Chabot Mathematics
§9.5a
Exponential Eqns
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Review § 9.4
MTH 55
 Any QUESTIONS About
• §9.4 → Logarithm Change-of-Base
 Any QUESTIONS About HomeWork
• §9.4 → HW-47
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BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Summary of Log Rules
 For any positive numbers M, N,
and a with a ≠ 1
log a ( MN )  log a M  log a N ;
log a M
p
 p log a M ;
M
log a
 log a M  log a N ;
N
k
log a a  k .
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BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Typical Log-Confusion
 Beware that Logs do NOT behave
Algebraically. In General:
log a ( MN )  (log a M )(log a N ),
M log a M
log a

,
N log a N
log a ( M  N )  log a M  log a N ,
log a ( M  N )  log a M  log a N .
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Solving Exponential Equations
 Equations with variables in
exponents, such as 3x = 5 and
73x = 90 are called
EXPONENTIAL EQUATIONS
 Certain exponential equations can
be solved by using the principle of
exponential equality
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BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Principle of Exponential Equality
 For any real number b, with
b ≠ −1, 0, or 1, then
bx = by is equivalent to x = y
 That is, Powers of the same base
are equal if and only if the
exponents are equal
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Exponential Equality
 Solve for x: 5x = 125
 SOLUTION
 Note that 125 = 53. Thus we can write
each side as a power of the same base:
5x = 53
 Since the base is the same, 5, the
exponents must be equal.
Thus, x must be 3. The solution is 3.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Exponential Equality
 Solve each Exponential Equation
x
x
x1
a. 25  125
b. 9  3
 SOLUTION
  5
a. 5
2 x
3
52 x  53
2x  3
3
x
2
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b.
3   3
2 x
x1
32 x  3x1
2x  x  1
2x  x  1
x 1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Principle of Logarithmic Equality
 For any logarithmic base a, and
for x, y > 0,
x = y is equivalent to logax = logay
 That is, two expressions are equal
if and only if the logarithms of those
expressions are equal
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Logarithmic Equality
 Solve for x: 3x+1 = 43
 SOLUTION
3 x +1 = 43
log 3 x +1 = log 43
Principle of
logarithmic equality
Power rule for logs
(x +1)log 3 = log 43
x +1 = log 43/log 3
x = (log 43/log 3) – 1
x  2.4236.
 The solution is (log 43/log 3) − 1, or
approximately 2.4236.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Logarithmic Equality
 Solve for t: e1.32t = 2000
 SOLUTION
e1.32t = 2000
ln
e1.32t
= ln 2000
1.32t = ln 2000
Note that we use the
natural logarithm
Logarithmic and exponential
functions are inverses of each other
t = (ln 2000)/1.32
t  5.7583.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
To Solve an Equation of the
Form at = b for t
1. Take the logarithm (either natural or
common) of both sides.
2. Use the power rule for exponents so that
the variable is no longer written as an
exponent.
3. Divide both sides by the coefficient of the
variable to isolate the variable.
4. If appropriate, use a calculator to find an
approximate solution in decimal form.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Solve by Taking Logs
 Solve each equation and approximate
the results to three decimal places.
a. 2 x  15
b. 5  2 x2  17
 SOLUTION a. 2  15
x
ln 2  ln15
x ln 2  ln15
ln15
x
 3.907
ln 2
x
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Solve by Taking Logs
 SOLUTION b. 5  2 x2  17
b. 5  2 x3  17
17
x3
2 
5
 17 
x3
ln 2  ln  
 5
 17 
x  3ln 2  ln  
5
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 17 
ln  
 5
x3
ln 2
 17 
ln  
 5
x
3
ln 2
x  4.766
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Different Bases
 Solve the equation 52x−3 = 3x+1 and
approximate the answer to 3 decimals
x1
2 x3
Take ln of both sides
 SOLUTION ln 5
 ln 3
2x  3ln 5  x  1ln 3
2x ln 5  3ln 5  x ln 3  ln 3
2x ln 5  x ln 3  ln 3  3ln 5
x 2 ln 5  ln 3  ln 3  3ln 5
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ln 3  3ln 5
 2.795
x
2 ln 5  ln 3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Eqn Quadratic in Form
 Solve for x: 3x − 8∙3−x = 2.
x
x
x
x
 SOLUTION 3 3  8  3  2 3

  
32 x  8  30  2  3x
32 x  8  2  3x
3  23  8  0
2x
x
 This equation is quadratic in form.
 Let y = 3x then y2 = (3x)2 = 32x. Then,
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Eqn Quadratic in Form
 Soln
cont.
32 x  2  3x  8  0
y 2  2y  8  0
y  2 y  4   0
y  2  0 or y  4  0
y  2
or y  4
3x  2 or 3x  4
 But 3x = −2 is not possible because
3x > 0 for all numbers x.
So, solve 3x = 4 to find the solution
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Eqn Quadratic in Form
 Soln
cont.
3 4
x
ln 3  ln 4
x ln 3  ln 4
ln 4
x
ln 3
x  1.262
x
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Population Growth
 The following table shows the
approximate population and annual
growth rate of the United States of
America and Pakistan in 2005
Country Population
USA
295 million
1.0%
Pakistan 162 million
3.1%
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Annual
Population
Growth Rate
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Population Growth
 Use the population model P = P0(1 + r)t
and the information in the table, and
assume that the growth rate for each
country stays the same.
 In this model,
• P0 is the initial population,
• r is the annual growth rate as a decimal
• t is the time in years since 2005
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Population Growth

Use P = P0(1 + r)t and the data table:
a. to estimate the population of each
country in 2015.
b. If the current growth rate continues, in
what year will the population of the United
States be 350 million?
c. If the current growth rate continues, in
what year will the population of Pakistan
be the same as the population of the
United States?
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Population Growth
 SOLUTION: Use model P = P0(1 + r)t
a. US population in 2005 is P0 = 295.
The year 2015 is 10 years from 2005.
P  295 1  0.01  325.86 million
10
Pakistan in 2005 is P0 = 162
P  162 1  0.31  219.84 million
10
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Population Growth
 SOLUTION b.: Solve for t to find when
the United States population will be 350.
t
 350 
350  295 1  0.01
ln 
 t ln 1.01

 295 
350
t
 1.01
 350 
295
ln 


295
t
 350 
t
 17.18
ln 

ln
1.01


ln 1.01
 295 
 Some time in yr 2022 (2005 + 17.18)
the USA population will be 350 Million
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Population Growth
 SOLUTION c.: Solve for t to find when
the population will be the same in both
countries. 295 1  0.01t  162 1  0.031t
295 1.01  162 1.031
t
t
295  1.031 


162  1.01 
t
 1.031 
 295 
 ln 
ln 


 1.01 
 162 
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t
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Example  Population Growth
 Soln c.
 295 
 1.031 
ln 
 t ln 
cont.

 162 
 1.01 
 295 
ln 
 162 
t
 29.13
 1.031 
ln 
 1.01 
 Some time year 2034 (2005 + 29.13)
the two populations will be the same.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
WhiteBoard Work
 Problems From §9.5 Exercise Set
• 16, 20, 32, 34, 36, 40
 logistic difference
equation by
Belgian Scientist
Pierre Francois
Verhulst
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
All Done for Today
EMP Widmark
BAC Eqn
Calculator
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BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-64_sec_9-5a_Exponential_Eqns.ppt