Quantitative Reasoning: Exponential Models
This material was prepared originally by Dr. Georgia Tolias for the ISP 120 course offered by the Quantitative
Reasoning Center and modified by Mr. Louis Aquila for the SNL online QR course.
There are numerous phenomena in nature, science, and finance that are modeled either exactly or approximately
by using exponential equations. In this course we will use exponential equations to model population change,
bank loans and investments and use these equations to make predictions.
Definition: A quantity increases exponentially when it increases by a constant percent for each fixed unit of
time. A quantity decreases exponentially when it decreases by a constant percent for each fixed unit of time.
Review of Percent Change: Recall that problems involving quantities that either increase or decrease by a
given percent can be dealt with in one of two equivalent ways. These are summarized in the formulas below:

To increase S by r %:
S + r·S = S(1 + r) where S is some initial quantity and r is a percent
increase expressed in decimal form

To decrease S by r %: S – r·S = S(1 – r) where S is some initial quantity and r is a percent
decrease expressed in decimal form
Example 1: Because of inflation, the value of the dollar decreased by 2% in each of the three successive years
from 1996 to 1999. This situation is illustrated in the table below. What was the percent decrease in the value
of the dollar over the entire 3-year period?
Year
1996
1997
1998
1999
Inflation
Rate
2%
2%
2%
Solution: From 1996 to 1997, the dollar lost 2% of its buying power. Thus, it retained 98% = 0.98 of its value
because 100% – 2% = 98%. So
The value of the 1997 dollar was 0.98 of its 1996 value.
From 1997 to 1998, the 1997 dollar lost 2% of its buying power. Thus, it retained 98% of its buying power. So,
the 1998 value of the dollar was 0.98 of its 1997 value. Hence we multiply the value of the 1997 dollar by 98%
= 0.98 to get that
the value of the 1998 dollar was (0.98)(0.98) = 0.982 of its 1996 value.
From 1998 to 1999, the 1998 dollar lost 2% of its buying power. Thus, it retained 98% of its buying power. So,
the value of the 1999 dollar was 0.98 of its 1998 value. Hence we multiply the value of the 1998 dollar by 98%
= 0.98 to get that
the value of the 1999 dollar was (0.98)(0.982) = 0.983 of its 1996 value.
Observe that 0.983 = 0.9412 = 94.12%. Thus the 1999 dollar retained 94.12% of its 1996 value.
We say that compared to its value in 1996, the dollar was worth only 94.12 cents in 1999.
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Note also that 100% - 94.12% = 5.88%. Hence, the dollar decreased in value by 5.88% from 1996 to 1999.
Important Note: To obtain this successive percent change we performed successive multiplication by the
constant factor 0.98.
Example 2a: A 100 gallon tank of water is drained at a rate of 5% per day.
 There are two variable: Water (W) in the tank and time elapsed (T)
 The amount of water in the tank is uniquely determined by the time elapsed.
 So W is the dependent variable and T is the independent variable.
Because the amount of water in the tank decreases by a fixed percentage each day, the amount of water in the
tank is decreasing exponentially. The following table illustrates the situation.
Time (days) T Water (gal) W
0
1
2
3
4
100
95.00
90.25
85.74
81.45
Computation
Percent Change in Water
for W
100 x .95
95 x .95
90.25 x .95
85.74 x .95
-5% =(95 -100)/100
-5% =(90.25 -95)/95
-5% =(85.74 -90.25)/90.25
-5% =(81.45 -85.74)/85.74
Note that since the values for W were rounded to two decimal places, the values for % change will not all be
exactly 5%.
Definition: An exponential equation involving two variables, say Y and T, has the form Y = A·BT, where B is
the constant growth or decay factor, A is the initial value (the value of Y when T=0) and T is the time.
 If there is a constant percent increase, then the growth factor is B = 1 + r, where r is the percent
expressed in decimal form.
 If there is a constant percent decrease, then the decay factor is B = 1 – r, where r is the percent
expressed in decimal form.
Example 2b: Find the exponential equation that describes the water draining from the tank and use the
equation to predict how much water is in the tank after 10 days.
Solution: Since 5% of the water is drained each day, this implies that 95% of the water remains in the tank.
The constant decay factor is
100% – 5% = 95% = 0.95.
Thus, the exponential equation that models the water draining from the tank is
W = 100·(0.95)T
To predict the amount of water after 10 days, we substitute 10 in for the T variable and find W.
W = 100·(0.95)10 = 59.974 gallons.
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The Graph of an Exponential Equation
If a problem describes exponential growth (a constant percent increase) of some initial quantity, then the graph
of the exponential equation used to model the problem is an increasing curve (As the graph is observed from
left-to-right). Similarly, the graph of an exponential equation which models something decreasing by a constant
percent is a decreasing curve (As the graph is observed from left-to-right). Below is the Excel graph of the
exponential equation that models the water draining from the tank during the first 10 days.
Water Drained at 5% per day
y=100(.95)^x
120
100
gallons
80
60
40
20
0
0
1
2
3
4
5
6
7
8
9
10
11
12
days
NOTE: If we look at only the first ten days, then the graph appears to be associated with a linear equation.
However, if we continue the graph over a greater number of days, the graph’s takes on a different look.
Water drained at 5% per day
y = 100*.95^x
120
gallons
100
80
60
40
20
0
1
10 19 28 37 46 55 64 73 82 91 100 109 118 127 136 145
days
These two graphs illustrate that we can be deceived by looking at a graph associated with a segment of the data.
The first graph gives the appearance that the data are following a decreasing line. Whereas, if we restricted
ourselves, in the second graph, to the data starting with day 82, it appears that the data are essentially located on
the horizontal axis. We need to be very cautious when analyzing data, or looking at the conclusions based on
someone else’s analysis.
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