
## SectionExponential Functions

###### Example166

Bill's parents started a college fund for Bill. When he was born they put $1,000 in a jar for Bill. Each year they add 50% of the jar's current value into the jar. Write a formula for the amount of money, A(t), in the jar when Bill is $t$ years old. Solution After 1 year, Bill's parents will add 50% of the original$1,000 into the jar. Thus, we have \begin{align*} A(1)\amp =1500\\ \amp=1000+500\\ \amp =1000(1)+1000(.5)\\ \amp=\alert{1000(1.5)^1}. \end{align*} After 2 years, Bill's parents add %50 of the \$1500 in the jar. We have \begin{align*} A(2) \amp =\alert{1000(1.5)^1}+\alert{1000(1.5)^1}(.5)\\ \amp =\alert{1000(1.5)^1}(1.5)\\ \amp =1000(1.5)^2. \end{align*} This trend will continue so that the amount of money in the jar after $t$ years is given by the formula

\begin{equation*} A(t)=1000(1.5)^t. \end{equation*}

The above example was an example of an exponential function. We define such a function as follows:

###### Exponential Function
\begin{equation*} P(t) = a(b)^t \text{, where } b \gt 0 \text{ and } b \ne 1 \text{, } a \ne 0. \end{equation*}

The constant $a$ is the $y$-value of the $y$-intercept of the function.

Some examples of exponential functions are

\begin{equation*} f (x) = 5^x \text{, } P(t) = 250(1.7)^t \text{, and } g(n) = 2.4(0.3)^n. \end{equation*}

The constant $a$ is the $y$-intercept of the graph because

\begin{equation*} f (0) = a \cdot b^0 = a \cdot 1 = a. \end{equation*}

For the examples above, we find that the $y$-intercepts are \begin{align*} f(0) \amp= 5^0 = 1 \text{,} \\ P(0) \amp= 250(1.7)^0 = 250\text{, and} \\ g(0) \amp= 2.4(0.3)^0 = 2.4. \end{align*}

The positive constant $b$ is called the base of the exponential function.

• We do not allow $b$ to be negative, because if $b \lt 0\text{,}$ then $b^x$ is not a real number for some values of $x\text{.}$ For example, if $b = -4$ and $f (x) = (-4)^x\text{,}$ then $f (1/2) = (-4)^{1/2}$ is an imaginary number.
• We also exclude $b = 1$ as a base because $1^x = 1$ for all values of $x\text{;}$ hence the function $f (x) = 1^x$ is actually the constant function $f (x) = 1\text{.}$

In general, exponential functions have the following properties.

###### Properties of Exponential Functions, $f(x) = a(b)^x\text{,}$ $a \gt 0$
1. Domain: all real numbers.

2. Range: all positive numbers.

3. If $b \gt 1\text{,}$ the function is increasing;

if $0 \lt b \lt 1\text{,}$ the function is decreasing.

4. The $y$-intercept is $(0, a)\text{.}$ There is no $x$-intercept.

If $b \gt 1\text{,}$ the function is said to have exponential growth. If $0 \lt b \lt 1\text{,}$ the function is said to have exponential decay.

### SubsectionExponential Equations

An exponential equation is one in which the variable is part of an exponent. For example, the equation

\begin{equation*} 3^x = 81 \end{equation*}

is exponential. Many exponential equations can be solved by writing both sides of the equation as powers with the same base. To solve the equation above, we write

\begin{equation*} 3^x = 3^4 \end{equation*}

which is true if and only if $x = 4\text{.}$ In general, if two equivalent powers have the same base, then their exponents must be equal also, as long as the base is not $0$ or $\pm 1\text{.}$

Sometimes the laws of exponents can be used to express both sides of an equation as single powers of a common base.

###### Example167

Solve the following equations.

1. $3^{x-2} = 9^3$
2. $27 \cdot 3^{-2x} = 9^{x+1}$
Solution
1. Using the fact that $9 = 3^2\text{,}$ we write each side of the equation as a power of $3\text{:}$ \begin{align*} 3^{x-2} \amp = 9^3 \\ 3^{x-2} \amp = \left(3^2\right)^3 \\ 3^{x-2} \amp = 3^6. \end{align*} Now we equate the exponents to obtain \begin{align*} x - 2 \amp = 6 \\ x \amp = 8. \end{align*}
2. We write each factor as a power of $3\text{.}$
\begin{equation*} \begin{aligned} 27 \cdot 3^{-2x} \amp= 9^{x+1} \\ 3^3 \cdot 3^{-2x} \amp= \left(3^2\right)^{x+1} \end{aligned} \end{equation*}
We use the laws of exponents to simplify each side:
\begin{equation*} 3^{3-2x} = 3^{2x+2}. \end{equation*}
Now we equate the exponents to obtain \begin{align*} 3 - 2x \amp = 2x + 2 \\ -4x =\amp -1. \end{align*} The solution is $x = \dfrac{1}{4}\text{.}$