chapter3.ppt

3
 Exponential Functions
 Logarithmic Functions
 Exponential Functions as
Mathematical Models
Exponential and Logarithmic Functions

3.1
Exponential Functions
x
y
– 2 2
4
2
f(x) = (1/2)x
f(x) = 2x

Exponential Function
 The function defined by
is called an exponential function with base b
and exponent x.
 The domain of f is the set of all real numbers.
( ) ( 0, 1)
x
f x b b b
  

Example
 The exponential function with base 2 is the function
with domain (– , ).
 The values of f(x) for selected values of x follow:
( ) 2x
f x 
(3)
f 
3
2
f
 

 
 
(0)
f 
3
2 8

3/2 1/2
2 2 2 2 2
  
0
2 1


Example
 The exponential function with base 2 is the function
with domain (– , ).
 The values of f(x) for selected values of x follow:
( ) 2x
f x 
( 1)
f  
2
3
f
 
 
 
 
1 1
2
2


2/3
2/3 3
1 1
2
2 4

 

Laws of Exponents
 Let a and b be positive numbers and let x
and y be real numbers. Then,
1.
2.
3.
4.
5.
x y x y
b b b 
 
x
x y
y
b
b
b


 
y
x xy
b b

 
x x x
ab a b

x x
x
a a
b b
 

 
 

Examples
 Let f(x) = 2
2x – 1
. Find the value of x for which f(x) = 16.
Solution
 We want to solve the equation
2
2x – 1
= 16 = 2
4
 But this equation holds if and only if
2x – 1 = 4
giving x = .
5
2

Examples
 Sketch the graph of the exponential function f(x) = 2x.
Solution
 First, recall that the domain of this function is the set of
real numbers.
 Next, putting x = 0 gives y = 20 = 1, which is the y-intercept.
(There is no x-intercept, since there is no value of x for
which y = 0)

Examples
Solution
 Now, consider a few values for x:
 Note that 2x approaches zero as x decreases without bound:
✦ There is a horizontal asymptote at y = 0.
 Furthermore, 2x increases without bound when x increases
without bound.
 Thus, the range of f is the interval (0, ).
x – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5
y 1/32 1/16 1/8 1/4 1/2 1 2 4 8 16 32

Examples
Solution
 Finally, sketch the graph:
x
y
– 2 2
4
2
f(x) = 2x

Examples
 Sketch the graph of the exponential function f(x) = (1/2)x.
Solution
 First, recall again that the domain of this function is the
set of real numbers.
 Next, putting x = 0 gives y = (1/2)0 = 1, which is the
y-intercept.
(There is no x-intercept, since there is no value of x for
which y = 0)

Examples
Solution
 Now, consider a few values for x:
 Note that (1/2)x increases without bound when x decreases
without bound.
 Furthermore, (1/2)x approaches zero as x increases without
bound: there is a horizontal asymptote at y = 0.
 As before, the range of f is the interval (0, ).
x – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5
y 32 16 8 4 2 1 1/2 1/4 1/8 1/16 1/32

Examples
Solution
 Finally, sketch the graph:
x
y
– 2 2
4
2
f(x) = (1/2)x

Examples
Solution
 Note the symmetry between the two functions:
x
y
– 2 2
4
2
f(x) = (1/2)x
f(x) = 2x

Properties of Exponential Functions
 The exponential function y = bx (b > 0, b ≠ 1) has
the following properties:
1. Its domain is (– , ).
2. Its range is (0, ).
3. Its graph passes through the point (0, 1)
4. It is continuous on (– , ).
5. It is increasing on (– , ) if b > 1 and
decreasing on (– , ) if b < 1.

The Base e
 Exponential functions to the base e, where e is an
irrational number whose value is 2.7182818…, play an
important role in both theoretical and applied problems.
 It can be shown that
1
lim 1
m
m
e
m

 
 
 
 

Examples
 Sketch the graph of the exponential function f(x) = ex.
Solution
 Since ex > 0 it follows that the graph of y = ex is similar to the
graph of y = 2x.
 Consider a few values for x:
x – 3 – 2 – 1 0 1 2 3
y 0.05 0.14 0.37 1 2.72 7.39 20.09

5
3
1
Examples
 Sketch the graph of the exponential function f(x) = ex.
Solution
 Sketching the graph:
x
y
– 3 – 1 1 3
f(x) = ex

