
annually.
semiannually.
quarterly.
monthly.
daily.
EXAMPLE 3
Using Different Compounding Periods to
Compare Future Values
Solution
In the following computations, P = 100,
r = 0.05 and t = 1. Only n, the number of times
interest is compounded each year, changes.
Since t = 1, nt = n(1) = n.
nt
(i) Annual
Compounding:
 r
A  P 1  
 n
A  100 1  0.05   $105.00
EXAMPLE 3
Using Different Compounding Periods to
Compare Future Values
(ii) Semiannual Compounding:
2
 r
A  P 1  
 n
2
 0.05 
A  100 1 
  $105.06
2 

(iii) Quarterly Compounding:
4
 r
A  P 1  
 4
4
 0.05 
A  100 1 
  $105.09
4 

EXAMPLE 3
Using Different Compounding Periods to
Compare Future Values
(iv) Monthly Compounding:
12
r 

A  P 1  
 12 
12
 0.05 
A  100 1 
  $105.12
12 

(v) Daily Compounding:
365
r 

A  P 1 

 365 
 0.05 
A  100 1 

365 

365
 $105.13
EXAMPLE 8
Bacterial Growth
A technician to the French microbiologist Louis
Pasteur noticed that a certain culture of bacteria
in milk doubles every hour. If the bacteria count
B(t) is modeled by the equation
B t   2000  2t ,
with t in hours, find
a. the initial number of bacteria,
b. the number of bacteria after 10 hours; and
c. the time when the number of bacteria will be
32,000.
EXAMPLE 8
Bacterial Growth
Solution
a. Initial size
B0  B  0   2000  20  2000  1  2000
b. B 10   2000  2  2, 048,000
10
c. Find t when B(t) = 32,000
t
4
t
32,000  2000  2
2 2
t
4t
16  2
After 4 hours, the number of bacteria will be
32,000.
THE VALUE OF e
As h gets larger and larger,
 1
1  
 h
h
gets closer and closer to a fixed number. This
irrational number is denoted by e and is sometimes
called the Euler number.
The value of e to 15 places is
e = 2.718281828459045.
CONTINUOUS COMPOUND
FORMULA
rt
A  Pe
A = amount after t years
P = principal
r = annual rate (expressed as a decimal)
t = number of years
EXAMPLE 4
Calculating Continuous Compound Interest
Find the amount when a principal of $8300 is
invested at a 7.5% annual rate of interest
compounded continuously for eight years and
three months.
Solution
P = $8300 and r = 0.075. Convert eight years
and three months to 8.25 years.
A  Pe rt
 0.0758.25
A  $8300e
 $15, 409.83
EXAMPLE 5
Calculating the Amount of Repaying a Loan
How much money did the government owe
DeHaven’s descendants for 213 years on a
$450,000 loan at the interest rate of 6%?
Solution
a. With simple interest,
A  P  Prt  P 1  rt 
A  $450,000 1   0.06  213 
 $6.201 million.
EXAMPLE 5
Calculating the Amount of Repaying a Loan
Solution continued
b. With interest compounded yearly,
t
213
A  P 1  r   $450,000 1  0.06 
A  $1.105  10
 $110.500 million.
c. With interest compounded quarterly,
4t
4 213
 r
 0.06 
A  P 1    $450, 000  1 

