Exponential Functions: Changing Bases

Objectives

At the end of this lesson, you should be able to:

Change the base of an exponential function to a power or root of the same base.
Use negative exponents to invert the base of decreasing exponential functions.

In the lesson Introduction to Exponential Functions, we learned that if periodic function outputs can be produced by multiplying by the same constant the previous result by the same factor, the function can be expressed in the exponential form, a(b)^x where a is the initial value (the value at x = 0) and b is a power (or root) of the constant multiplier.

In what follows, we will analyze various cases illustrating how to find an exponential function to model certain situations. We will produce tables of values based on the situation described and use the tables to derive the formulas for the corresponding functions. We will also examine the graphs of the functions derived in each case.

An investment begins with a single dollar and is doubled every year. Let the function f (x) represent the investment's value after x years. The following table shows the initial investment and its value at the end of each of the first four years.

x	0	1	2	3	4
f (x)	1	1×2=2	2×2=4	4×2=8	8×2=16

A formula for f (x) can be deduced from the previous table. The initial value is 1 (f (0) = 1) and every time the input increases by 1 the result is two times the prior result.

It follows that $f (x) = 1 \times 2^{x}$

x	0	1	2	3	4
f (x)	1=1×2⁰	2=1×2¹	4=1×2²	8=1×2³	16=1×2⁴

An investment begins with a single dollar and is tripled every 1.6 years. Let the function f (x) represent the investment's value after x years. The following table shows the initial investment and its value at the end of each of the first four 1.6 year periods.

x	0	1.6	3.2	4.8	6.4
f (x)	1	1×3=3	3×3=9	9×3=27	27×3=81

A formula for f (x) can be deduced from the previous table. The initial value is 1 (f (0) = 1) and every time the input increases by 1.6 the result is three times the prior result.

It follows that $f (x) = 1 \times 3^{\frac{x}{1.6}}$

x	0	1.6	3.2	4.8	6.4
f (x)	$1 = 1 \times 3^{\frac{0}{1.6}}$	$3 = 1 \times 3^{\frac{1.6}{1.6}}$	$9 = 1 \times 3^{\frac{3.2}{1.6}}$	$27 = 1 \times 3^{\frac{4.8}{1.6}}$	$81 = 1 \times 3^{\frac{6.4}{1.6}}$

An investment begins with a single dollar and is quadrupled every 2 years. Let the function f (x) represent the investment's value after x years. The following table shows the initial investment and its value at the end of each of the first four 2 year periods.

x	0	2	4	6	8
f (x)	1	1×4=4	4×4=16	16×4=64	64×4=256

A formula for f (x) can be deduced from the previous table. The initial value is 1 (f (0) = 1) and every time the input increases by 2 the result is four times the prior result.

It follows that $f (x) = 1 \times 4^{\frac{x}{2}}$

x	0	2	4	6	8
f (x)	$1 = 1 \times 4^{\frac{0}{2}}$	$4 = 1 \times 4^{\frac{2}{2}}$	$16 = 1 \times 4^{\frac{4}{2}}$	$64 = 1 \times 4^{\frac{6}{2}}$	$256 = 1 \times 4^{\frac{8}{2}}$

Note that the graphs for the first and third situations are the same.

In fact, the graph for the second situation is very similar to those for the first and third situations. We can confirm this similarity graphically, superimposing all three graphs as shown in the following figure:

2exp

We can also verify that the first and third functions are the same using their formulas.

The formula for the third situation: $4^{\frac{x}{2}} = 4^{(\frac{1}{2}) x} = 2^{x}$ is equivalent to the formula for the first situation.

Beginning with the formula for the first situation: $2^{x} = 2^{\frac{2 x}{2}} = {(2^{2})}^{\frac{x}{2}} = 4^{\frac{x}{2}}$ we can convert it to the formula for the third situation.