Examples
 Sketch the graph of the exponential function f(x) = e–x
.
Solution
 Since e–x
> 0 it follows that 0 < 1/e < 1 and so
f(x) = e–x
= 1/ex
= (1/e)x
is an exponential function with
base less than 1.
 Therefore, it has a graph similar to that of y = (1/2)x
.
 Consider a few values for x:
x – 3 – 2 – 1 0 1 2 3
y 20.09 7.39 2.72 1 0.37 0.14 0.05

5
3
1
Examples
 Sketch the graph of the exponential function f(x) = e–x
.
Solution
 Sketching the graph:
x
y
– 3 – 1 1 3
f(x) = e–x

3.2
Logarithmic Functions
1
x
y
1
y = ex
y = ln x
y = x

Logarithms
 We’ve discussed exponential equations of the form
y = bx (b > 0, b ≠ 1)
 But what about solving the same equation for y?
 You may recall that y is called the logarithm of x to the
base b, and is denoted logbx.
✦ Logarithm of x to the base b
y = logbx if and only if x = by (x > 0)

Examples
 Solve log3x = 4 for x:
Solution
 By definition, log3x = 4 implies x = 34 = 81.

Examples
 Solve log164 = x for x:
Solution
 log164 = x is equivalent to 4 = 16x = (42)x = 42x, or 41 = 42x,
from which we deduce that
2 1
1
2
x
x



Examples
 Solve logx8 = 3 for x:
Solution
 By definition, we see that logx8 = 3 is equivalent to
3 3
8 2
2
x
x
 


Logarithmic Notation
log x = log10 x Common logarithm
ln x = loge x Natural logarithm

Laws of Logarithms
 If m and n are positive numbers, then
1.
2.
3.
4.
5.
log log log
b b b
mn m n
 
log log log
b b b
m
m n
n
 
log log
n
b b
m n m

log 1 0
b 
log 1
b b 

Examples
 Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990,
use the laws of logarithms to find
log15 log3 5
log3 log5
0.4771 0.6990
1.1761
 
 
 


Examples
log7.5 log(15 / 2)
log(3 5 / 2)
log3 log5 log2
0.4771 0.6990 0.3010
0.8751

 
  
  


Examples
log81 4
log3
4log3
4(0.4771)
1.9084





Examples
log50 log5 10
log5 log10
0.6990 1
1.6990
 
 
 


Examples
 Expand and simplify the expression:
2 3
3
log x y 2 3
3 3
3 3
log log
2log 3log
x y
x y
 
 

Examples
2
2
1
log
2x
x 
 
 
 
2
2 2
2
2 2
2
2
log 1 log 2
log 1 log 2
log 1
x
x
x x
x x
  
  
  

Examples
2 2
1
ln x
x x
e
 2 2 1/2
2 2 1/2
2
2
( 1)
ln
ln ln( 1) ln
1
2ln ln( 1) ln
2
1
2ln ln( 1)
2
x
x
x x
e
x x e
x x x e
x x x


   
   
   

Examples
 Use the properties of logarithms to solve the equation for x:
3 3
log ( 1) log ( 1) 1
x x
   
3
1
log 1
1
x
x



1
1
3 3
1
x
x

 

1 3( 1)
x x
  
1 3 3
x x
  
4 2x

2
x 
Law 2
Definition of
logarithms

Examples
 Use the properties of logarithms to solve the equation for x:
log log(2 1) log6
x x
  
log log(2 1) log6 0
x x
   
(2 1)
log 0
6
x x 

0
(2 1)
10 1
6
x x 
 
(2 1) 6
x x  
2
2 6 0
x x
  
(2 3)( 2) 0
x x
  
2
x 
Laws 1 and 2
Definition of
logarithms
3
2
log
x
x
  is out of
the domain of ,
so it is discarded.

Logarithmic Function
 The function defined by
is called the logarithmic function with base b.
 The domain of f is the set of all positive numbers.
( ) log ( 0, 1)
b
f x x b b
  

Properties of Logarithmic Functions
 The logarithmic function
y = logbx (b > 0, b ≠ 1)
has the following properties:
1. Its domain is (0, ).
2. Its range is (– , ).
3. Its graph passes through the point (1, 0).
4. It is continuous on (0, ).
5. It is increasing on (0, ) if b > 1
and decreasing on (0, ) if b < 1.