4 
 4

11
A  $1.45305  10
 $145.305 billion.
11
EXAMPLE 5
Calculating the Amount of Repaying a Loan
Solution continued
d. With interest compounded continuously,
0.06 213
rt
A  Pe  $450,000e
A  $1.5977  1011
 $159.77 billion.
Notice the dramatic difference between
quarterly and continuous compounding and
the dramatic difference between simple
interest and compound interest.
THE NATURAL EXPONENTIAL FUNCTION
The exponential function
f  x  e
x
with base e is so prevalent in the sciences
that it is often referred to as the exponential
function or the natural exponential function.
EXAMPLE 6
Sketching a Graph
Use transformations to sketch the graph of
Solution
Start with the
graph of y = ex.
EXAMPLE 6
Sketching a Graph
Use transformations to sketch the graph of
Solution coninued
Shift the graph of
y = ex one unit
right.
EXAMPLE 6
Sketching a Graph
Use transformations to sketch the graph of
Solution continued
Shift the graph of
y = ex – 1 two units
up.
MODEL FOR EXPONENTIAL
GROWTH OR DECAY
A  t   A0 e
kt
A(t) = amount at time t
A0 = A(0), the initial amount
k = relative rate of growth (k > 0) or decay
(k < 0)
t = time
EXAMPLE 7
Modeling Exponential Growth and Decay
In the year 2000, the human population of the
world was approximately 6 billion and the
annual rate of growth was about 2.1%. Using the
model on the previous slide, estimate the
population of the world in the following years.
a. 2030
b. 1990
EXAMPLE 7
Modeling Exponential Growth and Decay
Solution
a. The year 2000 corresponds to t = 0. So
A0 = 6 (billion), k = 0.021, and 2030
corresponds to t = 30.
A  30   6 e
 0.021 30 
 11.265663
The model predicts that if the rate of growth is
2.1% per year, over 11.26 billion people will be
in the world in 2030.
EXAMPLE 7
Modeling Exponential Growth and Decay
Solution
b. The year 1990 corresponds to t = 10.
A  10   6e
 0.021 10 
 4.8635055
The model predicts that the world had over 4.86
billion people in 1990. (The actual population
in 1990 was 5.28 billion.)
SECTION 4.3
Logarithmic Functions
1
2
3
4
5
6
7
Define logarithmic functions.
Inverse Functions
Evaluate logarithms.
Rules of Logarithms
Find the domains of logarithmic functions.
Graph logarithmic functions.
Use logarithms to evaluate exponential equations.
DEFINITION OF THE
LOGARITHMIC FUNCTION
For x > 0, a > 0, and a ≠ 1,
y  log a x if and only if
xa .
y
The function f (x) = loga x, is called the
logarithmic function with base a.
The logarithmic function is the inverse
function of the exponential function.
Inverse Functions
Certain pairs of one-to-one functions
“undo” one another. For example, if
x 5
f ( x )  8 x  5 and g ( x ) 
,
8
then
f (10)  8(10)  5  85
(85)  5
g (85) 
 10
8
Inverse Functions
Starting with 10, we “applied” function 
and then “applied” function g to the
result, which returned the number 10.
Inverse Functions
As further examples, check that
f (3)  29 and g ( 29)  3,
f ( 5)  35 and g ( 35)  5,
3
g (2)  
8
 3
and g     2,
 8
Inverse Functions
In particular, for this pair of functions,
f ( g (2))  2 and g ( f ( 2))  2.
In fact, for any value of x,
f ( g ( x ))  x and g ( f ( x ))  x,
or ( f
g )( x )  x and (g
f )( x )  x.
Because of this property, g is called the
inverse of .
Inverse Function
Let  be a one-to-one function. Then g
is the inverse function of  if
(f
g )( x )  x
for every x in the
domain of g,
and
(g
f )( x )  x for every x in the
domain of .
EXAMPLE 1
Converting from Exponential to
Logarithmic Form
Write each exponential equation in logarithmic
form.
4
2
1
1