The previous example clearly shows that the exponential function $2^{x}$ can be expressed with base 4 rather than base 2 (technically the base is still 2, it is just expressed as a power of 4, i.e. $4^{\frac{1}{2}}$ .) Given the close similarity between the graphs for the first and second situations, it looks like the exponential function $2^{x}$ can be expressed as a function with base 3. Actually $3^{\frac{x}{1.6}}$ is quite close. With a small numerical adjustment in the divisor 1.6, it can be expressed as an exponential function with base 3 or any positive number other than 1 as base. In order to make these changes, we will need logarithms. For this lesson, we will learn how to express the base of an exponential function with different powers or roots of that base.

Graphs of Exponential Functions with different Bases

Observe and Learn:

Did you know that:

Given an exponential function

f (x) = a^{x}

and a real positive number b other than 1, there is a real number r such that

a^{x} = b^{rx}

for all real numbers x.

The following applet will let you compare the behavior of various exponential functions

Move the slider to set the value k and select one of four exponential functions offered to view its graph.
Find the input (the x-value) for the selected exponential function that produces the output k (the y-value).
Click on the button and watch how the output (the y-value) of each exponential function changes when the input (the x-value) increases by one. Use the zoom slider to get a better view of the graph.
Discover the common ratio between outputs for consecutive inputs $(\frac{f (x + 1)}{f (x)})$ .

The k in the exponential function $f (x) = k \times b^{x}$ is called the initial value (k = f (0)) and b is called the base or the common ratio.

The period is 1 in this case, i.e. the value of f (x+1) = b×f (x).

To change the period of an exponential function from 1 to p, we have to divide the exponent x by p.

The initial value of the exponential function $f (x) = k \times b^{\frac{x}{p}}$ is still k and we call b the base, but now we multiply by b every period p, i.e. f (x+p) = b×f (x).

Decreasing Exponential Functions

The decreasing exponential function examples in the Introduction to Exponential Functions lesson were expressed as:

$f (x) = a \times b^{x}$

where 0 < b < 1.

However,: ${\frac{1}{2}}^{x} = {(2^{−1})}^{x} = 2^{−x}$

So, a decreasing exponential function can be expressed with a base larger than 1 using a negative exponent. Let's consider some depreciation functions:

Example:

An industrial machine that initially cost 3 million dollars depreciates at a rate that loses half its value every three years. Let t represent the years since the machine's initial purchase and f (t) the machine's value in millions of dollars.

What is the formula for the function f (t)?
What would be the formula for the machine's value be if it lost a third of its value every 5 years?
What would the formula be if the machine lost half its value every 6 years?
How would we express the depreciation rate in words if the formula for its value after t years were $f (t) = 3 {(\frac{10}{9})}^{- \frac{t}{5}}$ ?
How would we express the depreciation rate in words if the formula for its value after t years were $f (t) = 3 {(\frac{5}{4})}^{- \frac{t}{10}}$ ?
How would we express the depreciation rate in words if the formula for its value after t years were $f (t) = 3 {(\frac{7}{3})}^{- \frac{t}{2}}$ ?

Solution:

What is the formula for the function f (t)?
Analyzing the situation the way we learned in the Introduction to Exponential Functions lesson: we have the initial value is 3, the base (depreciation factor) is $1 - \frac{1}{2} = \frac{1}{2}$ , and the time per depreciation period is 3 years. So the decreasing exponential function that represents the value of the machine after t years is:

$f (t) = 3 {(\frac{1}{2})}^{\frac{t}{3}} = 3 {(2^{-1})}^{\frac{t}{3}} = 3 {(2)}^{- \frac{t}{3}}$
What would be the formula for the machine's value be if it lost a third of its value every 5 years?
Analyzing the situation as before, we have the initial value is 3, the base (depreciation factor) is $1 - \frac{1}{3} = \frac{2}{3}$ , and the time per depreciation period is 5 years. So the decreasing exponential function that represents the value of the machine after t years is:

$f (t) = 3 {(\frac{2}{3})}^{\frac{t}{5}} = 3 {({(\frac{3}{2})}^{-1})}^{\frac{t}{5}} = 3 {(\frac{3}{2})}^{- \frac{t}{5}}$
What would the formula be if the machine lost half its value every 6 years?
In this case, the initial value is 3, the base (depreciation factor) is $1 - \frac{1}{2} = \frac{1}{2}$ , and the time per depreciation period is 6 years. So the value of the machine after t years is:

$f (t) = 3 {(\frac{1}{2})}^{\frac{t}{6}} = 3 {(2^{-1})}^{\frac{t}{6}} = 3 {(2)}^{- \frac{t}{6}}$
How would we express the depreciation rate in words if the formula for its value after t years were $f (t) = 3 {(\frac{10}{9})}^{- \frac{t}{5}}$ ?
$f (t) = 3 {(\frac{10}{9})}^{- \frac{t}{5}} = 3 {({(\frac{10}{9})}^{-1})}^{\frac{t}{5}} = = 3 {(\frac{9}{10})}^{\frac{t}{5}}$

Analyzing from the formula: the initial value is 3, the base (depreciation factor) is $\frac{9}{10} = 1 - \frac{1}{10}$ , and the time per depreciation period is 5 years.

So the statement corresponding to this depreciation formula would be: An industrial machine whose initial value is 3 million dollars loses a tenth of its value every five years.
How would we express the depreciation rate in words if the formula for its value after t years were $f (t) = 3 {(\frac{5}{4})}^{- \frac{t}{10}}$ ?
$f (t) = 3 {(\frac{5}{4})}^{- \frac{t}{10}} = 3 {({(\frac{5}{4})}^{-1})}^{\frac{t}{10}} = = 3 {(\frac{4}{5})}^{\frac{t}{10}}$

Analyzing from the formula: the initial value is 3, the base (depreciation factor) is $\frac{4}{5} = 1 - \frac{1}{5}$ , and the time per depreciation period is 10 years.

So the statement corresponding to this formula would be: An industrial machine whose initial value is 3 million dollars loses a fifth of its value every ten years.
How would we express the depreciation rate in words if the formula for its value after t years were $f (t) = 3 {(\frac{7}{3})}^{- \frac{t}{2}}$ ?
$f (t) = 3 {(\frac{7}{3})}^{- \frac{t}{2}} = 3 {({(\frac{7}{3})}^{-1})}^{\frac{t}{2}} = = 3 {(\frac{3}{7})}^{\frac{t}{2}}$

Analyzing from the formula: the initial value is 3, the base (depreciation factor) is $\frac{3}{7} = 1 - \frac{4}{7}$ , and the time per depreciation period is 2 years.

So the statement corresponding to this formula would be: An industrial machine whose initial value is 3 million dollars loses four sevenths of its value every two years.

Summary

We have seen that we can express any exponential function with a base larger than 1. We have seen that we can change the base of an exponential function such as 4 to a power or root of that base. We have also seen that we can change base smaller than 1 to a base larger than 1 using reciprocals and the opposites of the exponents. For example,

${(\frac{1}{4})}^{x} = 4^{- x} = 2^{- 2 x} = 16^{- \frac{x}{2}}$

It is actually possible to change from any given positive real number base other than 1 to any positive real number base that we want other than 1. However, to do this we will need logarithms. The base e will also be helpful for changing bases and it will give us still another way to express and to think about exponential functions. The number e ≈ 2.71828182846 is a fascinating number that is best explained as a limit in calculus. We will use it as a base because it turns out to be the most efficient exponential function base to calculate values for using technology as well as the most commonly used base.

In a future lesson we will learn how to express $2^{x}$ , $3^{x}$ and $10^{x}$ using the base e.

We will also learn how to express ${(\frac{1}{2})}^{x}$ , ${(\frac{1}{3})}^{x}$ and ${(\frac{1}{10})}^{x}$ with the base e.

Now that you have completed this lesson, you should be able to:

Change the base of an exponential function to a power or root of the same base.
Use negative exponents to invert the base of decreasing exponential functions.