Answer
The doubling time is about $0.5\text{ hours}$.
Work Step by Step
The exponential growth model $P\left( t \right)={{P}_{0}}{{e}^{kt}},k>0$.
Here, ${{P}_{0}}$ is the investment at time 0, $P\left( t \right)$ is the balance at time $t$ years, $k$ is the interest rate compounded continuously, and doubling time is the amount of time necessary for the investment to double in size.
It is given that the number of bacteria grows at an exponential growth rate of $139\%$ per hour.
Therefore, $k=1.39$.
Now, doubling time is the amount of time necessary for the population to double in size.
If ${{P}_{0}}$ is the population at time 0, then the double of ${{P}_{0}}$ is $2{{P}_{0}}$
Thus, to find the doubling time, replace $P\left( t \right)$ with $2{{P}_{0}}$ in $P\left( t \right)={{P}_{0}}{{e}^{kt}}$; substitute $k=1.39$ and solve for $t$:
$\begin{align}
& P\left( t \right)={{P}_{0}}{{e}^{kt}} \\
& 2{{P}_{0}}={{P}_{0}}{{e}^{1.39t}}
\end{align}$
Divide by ${{P}_{0}}$ on both the sides:
$\begin{align}
& \frac{2{{P}_{0}}}{{{P}_{0}}}=\frac{{{P}_{0}}{{e}^{1.39t}}}{{{P}_{0}}} \\
& 2={{e}^{1.39t}} \\
\end{align}$
Take the natural logarithm on both the sides:
$\ln 2=\ln {{e}^{1.39t}}$
Apply the power rule of logarithms $\ln {{m}^{n}}=n\ln m$ in $\ln {{e}^{1.39t}}$:
$\begin{align}
& \ln 2=\ln {{e}^{1.39t}} \\
& \ln 2=1.39t\ln e \\
\end{align}$
$\begin{align}
& \ln 2=1.39t\ln e \\
& 0.6931=1.39t
\end{align}$
Divide by $1.39$on both sides,
$\begin{align}
& \frac{0.6931}{1.39}=\frac{1.39t}{1.39} \\
& 0.49\approx t
\end{align}$