3
c.
a
7
a. 4  64
b.   
 2
16
Solution
a. 4  64  log 4 64  3
3
4
1
1
1
b.     log1 2  4
16
 2  16
c. a 2  7  log a 7  2
EXAMPLE 2
Converting from Logarithmic Form to
Exponential Form
Write each logarithmic equation in exponential
form.
a. log 3 243  5 b. log 2 5  x c. log a N  x
Solution
a. log3 243  5  243  3
5
b. log 2 5  x  5  2 x
c. log a N  x  N  a x
EXAMPLE 3
Evaluating Logarithms
Find the value of each of the following
logarithms.
c. log1 3 9
a. log 5 25
b. log 2 16
1
d. log 7 7
e. log 6 1
f. log 4
2
Solution
a. log 5 25  y  25  5 or 5  5
y
y2
b. log 2 16  y  16  2 or 2  2
y
y4
y
y
2
4
EXAMPLE 3
Evaluating Logarithms
Solution continued
y
1
c. log1 3 9  y  9    or 32  3 y
3
d. log 7 7  y  7  7 y or 71  7 y
e. log 6 1  y  1  6 or 6  6
y
0
y
y  2
y 1
y0
1
1
y
1
2y
f. log 4  y   4 or 2  2
2
2
1
y
2
EXAMPLE 4
Using the Definition of Logarithm
Solve each equation.
a. log 5 x  3
c. log z 1000  3
Solution
a. log 5 x  3
x  53
1
1
x 3 
5 125
1
b. log 3
y
27
d. log 2  x 2  6 x  10   1
EXAMPLE 4
Using the Definition of Logarithm
Solution continued
1
b. log 3
y
27
1
 3y
27
3
y
3 3
3  y
c. log z 1000  3
1000  z 3
103  z 3
10  z
EXAMPLE 4
Using the Definition of Logarithm
Solution continued
d. log 2  x  6 x  10   1
2
x  6 x  10  2  2
2
1
x  6x  8  0
2
 x  2  x  4   0
x  2  0 or x  4  0
x  2 or x  4
Rules of Logarithms
Rules of Logarithms with Base a
If M, N, and a are positive real numbers with a ≠ 1, and x is
any real number, then
1. loga(a) = 1
3. loga
(ax)
=x
2. loga(1) = 0
These relationships are
used to solve exponential
or logarithmic equations
4.
a loga ( N )  N
5. loga(MN) = loga(M) + loga(N)
6. loga(M/N) = loga(M) – loga(N)
7. loga(Mx) = x · loga(M)
8. loga(1/N) = – loga(N)
COMMON LOGARITHMS
The logarithm with base 10 is called the
common logarithm and is denoted by
omitting the base: log x = log10 x. Thus,
y = log x if and only if x = 10 y.
Applying the basic properties of logarithms
1. log 10 = 1
2. log 1 = 0
3. log 10x = x
log x
4. 10  x
NATURAL LOGARITHMS
The logarithm with base e is called the
natural logarithm and is denoted by ln x.
That is, ln x = loge x. Thus,
y = ln x if and only if x = e y.
Applying the basic properties of logarithms
1. ln e = 1
2. ln 1 = 0
3. log ex = x
ln x
4. e  x
DOMAIN OF LOGARITHMIC
FUNCTION
Domain of y = loga x is (0, ∞)
Range of y = loga x is (–∞, ∞)
Logarithms of 0 and negative numbers
are not defined.
EXAMPLE 5
Finding the Domain
Find the domain of f  x   log3  2  x 
Solution
Domain of a logarithmic function must be
positive, that is,
2 x 0
2x
The domain of f is (–∞, 2).
EXAMPLE 6
Sketching a Graph
Sketch the graph of y = log3 x.
Solution by plotting points (Method 1)
Make a table of values.
EXAMPLE 6
Sketching a Graph
Solution continued
Plot the
ordered pairs
and connect
with a smooth
curve to
obtain the
graph of
y = log3 x.
EXAMPLE 6
Sketching a Graph
Solution by using the inverse function (Method 2)
Graph y = f (x) = 3x.
Reflect the graph of
y = 3x in the line
y = x to obtain the
graph of
y = f –1(x) = log3 x.
66
GRAPHS OF LOGARITHMIC
FUNCTIONS
PROPERTIES OF EXPONENTIAL AND
LOGARITHMIC FUNCTIONS
Exponential Function
f (x) = ax
Logarithmic Function
f (x) = loga x
1. Domain (–∞, ∞)
Range (0, ∞)
Domain (0, ∞)
Range (–∞, ∞)
2. y-intercept is 1
No x-intercept
x-intercept is 1
No y-intercept
3. x-axis (y = 0) is the
y-axis (x = 0) is the
horizontal asymptote vertical asymptote
PROPERTIES OF EXPONENTIAL AND
LOGARITHMIC FUNCTIONS
Exponential Function
f (x) = ax
4. The graph is a
continuous smooth
curve that passes
through the points
1

