Answer
a)
The exponential growth rate $k$ is approximately equal to 0.288, and the equation for the exponential growth function is $P\left( t \right)=4.2{{e}^{0.288t}}$.
b)
The Conflicker worm takes 4.96 days to infect 10 million computers.
Work Step by Step
a)
Consider the exponential growth function, $P\left( t \right)={{P}_{0}}{{e}^{kt}}$ …… (1)
Now, put ${{P}_{0}}=2.4$ in the exponential growth function as,
$P\left( t \right)=2.4{{e}^{kt}}$ …… (2)
Here, $k$ is the exponential growth rate,
Put $t=1$ in equation-2 and calculate the value of $k$:
$\begin{align}
& P\left( t \right)=2.4{{e}^{k\cdot 1}} \\
& 3.2=2.4{{e}^{k\cdot 1}} \\
& \frac{3.2}{2.4}={{e}^{k\cdot 1}} \\
& 1.33={{e}^{k\cdot 1}}
\end{align}$
$\begin{align}
& 1.33={{e}^{k\cdot 1}} \\
& \ln 1.33=\ln {{e}^{k\cdot 1}} \\
& 0.288\approx k
\end{align}$
Then, the exponential growth rate is $k\approx 0.288$.
Now, put the value of $k$ in equation-2,
$P\left( t \right)=2.4{{e}^{0.288t}}$
Therefore, $P\left( t \right)=2.4{{e}^{0.288t}}$ is the exponential growth function.
b)
From part-(a), it is clear that the exponential growth function is $P\left( t \right)=2.4{{e}^{0.288t}}$. Thus:
$\begin{align}
& P\left( t \right)=2.4{{e}^{0.288t}} \\
& 10=2.4{{e}^{0.288t}} \\
& \frac{10}{2.4}={{e}^{0.288t}} \\
& 4.166={{e}^{0.288t}}
\end{align}$
$\begin{align}
& 4.166={{e}^{0.288t}} \\
& \ln 4.166=\ln {{e}^{0.288t}} \\
& 1.426=0.288t \\
& 4.96\approx t
\end{align}$
