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.
SolutionAfter 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, \(f(x) = a(b)^x\text{,}\) \(a \gt 0\)
Domain: all real numbers.
Range: all positive numbers.

If \(b \gt 1\text{,}\) the function is increasing;
if \(0 \lt b \lt 1\text{,}\) the function is decreasing.
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.
 \(3^{x2} = 9^3\)
 \(27 \cdot 3^{2x} = 9^{x+1}\)
Solution
 Using the fact that \(9 = 3^2\text{,}\) we write each side of the equation as a power of \(3\text{:}\) \begin{align*} 3^{x2} \amp = 9^3 \\ 3^{x2} \amp = \left(3^2\right)^3 \\ 3^{x2} \amp = 3^6. \end{align*} Now we equate the exponents to obtain \begin{align*} x  2 \amp = 6 \\ x \amp = 8. \end{align*}
 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^{32x} = 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 \(x = \dfrac{1}{4}\text{.}\)