 1,  , (0, 1), and
a

(1, a).
Logarithmic Function
f (x) = loga x
The graph is a
continuous smooth
curve that passes
through the points
1
 (1, 0), and
 ,  1 ,
a

(a, 1).
PROPERTIES OF EXPONENTIAL AND
LOGARITHMIC FUNCTIONS
Exponential Function
f (x) = ax
5. Is one-to-one , that
is, au = av if and
only if u = v.
Logarithmic Function
f (x) = loga x
Is one-to-one, that is,
logau = logav
if and only if u = v.
6. Increasing if a > 1
Increasing if a > 1
Decreasing if 0 < a < 1 Decreasing if 0 < a < 1
EXAMPLE 7
Using Transformations
Start with the graph of f (x) = log3 x and use
transformations to sketch the graph of each
function.
a. f x   log 3 x  2
b. f x   log 3 x  1
c. f x    log 3 x
d. f x   log 3 x 
State the domain and range and the vertical
asymptote for the graph of each function.
EXAMPLE 7
Using Transformations
Solution
a. f  x   log3 x  2
Shift up 2
Domain (0, ∞)
Range (–∞, ∞)
Vertical
asymptote x = 0
EXAMPLE 7
Using Transformations
Solution continued
b. f  x   log3  x  1
Shift right 1
Domain (1, ∞)
Range (–∞, ∞)
Vertical
asymptote x = 1
EXAMPLE 7
Using Transformations
Solution continued
c. f  x    log3 x
Reflect graph
of y = log3 x in
the x-axis
Domain (0, ∞)
Range (–∞, ∞)
Vertical
asymptote x = 0
EXAMPLE 7
Using Transformations
Solution continued
d. f  x   log3   x 
Reflect graph
of y = log3 x in
the y-axis
Domain (∞, 0)
Range (–∞, ∞)
Vertical
asymptote x = 0
EXAMPLE 8
Using Transformations to Sketch a Graph
Sketch the graph of y  2  log x  2 .
Solution
Start with the graph
of f (x) = log x.
Step 1:
Replacing x with
x – 2 shifts the
graph two units
right.
EXAMPLE 8
Using Transformations to Sketch a Graph
Solution continued
Step 2: Multiplying
by 1 reflects the graph
in the x-axis.
Step 3: Adding 2
shifts the graph
two units
up.
Rules of Logarithms
Rules of Logarithms with Base a
If M, N, and a are positive real numbers with a ≠ 1, and x is
any real number, then
1. loga(a) = 1
2. loga(1) = 0
3. loga(ax) = x
4.
a loga ( N )  N
5. loga(MN) = loga(M) + loga(N)
6. loga(M/N) = loga(M) – loga(N)
7. loga(Mx) = x · loga(M)
8. loga(1/N) = – loga(N)
EXAMPLE 1
Using Rules of Logarithms to Evaluate
Expressions
Given that log 5 z = 3 and log 5 y = 2, evaluate
each expression.
7
b. log 5 125y 
a. log5  yz 
c. log 5
d. log 5  z1/30 y 5 
z
y
Solution
a. log 5  yz   log 5 y  log 5 z
 23
5
EXAMPLE 1
Using Rules of Logarithms to Evaluate
Expressions
Solution continued
b. log 5 125 y 7   log 5 125  log 5 y 7
 log 5 53  7 log 5 y
 3  7  2   17
c. log 5

