Answer
a.
After about $\text{14 years}$ , the e-book net sales were $\$\ 8\text{ billion}$.
b.
The doubling time is about $\text{1}\text{.2 years}$.
Work Step by Step
a.
Consider the model $S\left( t \right)=2.05{{\left( 1.8 \right)}^{t}}$
Here, $S\left( t \right)$ is the net sales in period $t$ and the variable $t$ represents the exponential growth rate.
Replace $S\left( t \right)$ with 8000 ($\left( \$8\text{billion}=\$8000\text{million}\right)$and solve for $t$ to calculate the time $t$when the net sales reach to $\$8\text{billion}$ .
$\begin{align}
& S\left( t \right)=2.05{{\left( 1.8 \right)}^{t}} \\
& 8000=2.05{{\left( 1.8 \right)}^{t}}
\end{align}$
Divide both the sides by $2.05$:
$\begin{align}
& \frac{8000}{2.05}=\frac{2.05{{\left( 1.8 \right)}^{t}}}{2.05} \\
& 3902.439={{\left( 1.8 \right)}^{t}} \\
\end{align}$
$\log 3902.439=\log {{\left( 1.8 \right)}^{t}}$
Apply the power rule of logarithms $\log {{m}^{n}}=n\log m$ in $\log {{\left( 1.8 \right)}^{t}}$:
$\begin{align}
& \log 3902.439=\log {{\left( 1.8 \right)}^{t}} \\
& \log 3902.439=t\log 1.8 \\
\end{align}$
Divide by $\log 1.8$ on both the sides:
$\begin{align}
& \frac{\log 3902.439}{\log 1.8}=\frac{t\log 1.8}{\log 1.8} \\
& \frac{\log 3902.439}{\log 1.8}=t
\end{align}$
Therefore,
$\begin{align}
& \frac{\log 3902.439}{\log 1.8}=t \\
& \frac{3.591}{0.255}\approx t \\
& 14.08\approx t
\end{align}$
b.
Consider the model $S\left( t \right)=2.05{{\left( 1.8 \right)}^{t}}$
First calculate the net sales in the beginning, $S\left( 0 \right)$:
$\begin{align}
& S\left( t \right)=2.05{{\left( 1.8 \right)}^{t}} \\
& S\left( 0 \right)=2.05{{\left( 1.8 \right)}^{0}} \\
& =2.05
\end{align}$
Now, doubling time is the amount of time necessary for the net sales to double in size.
If the initial sales $S\left( 0 \right)=2.05$, then $2\left( 2.05 \right)=4.10$ is the doubled net sales.
Therefore, to find the doubling time, replace $S\left( t \right)$ with $4.10$ in $S\left( t \right)=2.05{{\left( 1.8 \right)}^{t}}$and then solve for $t$:
$\begin{align}
& S\left( t \right)=2.05{{\left( 1.8 \right)}^{t}} \\
& 4.10=2.05{{\left( 1.8 \right)}^{t}}
\end{align}$
$\begin{align}
& \frac{4.10}{2.05}=\frac{2.05{{\left( 1.8 \right)}^{t}}}{2.05} \\
& 2={{\left( 1.8 \right)}^{t}} \\
\end{align}$
Take common logarithm on both the sides:
$\log 2=\log {{\left( 1.8 \right)}^{t}}$
Apply the power rule of logarithms $\log {{m}^{n}}=n\log m$ in $\log {{\left( 1.8 \right)}^{t}}$:
$\begin{align}
& \log 2=\log {{\left( 1.8 \right)}^{t}} \\
& \log 2=t\log 1.8 \\
\end{align}$
Divide by $\log 1.8$ on both the sides:
$\begin{align}
& \log 2=t\log 1.8 \\
& \frac{\log 2}{\log 1.8}=\frac{t\log 1.8}{\log 1.8} \\
& \frac{\log 2}{\log 1.8}=t \\
\end{align}$
Thus:
$\begin{align}
& \frac{\log 2}{\log 1.8}=t \\
& \frac{0.3010}{0.255}\approx t \\
& 1.18\approx t
\end{align}$
