#### Example 7.5.1.

Find an exponential function \(f(x)=ab^x\) that has the values \(f(2) = 4.5\) and \(f(5) = 121.5\text{.}\)

#### Solution.

We would like to find values of \(a\) and \(b\) so that the given function values satisfy \(f(x) = ab^x\) . By substituting the function values into the formula, we can write two equations.

\begin{align*}
f(2) \amp= 4.5~~~ \text{ means }~~~x = 2, y = 4.5, \amp\amp\text{so}~~~~~ab^2 = 4.5\\
f(5) \amp= 121.5~~~ \text{ means }~~~x = 5, y = 121.5, \amp\amp\text{so}~~~~~ab^5 = 121.5
\end{align*}

We now have a system of equations in the two unknowns \(a\) and \(b\text{:}\)

\begin{gather*}
ab^2 = 4.5\\
ab^5 = 121.5
\end{gather*}

but it is not a linear system. We can solve the system by the method of elimination, but instead of adding the equations, we will divide one of the equations by the other.

\begin{align*}
\frac{ab^5}{ab^2}\amp= \frac{121.5}{4.5}\\
b^3 \amp = 27
\end{align*}

Note that by dividing the two equations, we eliminated \(a\text{,}\) and we can now solve for \(b\text{.}\)

\begin{align*}
b^3 \amp = 27 \\
b \amp = \sqrt[3]{27} = 3
\end{align*}

Next we substitute \(b = 3\) into either of the two equations and solve for \(a\text{.}\)

\begin{align*}
a(3)^2 \amp = 4.5 \\
a \amp= \frac{4.5}{9} = 0.5
\end{align*}

Thus, \(a = 0.5\) and \(b = 3\text{,}\) and the function is \(f(x) = 0.5(3^x)\text{.}\)