z
 log 5 
y

1/2
z
y 
1
  log 5 z  log 5 y 
2
1
1
 3  2 
2
2
EXAMPLE 1
Using Rules of Logarithms to Evaluate
Expressions
Solution continued
d. log 5  z1/30 y 5   log 5 z1/30  log 5 y 5
1
 log 5 z  5 log 5 y
30
1
  3  5  2 
30
 0.1  10
 10.1
EXAMPLE 2
Writing Expressions In Expanded Form
Write each expression in expanded form.
x  x  1
2
a. log 2
 2 x  1
3
3
2 5
b. ln x y z
4
Solution
x  x  1
2
a. log 2
 2 x  1
3
4
 log 2 x  x  1  log 2  2 x  1
2
3
 log 2 x  log 2  x  1  log 2  2 x  1
2
3
4
4
 2log 2 x  3log 2  x  1  4log 2  2 x  1
EXAMPLE 2
Writing Expressions In Expanded Form
Solution continued
b. ln x y z  ln  x y z
3
2 5
3

2 5 1/2
1
3 2 5
 ln  x y z 
2
1
  ln x 3  ln y 2  ln z 5 
2
1
  3ln x  2ln y  5ln z 
2
3
5
 ln x  ln y  ln z
2
2
EXAMPLE 3
Writing Expressions in Condensed Form
Write each expression in condensed form.
a. log3 x  log 4 y
1
b. 2ln x  ln  x 2  1
2
c. 2log 2 5  log 2 9  log 2 75
1
2

d. ln x  ln  x  1  ln  x  1 
3
EXAMPLE 3
Writing Expressions in Condensed Form
Solution
 3x 
a. log3 x  log 4 y  log  
 4y 
1/2
1
2
2
2
b. 2 ln x  ln  x  1  ln x  ln  x  1
2

 ln x 2 x 2  1

EXAMPLE 3
Writing Expressions in Condensed Form
Solution continued
c. 2log 2 5  log 2 9  log 2 75
 log 2 5  log 2 9  log 2 75
2
 log 2  25  9   log 2 75
25  9
 log 2
75
 log 2 3
EXAMPLE 3
Writing Expressions in Condensed Form
Solution continued
1
d. ln x  ln  x  1  ln  x 2  1 
3
1
2

 ln x  x  1  ln  x  1 
3
1  x  x  1 
 ln  2

3  x 1 
 ln 3
x  x  1
x 1
2
CHANGE-OF-BASE FORMULA
Let a, b, and x be positive real numbers
with a ≠ 1 and b ≠ 1. Then logb x can be
converted to a different base as follows:
log a x
log b x 
log a b
(base a )

