Chapter 3: Exponential and Logarithmic Functions
10/8/2017
Econ 506, by Dr. M. Zainal 1
Mathematics for Economists
Department of Economics
Dr. Mohammad ZainalEcon 506
Chapter 3Exponential and Logarithmic
Functions
3
•Exponential Functions
•Logarithmic Functions
•Exponential Functions as Mathematical Models
Econ 506, by Dr. M. Zainal Ch 3 - 2
Chapter Outline
Chapter 3: Exponential and Logarithmic Functions
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3.1Exponential Functions
Econ 506, by Dr. M. Zainal Ch 3 - 3
x
y
4
2
f(x) = (1/2)x
f(x) = 2x
– 2 2
Exponential Function• The function defined by
is called an exponential function with base b and exponentx.
• The domain of f is the set of all real numbers.
( ) ( 0, 1) xf x b b b ( ) ( 0, 1) xf x b b b
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Chapter 3: Exponential and Logarithmic Functions
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Example• The exponential function with base 2 is the function
with domain (–, ).
• The values of f(x) for selected values of x follow:
( ) 2xf x ( ) 2xf x
(3)f (3)f
3
2f
3
2f
(0)f (0)f
32 832 8
3/2 1/22 2 2 2 2 3/2 1/22 2 2 2 2
02 102 1
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Example• The exponential function with base 2 is the function
with domain (–, ).
• The values of f(x) for selected values of x follow:
( 1)f ( 1)f
2
3f
2
3f
1 12
2 1 1
22
2/32/3 3
1 12
2 4 2/3
2/3 3
1 12
2 4
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( ) 2xf x ( ) 2xf x
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• Let a and b be positive numbers and let x and y be real numbers. Then,
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Laws of Exponents
• Let f(x) = 22x – 1. Find the value of x for which f(x) = 16.
Solution
• We want to solve the equation
22x – 1 = 16 = 24
• But this equation holds if and only if
2x – 1 = 4
giving x = .5
2
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Examples
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• 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)
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Examples
• Sketch the graph of the exponential function f(x) = 2x.
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
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Examples
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• Sketch the graph of the exponential function f(x) = 2x.
Solution
• Finally, sketch the graph:
x
y
– 2 2
4
2
f(x) = 2x
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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 realnumbers.
• 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)
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Examples
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• Sketch the graph of the exponential function f(x) = (1/2)x.
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, ).
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Examples
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
• Sketch the graph of the exponential function f(x) = (1/2)x.
Solution
• Finally, sketch the graph:
x
y
– 2 2
4
2
f(x) = (1/2)x
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Examples
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Examples• Sketch the graph of the exponential function f(x) = (1/2)x.
Solution
• Note the symmetry between the two functions:
x
y
4
2
f(x) = (1/2)x
f(x) = 2x
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– 2 2
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.
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Chapter 3: Exponential and Logarithmic Functions
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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
1lim 1
m
me
m
1lim 1
m
me
m
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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
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Chapter 3: Exponential and Logarithmic Functions
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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
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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 than1.
• 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
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Chapter 3: Exponential and Logarithmic Functions
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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
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3.2Logarithmic Functions
1
x
y
1
y = ex
y = ln x
y = x
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Chapter 3: Exponential and Logarithmic Functions
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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)
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Examples• Solve log3x = 4 for x:
Solution
• By definition, log3x = 4 implies x = 34 = 81.
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• 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
2 1
1
2
x
x
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Examples• Solve logx8 = 3 for x:
Solution
• By definition, we see that logx8 = 3 is equivalent to
3 38 2
2
x
x
3 38 2
2
x
x
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Logarithmic Notation
log x = log10 x Common logarithm
ln x = loge x Natural logarithm
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Laws of Logarithms
• If m and n are positive numbers, then
1.
2.
3.
4.
5.
