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 or the growth factor of the exponential function. We can also definte the growth rate of the function to be the qunantity \(r=b1\text{,}\) which is usually expressed as a percentage value. Note that \(r\) can be positive or negative, and represents the percent change in the value of \(f(x)\) between any two \(x\)values spaced one unit apart.
 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 a complex number.
 Note that if \(b = 1\) is the growth factor, then \(1^x = 1\) for all values of \(x\text{;}\) hence the function \(f (x) = 1^x\) is actually the constant function \(f (x) = 1\text{.}\) Thus, constant functions are exponential functions!
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 \(b \gt 1\) (equivalently \(r\gt 0\)), the function is increasing;
if \(0 \lt b \lt 1\) (equivalently \(1\lt r\lt 0\)), the function is decreasing.
The \(y\)intercept is \((0, a)\text{.}\) There is no \(x\)intercept.
If \(b \gt 1\) (equivalently \(r\gt 0\)), the function is said to have exponential growth. If \(0 \lt b \lt 1\) (equivalently \(1\lt r\lt 0\)), 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 \(x = \dfrac{1}{4}\text{.}\)