log x
log b

ln x
ln b
(base 10) (base e)
EXAMPLE 4
Using a Change of Base to Compute
Logarithms
Compute log513 by changing to a. common
logarithms and b. natural logarithms.
Solution
log13
a. log 5 13 
log 5
 1.59369
ln 13
b. log 5 13 
ln 5
 1.59369
EXAMPLE 9
Evaluating the Natural Logarithm
Evaluate each expression.
1
4
b. ln 2.5
a. ln e
e
c. ln 3
Solution
a. ln e  4
4
1
2.5
b. ln 2.5  ln e  2.5
e
c. ln 3  1.0986123
Use a calculator.
EXAMPLE 10 Doubling Your Money
a. How long will it take to double your money if it
earns 6.5% compounded continuously?
b. At what rate of return, compounded continuously,
would your money double in 5 years?
Solution
a. If P is the original
amount invested,
A = 2P.
It will take 11 years
to double your
money.
EXAMPLE 10 Doubling Your Money
Solution continued
b.
Your investment will double
in 5 years at the rate of
13.86%.
Solving Exponential Or
Logarithmic Equations
To solve an exponential or logarithmic equation,
change the given equation into one of the following
forms, where a and b are real numbers, a > 0 and
a ≠ 1, and follow the guidelines.
1.ax = b
Solve by taking logarithms on both sides.
2. Loga x = b
Solve by changing to exponential form ab = x.
SOLVING AN EXPONENTIAL
EQUATION
Solve 7x = 12. Give the solution to the
nearest thousandth.
Solution
While any appropriate base b can be used, the best
practical base is base 10 or base e. We choose
base e (natural) logarithms here.
SOLVING AN EXPONENTIAL
EQUATION
Solve 7x = 12. Give the solution to the
nearest thousandth.
Solution x
7  12
In 7  In 12
x
Property of logarithms
xIn 7  In 12 Power of logarithms
In 12
x
Divide by In 7.
In 7
x  1.277 Use a calculator.
The solution set is {1.277}.
SOLVING AN EXPONENTIAL
EQUATION
Solve 32x – 1 = .4x+2 . Give the solution to the
nearest thousandth.
Solution
32 x 1  .4 x 2
In 3
2 x 1
 In .4
x 2
(2 x  1) In 3  ( x  2) In .4
2 x In 3  In 3  x In .4  2 In .4
Take natural
logarithms on both
sides.
Property power
Distributive property
SOLVING AN EXPONENTIAL
EQUATION
Solve 32x – 1 =.4x+2 . Give the solution to the
nearest thousandth.
Solution
2 x In 3  x In .4  2 In .4  In 3
Write the terms
with x on one side
x(2 In 3  In .4)  2 In .4  In 3
Factor out x.
2 In .4  3
x
2 In 3  .4
In .42  In 3
x
In 32  In .4
Divide by
2 In 3 – In .4.
Power property
SOLVING AN EXPONENTIAL
EQUATION
Solve 32x – 1 =.4x+2 . Give the solution to the
nearest thousandth.
Solution
This is exact.
In .16  In 3
x
In 9  In .4
In .48
x
9
In
.4
x  .236
Apply the
exponents.
Product property;
Quotient property
This is approximate.
The solution set is { –.236}.
SOLVING BASE e EXPONENTIAL
EQUATIONS
Solve the equation. Give solutions to the nearest
thousandth.
a. e
x2
 200
Solution
e
In e
x2
x2
 200
 In 200
x  In 200
2
Take natural logarithms
on both sides.
In
e
x2
= x2
SOLVING BASE e EXPONENTIAL
EQUATIONS
Solve the equation. Give solutions to the nearest
thousandth.
a. e
x2
 200
Remember
both roots.
Solution
x   In 200
Square root property
x  2.302
Use a calculator.
The solution set is { 2.302}.
SOLVING BASE e EXPONENTIAL
EQUATIONS
Solve the equation. Give solutions to the nearest
thousandth.
b. e 2 x 1  e 4 x  3e
Solution
e
2 x 1
e
e
4 x
2 x 1
e
 3e
 3e
am an  amn
2 x
3
am
Divide by e; n  a m n .
a
2 x
 In 3
Take natural logarithms
on both sides.
In e
2 x In e  In 3
Power property
SOLVING BASE e EXPONENTIAL
EQUATIONS
Solve the equation. Give solutions to the nearest
thousandth.
2 x 1
4 x
e

