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5
Exponential and Logarithmic Functions
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
5.2
Applications of Exponential Functions
Objectives
• Solve compound interest problems.
• Solve exponential growth and decay problems.
Compound Interest
Compound interest provides another illustration of
exponential growth. Suppose that $500 (called the
principal) is invested at an interest rate of 8%
compounded annually. The interest earned the first year is
$500(0.08) = $40, and this amount is added to the original
$500 to form a new principal of $540 for the second year.
The interest earned during the second year is $540(0.08) =
$43.20, and this amount is added to $540 to form a new
principal of $583.20 for the third year. Each year a new
principal is formed by reinvesting the interest earned during
that year.
Compound Interest
In general, suppose that a sum of money P (called the
principal) is invested at an interest rate of r% compounded
annually. The interest earned the first year is Pr, and the
new principal for the second year is P + Pr or P(1 + r). Note
that the new principal for the second year can be found by
multiplying the original principal P by (1 + r). In like fashion,
we can find the new principal for the third year by multiplying
the previous principal, P(1 + r), by 1 + r, thus obtaining
P(1 + r)2. If this process is continued, then after t years the
total amount of money accumulated, A, is given by
A  P (1  r )
t
Compound Interest
Problem 1
What rate of interest is needed for an investment
of $1000 to yield $3000 in 10 years if the interest
is compounded annually?
Compound Interest
Problem 1
Solution:
Let’s substitute $1000 for P, $3000 for A, and 10 years for t in
the compound interest formula and solve for r:
A = P(1 + r)t
3000 = 1000(1 + r)10
3 = (1 + r)10
30.1 = [(1 + r)10]0.1
1+r
0.116123174  r
Raise both sides to the 0.1 power
1.116123174
r = 11.6% to the nearest tenth of a percent
Therefore a rate of interest of approximately 11.6% is needed.
Compound Interest
If money invested at a certain rate of interest is to be
compounded more than once a year, then the basic formula
A = P(1 + r)t can be adjusted according to the number of
compounding periods in a year. For example, for
compounding semiannually, the formula becomes
r

A  P 1 
 2
2t
and for compounding quarterly, the formula becomes
r

A  P 1 
4

4t
In general, if n represents the number of compounding
periods in a year, then the formula becomes
r

A  P 1 
 n
nt
Exponential Decay
An example of exponential decay involves radioactive
substances. The rate of decay can be described
exponentially and is based on the half-life of a substance.
The half-life of a radioactive substance is the amount of
time that it takes for one-half of an initial amount of the
substance to disappear as the result of decay.
Suppose there is an initial amount, Q0, of a radioactive
substance with a half-life of h. The amount of substance
remaining, Q, after a time period of t is given by the formula
 1
Q  Q0  
2
t /h
The units of measure for t and h must be the same.
Exponential Decay
Problem 2
Barium-140 has a half-life of 13 days. If there are
500 milligrams of barium initially, how many
milligrams remain after 26 days? After 100 days?
Exponential Decay
Problem 2
Solution:
With Q0 = 500 and h = 13, the half-life formula becomes
 1
Q  500  
2
t /13
If t = 26, then
 1
Q  500  
2
26/13
2
 1
 500  
2
 1
 500  
4
 125
Exponential Decay
Problem 2
Thus 125 milligrams remain after 26 days. If t = 100, then
100/13
 1
Q  500  
2
 500(0.5)
 2.4 to the nearest tenth of a milligram
100/13
Approximately 2.4 milligrams remain after 100 days.
Number e
The function defined by the equation f (x) = ex is the
natural exponential function.
Back to Compound Interest
Let’s return to the concept of compound interest. If the
number of compounding periods in a year is increased
indefinitely, we arrive at the concept of compounding
continuously. Mathematically, we can do this by applying
the limit concept to the expression
r

P 1 
n

nt
The formula
A  Pe rt
yields the accumulated value, A, of a sum of money, P, that
has been invested for t years at a rate of r% compounded
continuously.
Law of Exponential Growth
The law of exponential growth,
Q(t )  Q0e
kt
is used as a mathematical model for numerous growth-anddecay applications. In this equation, Q(t) represents the
quantity of a given substance at any time t; Q0 is the initial
amount of the substance (when t = 0), and k is a constant that
depends on the particular application. If k < 0, then Q(t)
decreases as t increases, and we refer to the model as the
law of decay.
Law of Exponential Growth
Problem 3
Suppose that in a certain culture, the equation
Q(t) = 15,000e0.3t expresses the number of bacteria
present as a function of the time t, where t is
expressed in hours. Find (a) the initial number of
bacteria, and (b) the number of bacteria after
3 hours.
Law of Exponential Growth
Problem 3
Solutions:
a. The initial number of bacteria is produced when t = 0.
Q(0) = 15,000e0.3(0)
= 15,000e0
= 15,000
b.
e0 = 1
Q(3) = 15,000e0.3(3)
= 15,000e0.9
= 36,894 to the nearest whole number
There should be approximately 36,894 bacteria present after
3 hours.