Haberman / Kling
MTH 111c
Section II: Exponential and Logarithmic Functions
Module 2: Exponential Models
m(t )  50  0.945 models the population of a
certain species of monkeys in South America t years after 1998. Describe
EXAMPLE: Suppose that the function
t
this monkey population.
SOLUTION:
Based on our study in Module 1, we know that an exponential function of the form
f ( x)  a  (1  r ) x has initial value a and growth rate r per unit of x. Thus, we can
determine immediately that the monkey population modeled by m(t )  50  0.945 has
t
initial value 50 and, since 0.945  1  ( 0.055) , the growth rate is 5.5% per year (since
the growth rate is negative, the population is decreasing).
50 monkeys in South America in 1998, and the population is
decreasing at the rate of 5.5% per year.
CONCLUSION: There were
EXAMPLE: Suppose that the population of a certain species of snakes in South America
was 136 in 1998. If this snake population is increasing at the rate of 2.7% per
year, find a function that models this snake population.
SOLUTION:
Based on our study in Module 1, we know that an exponential function of the form
f ( x)  a  (1  r ) x has initial value a and growth rate r per unit of x. Thus, we can
determine immediately that this snake population is modeled by the exponential function
s(t )  136 1.027  where t is years after 1998.
t
2
EXAMPLE: Suppose that the population of a certain species of rodent in South America
was 6996 in 1998. Suppose further that in 2002 the population of this same
species of rodent in South America was 8077. If this rodent population is
increasing exponentially, find a function that models this rodent population t
years after 1998.
SOLUTION:
Since we are told that the rodent population is increasing exponentially, we know that an
exponential function will model the population. So we know that the function we are
looking for has form r (t )  a  bt . Since t  0 represents 1998, we know that the “initial
value” is the rodent population in 1998. So we can immediately see that a  6996 , and
that r (t )  6996  bt . To find b we can use the fact that in 2002 the population of this
same species of rodent in South America was 8077. Since 2002 corresponds with t  4
we see that:
r (4)  8077
 6996  b 4  8077



b 4  8077
6996
b 
4
 6996 
b   8077 
6996
1 4
1 4
 8077
1 4
Thus,


r (t )  6996   8077
 6996
 6996 

t
 6996  8077
1 4



t
4
(Note that we cannot cancel the two 6996’s in the algebraic rule for r since one of them is
under the exponent and the other one isn’t!)
3
EXAMPLE: Find the rule for an exponential function passing through the two points (1, 6)
and (3, 24) .
SOLUTION:
Let’s name the desired function f, and use function notation to translate the given ordered
pairs into equations involving f:
(1, 6) 
(3, 24) 
f (1)  6
f (3)  24
Since we know that the desired function is exponential, we know that it has form
f ( x)  a  b x .
We can now use this definition of f to build on the equations we found above:
f (1)  a  b 1  6
f (3)  a  b3  24
Let’s use the first equation and solve for a:
a b  6  a 
6
b
Now we’ll substitute this expression for a into the second equation and solve for b.
a  b3  24
6 3
 b  24
b
 6b 2  24


b2  4

b 2
Note that although both 2 and  2 solve b 2  4,
only 2 can represent b since the base of an
exponential function is always positive.
Since b  2 , we know that the desired function is f ( x)  a  b x . We can use either one of
the given ordered pairs to find a.
(1, 6) 
f (1)  a  21  6

a3
Thus, f ( x)  3  2 x . [Check this yourself.]
4
Try this one yourself and check your answer.
is an exponential function and you know that f (3)  5
8
f (2)  20 , find a rule for f.
If you know that
f
and
SOLUTION:
Since we know that the desired function is exponential, we know that it has form
f ( x)  a  b x .
We can now use this definition of f to create equations base on the given information:
f (3)  5  a  b3
8
f (2)  20  a  b2
We can use the first equation to solve for a:
a  b 3  5  a  b 3  b3  5  b3
8
8
a  5 b3
8

Now we can substitute this expression for a into the second equation and solve for b.
ab 2  20  5 b3  b 2  20
8
5
8
 8 5
5 8

b5  20
b5  8  20
5

b5  8  4


b5  32
b2
Since b  2 , we know that the desired function is f ( x)  a  2x . We now use the fact that
f (2)  20 to find a.
f (2)  20  a  2 2

20  4a

a5
5
So it looks like f ( x)  5  2 .
x
Check:
f (3)  5  23
1
 5
2

and
3
f (2)  5  22
 5 4
 20
5
8
Both function values check. Therefore, f ( x)  5  2 x .
Try this one yourself and check your answer.
Find the equation of the exponential function, g (x ) that passes through (2, 5) and
(5, 63) .
SOLUTION:
Since both of these points should satisfy the equation g ( x)  ab x , we can use
them to obtain a system of equations:
(2, 5)  g (2)  5
 ab 2  5
(5, 63)  g (5)  63  ab5  63
Solving the second equation
2 for a, we obtain a 
equation to substitute for a in the first equation 1 :
ab5  63

5 5
b  63
b2
We can now solve this equation for b:
5 5
b  63
b2
5b3  63
63
b3 
5
1/ 3
 63 
b 
 5 
b  2.326967
1
2
5
and we can use this
b2
6
So 2.326967 can be substituted for the constant b in the equation g ( x)  ab x :
g ( x)  a(2.326967) x .
We can now find the value of the constant b by using the fact that the ordered pair
(2, 5) satisfies the function g ( x)  a(2.326967) x .
5  a(2.326967) 2
5  5.414775a
5
a
5.414775
a  0.923399.
So an approximation for the exponential function that passes through (2, 5) and
(5, 63) is g ( x)  0.923399(2.326867) x .
Click here some practice problems for this module