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Exponential Functions with Base e
Any positive number can be used as the base for an exponential function, but some bases are
more useful than others. For instance, in computer science applications, the base 2 is convenient.
The most important base though is the number denoted by the letter e.
The number e is irrational, so we cannot write its exact value; the approximate value to 20
decimal places is
e โ 2.
It may seem at first that a base such as 10 is easier to work with, but in certain applications, such
as compound interest or population growth, the number e is the best possible base.
The Natural Exponential Function:
The natural exponential function is the exponential function
x f x = e
with base e. It is often referred to as the exponential function.
Since 2 < e < 3, the graph of the natural exponential function lies between the graphs of y = 2
x
and y = 3
x , as shown below.
Example 1: Graph the function y = โ e
x โ , not by plotting points, but by starting from the
graph of y = e
x in the above figure. State the domain, range, and asymptote.
Solution:
Step 1: We will use transformation techniques to obtain the graph of
y = โ e
x โ
. Start with the graph of y = e
x , reflect it in the x -axis and
shift it rightward 1 unit.
Step 2: Since our transformation does not involve a vertical shift of the graph,
the horizontal asymptote of y = โ e
x โ is the same as that of y = e
x ;
that is, the horizontal asymptote is the x -axis, y = 0.
Looking at the graph, we see that the domain of y = โ e
x โ is all real
numbers (-โ,โ), and the range is (-โ, 0).
Continuously compounded interest is calculated by the formula
rt A t = Pe
where A ( t ) = amount after t years
P = principal
r = interest rate
t = number of years
Example 3: The number of bacteria in a culture is given by the function
n(t) = 10 e
0.22t
(a) What is the relative rate of growth of this bacterium population? Express your
answer as a percentage.
(b) What is the initial population of the culture (at t = 0)?
(c) How many bacteria will be in the culture at time t = 15?
Solution (a):
Population is modeled using the exponential growth model:
n(t) = n 0 e
rt
where r is the relative rate of growth. By inspecting the equation we are given,
we see that r = 0.22, or 22%.
Solution (b):
To find the initial population, we find the population at time t = 0. We do this by
substituting t = 0 into the equation for n ( t ).
n(t) = 10 e
0.22t
n(0) = 10 e
0.22(0)
n(0) = 10(1)
n(0) = 10
Thus, the initial population is 10 bacteria.
Note: We could have solved this problem by inspecting the given equation and
noticing that n o
, the initial population size, is 10.
Solution (c):
The number of bacteria at time 15 can be found by substituting t = 15 into the
equation for n ( t ).
n(t) = 10 e
0.22t
n(15) = 10 e
0.22(15)
n(15) = 10 e
33
n(15) โ 10(27.1126)
n(15) โ 271.
Thus, the population at time 15 is 271 bacteria.
Example 4: The Ewok population on the planet Endor has a relative growth rate of 3% per
year, and it is estimated that the population is 6,500.
(a) Find a function that models the population t years from now.
(b) Use the function from part (a) to estimate the Ewok population in 8 years.
(c) Sketch the graph of the population function.
Solution (a):
Step 1: To find a function that models the Ewok population, we will use the
exponential growth model
n(t) = n 0 e
rt
To use the model, we will need to determine what the values n o
and r.
Step 2: Since we are not explicitly told when our time starts, we can assume the
population was estimated to be 6,500 at time t = 0. Thus our initial
population is n o
Step 3: We are told in the problem that the relative growth rate is 3% per year, so
r = 0.03.
Step 4: Now we will substitute the values n o
= 6500 and r = 0.03 into the
formula for the exponential growth model to find the function that
models the population t years from now.
n(t) = n 0 e
rt
n(t) = 6500 e
0.03t
Solution (b):
An estimate of the Ewok population in 8 years can be found by substituting t = 8
into the equation for n ( t ).
n(t) = 6500 e
0.03t
n(8) = 6500 e
0.03(8)
n(8) = 6500 e
n(8) โ 6500(1.2710)
n(8) โ 8263.
Thus, the population in 8 years will be 8263.