Example
 Sketch the graph of the function y = ln x.
Solution
 We first sketch the graph of y = ex.
1
x
y
1
y = ex
y = ln x
y = x
 The required graph is
the mirror image of the
graph of y = ex with
respect to the line y = x:

Properties Relating
Exponential and Logarithmic Functions
 Properties relating ex and ln x:
eln x = x (x > 0)
ln ex = x (for any real number x)

Examples
 Solve the equation 2ex + 2 = 5.
Solution
 Divide both sides of the equation by 2 to obtain:
 Take the natural logarithm of each side of the equation
and solve:
2 5
2.5
2
x
e 
 
2
ln ln2.5
( 2)ln ln2.5
2 ln2.5
2 ln2.5
1.08
x
e
x e
x
x
x


 
 
  
 

Examples
 Solve the equation 5 ln x + 3 = 0.
Solution
 Add – 3 to both sides of the equation and then divide both
sides of the equation by 5 to obtain:
and so:
5ln 3
3
ln 0.6
5
x
x
 
   
ln 0.6
0.6
0.55
x
e e
x e
x






3.3
Exponential Functions as Mathematical Models
1. Growth of bacteria
2. Radioactive decay
3. Assembly time

Applied Example: Growth of Bacteria
 In a laboratory, the number of bacteria in a culture grows
according to
where Q0 denotes the number of bacteria initially present
in the culture, k is a constant determined by the strain of
bacteria under consideration, and t is the elapsed time
measured in hours.
 Suppose 10,000 bacteria are present initially in the culture
and 60,000 present two hours later.
 How many bacteria will there be in the culture at the end
of four hours?
0
( ) kt
Q t Q e


Solution
 We are given that Q(0) = Q0 = 10,000, so Q(t) = 10,000ekt.
 At t = 2 there are 60,000 bacteria, so Q(2) = 60,000, thus:
 Taking the natural logarithm on both sides we get:
 So, the number of bacteria present at any time t is given by:
0
2
2
( )
60,000 10,000
6
kt
k
k
Q t Q e
e
e



2
ln ln 6
2 ln 6
0.8959
k
e
k
k



0.8959
( ) 10,000 t
Q t e


Solution
 At the end of four hours (t = 4), there will be
or 360,029 bacteria.
0.8959(4)
(4) 10,000
360,029
Q e



Applied Example: Radioactive Decay
 Radioactive substances decay exponentially.
 For example, the amount of radium present at any time t
obeys the law
where Q0 is the initial amount present and k is a suitable
positive constant.
 The half-life of a radioactive substance is the time required
for a given amount to be reduced by one-half.
 The half-life of radium is approximately 1600 years.
 Suppose initially there are 200 milligrams of pure radium.
a. Find the amount left after t years.
b. What is the amount after 800 years?
0
( ) (0 )
kt
Q t Q e t

   

Solution
The initial amount is 200 milligrams, so Q(0) = Q0 = 200, so
Q(t) = 200e–kt
The half-life of radium is 1600 years, so Q(1600) = 100, thus
1600
1600
100 200
1
2
k
k
e
e





Solution
Taking the natural logarithm on both sides yields:
Therefore, the amount of radium left after t years is:
1600 1
ln ln
2
1
1600 ln ln
2
1
1600 ln
2
1 1
ln 0.0004332
1600 2
k
e
k e
k
k


 
 
  
0.0004332
( ) 200 t
Q t e


Solution
b. What is the amount after 800 years?
In particular, the amount of radium left after 800 years is:
or approximately 141 milligrams.
0.0004332(800)
(800) 200
141.42
Q e



Applied Example: Assembly Time
 The Camera Division of Eastman Optical produces a single
lens reflex camera.
 Eastman’s training department determines that after
completing the basic training program, a new, previously
inexperienced employee will be able to assemble
model F cameras per day, t months after the employee starts
work on the assembly line.
a. How many model F cameras can a new employee assemble
per day after basic training?
b. How many model F cameras can an employee with one
month of experience assemble per day?
c. How many model F cameras can the average experienced
employee assemble per day?
0.5
( ) 50 30 t
Q t e
 

Applied Example: Assembly Time
Solution
a. The number of model F cameras a new employee can
assemble is given by
b. The number of model F cameras that an employee with
1, 2, and 6 months of experience can assemble per day is
given by
or about 32 cameras per day.
c. As t increases without bound, Q(t) approaches 50.
Hence, the average experienced employee can be expected
to assemble 50 model F cameras per day.
(0) 50 30 20
Q   
0.5(1)
(1) 50 30 31.80
Q e
  

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