Question

A bacteria population increases sixfold in $10$ hours. Assuming normal growth, how long did it take for their population to double?

• A. $3.93$ hrs
• B. $3.87$ hrs
• C. $3.72$ hrs
• D. $3.54$ hrs

Answer 1

Answer: (B)

$3.87$ hrs

If $t$ is time in hours and $P\left(t\right)$ is the population at time $t$, then we know that we have $P\left(t\right)=C{e}^{kt}$ for some constants $C$ and $k$.

The fact that the bacteria increased sixfold in $10$ hours means that $P\left(10\right)=6P\left(0\right)$.

Using the formula for $P\left(t\right)$, this gives $C{e}^{10k}=P\left(10\right)=6P\left(0\right)=6C{e}^{0k}=6C$.

This gives $e^{10k}=6\Rightarrow 10k=ln6\Rightarrow k=10\times ln6$.

We are asked to find how long it took the population to double.

In other words, we want to find the value of $t$ for which $P\left(t\right)=2P\left(0\right)=2C$.

Thus we have

$C{e}^{\frac{tln6}{10}}=2C$

$\Rightarrow \frac{tln6}{10}=ln2$

$\Rightarrow t=\frac{10\times ln2}{ln6}$

$\therefore t=3.868\approx 3.87hrs$