e
 3e
b.
Solution
2 x  In 3
1
x   In 3
2
In e = 1
Multiply by – ½
x  .549
The solution set is {–.549}.
SOLVING A LOGARITHMIC
EQUATION
Solve log(x + 6) – log(x + 2) = log x.
Solution
log( x  6)  log( x  2)  log x
x6
log
 log x
x2
x6
x
x2
x  6  x ( x  2)
Quotient property
Property of
logarithms
SOLVING A LOGARITHMIC
EQUATION
Solve log(x + 6) – log(x + 2) = log x.
Solution
x  6  x  2x
2
x  x 60
2
Standard form
( x  3)( x  2)  0
x  3
or
Distributive property
Factor.
x2
Zero-factor property
The proposed negative solution (x = – 3) is not in
the domain of the log x in the original equation, so
the only valid solution is the positive number 2,
giving the solution set {2}.
SOLVING A LOGARITHMIC
EQUATION
Solve log(3x + 2) + log(x – 1 ) = 1. Give the
exact value(s) of the solution(s).
Solution
log(3 x  2)  log( x  1)  1
log(3 x  2)  log( x  1)  log10
log[(3 x  2)( x  1)]  log10
(3 x  2)( x  1)  10
Substitute.
Product property
Property of
logarithms
SOLVING A LOGARITMIC
EQUATION
Solve log(3x + 2) + log(x – 1 ) = 1. Give the
exact value(s) of the solution(s).
Solution
3 x  x  2  10
2
3 x 2  x  12  0
1  1  144
x
6
Multiply.
Subtract 10.
Quadratic formula
SOLVING A LOGARITMIC
EQUATION
Solve log(3x + 2) + log(x – 1 ) = 1. Give the
exact value(s) of the solution(s).
Solution
1  145
6
The number
is negative, so x – 1 is
negative. Therefore, log(x – 1) is not defined
and this proposed solution must be discarded.
1  145
6
Since
> 1, both 3x + 2 and x – 1 are
positive and the solution set is 1  145 .

6

NEWTON’S LAW OF COOLING
Newton’s Law of Cooling states that
T  Ts  T0  Ts e ,
 kt
where T is the temperature of the object at
time t, Ts is the surrounding temperature,
and T0 is the value of T at t = 0.
EXAMPLE 11 McDonald’s Hot Coffee
The local McDonald’s franchise has discovered
that when coffee is poured from a
coffeemaker whose contents are 180ºF into a
noninsulated pot, after 1 minute, the coffee
cools to 165ºF if the room temperature is 72ºF.
How long should the employees wait before
pouring the coffee from this noninsulated pot
into cups to deliver it to customers at 125ºF?
EXAMPLE 11 McDonald’s Hot Coffee
Solution
Use Newton’s Law of Cooling with T0 = 180
and Ts = 72 to obtain
 kt
T  72  180  72  e
T  72  108e
 kt
We have T = 165 and t = 1.
165  72  108e
93
k
e
108
k
 93 
ln 
  k
 108 
k  0.1495317
EXAMPLE 11 McDonald’s Hot Coffee
Solution continued
Substitute this value for k.
0.1495317t
T  72  108e
Solve for t when T = 125.
0.1495317 t
125  72  108e
1
 53 
t
ln 

125  72
0.1495317 t
0.1495317  108 
e
108
t  4.76
 53 
ln 
  0.1495317t The employee should
 108 
wait about 5 minutes.
GROWTH AND DECAY MODEL
A  A0 e
rt
A is the quantity after time t.
A0 is the initial (original) quantity (when t = 0).
r is the growth or decay rate per period.
t is the time elapsed from t = 0.
EXAMPLE 12 Chemical Toxins in a Lake
A chemical spill deposits 60,000 cubic meters
of soluble toxic waste into a large lake. If 20%
of the waste is removed every year, how many
years will it take to reduce the toxin to 1000
cubic meters?
Solution
In the equation A = A0ert, we need to find A0, r,
and the time when A = 1000.
EXAMPLE 12 Chemical Toxins in a Lake
Solution continued
1. Find A0. Initially (t = 0), we are given
A0 = 60,000. So
A  60,000e
rt
2. Find r. When t = 1 year, the amount of toxin
will be 80% of its initial value, or
EXAMPLE 12 Chemical Toxins in a Lake
Solution continued
2. continued So
EXAMPLE 12 Chemical Toxins in a Lake
Solution continued
3. Find t when A = 1000.
It will take approximately 18 years to reduce
toxin to 1000 m3.