Question

These exercises involve a difference quotient for an exponential function.

If $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathbf{10}}^{\mathbf{x}}$, show that

$\frac{f(x+h)-f\left(x\right)}{h}={10}^x\left(\frac{{10}^h-1}{h}\right)$.

Short answer

We have shown that $\frac{f(x+h)-f\left(x\right)}{h}=\frac{{10}^x\left({10}^h-1\right)}{h}$.

Step-by-step solution

Step 1. Given.

The given function is $f\left(x\right)={10}^x$.

Step 2. To determine.

We have to find a difference quotient for the given function.

Step 3. Calculation.

We first find$f(x+h)$ by plugging$x+h$ in place of x in $f\left(x\right)$.

$$\begin{gathered} f\left(x\right)={10}^x \\ f(x+h)={10}^{(x+h)} \end{gathered}$$

Then we find the difference quotient:

$$\begin{gathered} \frac{f(x+h)-f\left(x\right)}{h}=\frac{{10}^{(x+h)}-{10}^x}{h} \\ =\frac{{10}^x\cdot {10}^h-{10}^x}{h}[\text{using\hspace{0.17em}\hspace{0.17em}property\hspace{0.17em}\hspace{0.17em}}{a}^{(m+n)}=a^m\cdot a^n] \\ =\frac{{10}^x\left({10}^h-1\right)}{h}\left[\text{Factor\hspace{0.17em}\hspace{0.17em}out\hspace{0.17em}\hspace{0.17em}}{10}^x\right] \\ ={10}^x\left(\frac{{10}^h-1}{h}\right) \end{gathered}$$

Hence, it is proved that $\frac{f(x+h)-f\left(x\right)}{h}=\frac{{10}^x\left({10}^h-1\right)}{h}$.