log log logb b bmn m n log log logb b bmn m n
log log logb b b
mm n
n log log logb b b
mm n
n
log lognb bm n mlog lognb bm n m
log 1 0b log 1 0b
log 1b b log 1b b
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Examples• Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990, use the laws
of logarithms to find
log15log15 log 3 5
log 3 log 5
0.4771 0.6990
1.1761
log 3 5
log 3 log 5
0.4771 0.6990
1.1761
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• Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990, use the laws of logarithms to find
log81log81 4log 3
4log 3
4(0.4771)
1.9084
4log 3
4log 3
4(0.4771)
1.9084
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Examples• Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990, use the laws
of logarithms to findlog 7.5log 7.5 log(15 / 2)
log(3 5 / 2)
log 3 log 5 log 2
0.4771 0.6990 0.3010
0.8751
log(15 / 2)
log(3 5 / 2)
log 3 log 5 log 2
0.4771 0.6990 0.3010
0.8751
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• Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990, use the laws of logarithms to find
log50log50 log 5 10
log 5 log10
0.6990 1
1.6990
log 5 10
log 5 log10
0.6990 1
1.6990
Examples• Expand and simplify the expression:
2 33log x y2 33log x y 2 3
3 3
3 3
log log
2log 3log
x y
x y
2 33 3
3 3
log log
2log 3log
x y
x y
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• Expand and simplify the expression:
2
2
1log
2x
x 2
2
1log
2x
x
22 2
22 2
22
log 1 log 2
log 1 log 2
log 1
xx
x x
x x
22 2
22 2
22
log 1 log 2
log 1 log 2
log 1
xx
x x
x x
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Examples• Expand and simplify the expression:
2 2 1ln
x
x x
e
2 2 1ln
x
x x
e
2 2 1/2
2 2 1/2
2
2
( 1)ln
ln ln( 1) ln
12ln ln( 1) ln
21
2ln ln( 1)2
x
x
x x
e
x x e
x x x e
x x x
2 2 1/2
2 2 1/2
2
2
( 1)ln
ln ln( 1) ln
12ln ln( 1) ln
21
2ln ln( 1)2
x
x
x x
e
x x e
x x x e
x x x
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Examples• Use the properties of logarithms to solve the equation for x:
3 3log ( 1) log ( 1) 1x x 3 3log ( 1) log ( 1) 1x x
3
1log 1
1
x
x
3
1log 1
1
x
x
113 3
1
x
x
11
3 31
x
x
1 3( 1)x x 1 3( 1)x x
1 3 3x x 1 3 3x x
4 2x4 2x
2x 2x
Law 2
Definition of logarithms
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Examples• Use the properties of logarithms to solve the equation for x:
log log(2 1) log 6x x log log(2 1) log 6x x
log log(2 1) log 6 0x x log log(2 1) log 6 0x x
(2 1)log 0
6
x x
(2 1)log 0
6
x x
0(2 1)10 1
6
x x 0(2 1)
10 16
x x
(2 1) 6x x (2 1) 6x x 22 6 0x x 22 6 0x x
(2 3)( 2) 0x x (2 3)( 2) 0x x
Laws 1 and 2
Definition of logarithms
3
2log
x
x
is out of
the domain of , so it is discarded.
3
2log
x
x
is out of
the domain of , so it is discarded.
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2x 2x
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) bf x x b b ( ) log ( 0, 1) bf x x b b
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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.
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Example• Sketch the graph of the function y = ln x.
Solution
• We first sketch the graph of y = ex.
1
x
y 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
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1
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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)
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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 52.5
2xe 2 5
2.52
xe
2ln ln 2.5
( 2) ln ln 2.5
2 ln 2.5
2 ln 2.5
1.08
xe
x e
x
x
x
2ln ln 2.5
( 2) ln ln 2.5
2 ln 2.5
2 ln 2.5
1.08
xe
x e
x
x
x
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Chapter 3: Exponential and Logarithmic Functions
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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
3ln 0.6
5
x
x
5ln 3
3ln 0.6
5
x
x
ln 0.6
0.6
0.55
xe e
x e
x
ln 0.6
0.6
0.55
xe e
x e
x
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3.3Exponential Functions as Mathematical Models
1. Growth of bacteria2. Radioactive decay3. Assembly time
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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( ) ktQ t Q e 0( ) ktQ t Q e
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Applied Example: Growth of BacteriaSolution
• 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:
02
2
( )60,000 10,000
6
kt
k
k
Q t Q ee
e
02
2
( )60,000 10,000
6
kt
k
k
Q t Q ee
e
2ln ln 6
2 ln 6
0.8959
ke
k
k
2ln ln 6
2 ln 6
0.8959
ke
k
k
0.8959( ) 10,000 tQ t e 0.8959( ) 10,000 tQ t e
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Applied Example: Growth of BacteriaSolution
• 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
0.8959(4)(4) 10,000
360,029
Q e
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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 )ktQ t Q e t 0( ) (0 )ktQ t Q e t
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Applied Example: Radioactive DecaySolution
a. Find the amount left after t years.
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
1600
1600
100 200
1
2
k
k
e
e
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Applied Example: Radioactive DecaySolution
a. Find the amount left after t years. Taking the natural logarithm on both sides yields:
Therefore, the amount of radium left after t years is:
1600 1ln ln
21
1600 ln ln21
1600 ln21 1
ln 0.00043321600 2
ke
k e
k
k
1600 1ln ln
21
1600 ln ln21
1600 ln21 1
ln 0.00043321600 2
ke
k e
k
k
0.0004332( ) 200 tQ t e 0.0004332( ) 200 tQ t e
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Chapter 3: Exponential and Logarithmic Functions
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Applied Example: Radioactive DecaySolution
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
0.0004332(800)(800) 200
141.42
Q e
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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 basictraining?
b. How many model F cameras can an employee with one month of experienceassemble per day?
c. How many model F cameras can the average experienced employee assemble per day?
0.5( ) 50 30 tQ t e 0.5( ) 50 30 tQ t e
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Applied Example: Assembly TimeSolution
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 6months 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 assemble50 model F cameras per day.
(0) 50 30 20Q (0) 50 30 20Q
0.5(1)(1) 50 30 31.80Q e 0.5(1)(1) 50 30 31.80Q e
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END OF CHAPTER
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