Question

For 10-11, a colony of bacteria with an initial population of 5000 grows over time t (in hours) at a rate of 16% per hour.11. How long does it take for the population to double?

Answer 1

11. The exponential function that models the growth is as follows:

$y=5000(1.16)^t$

Where t is the time in hours. In this case, if we double the population we have to:

y = 5000 x 2 = 10000

Therefore, substitute y = 10000 in the function and solve for t:

$10000=5000\left(1.16\right)^t$

Divide both sides by 5000:

$\begin{gathered} \frac{10000}{5000}=\frac{5000(1.16)^t}{5000} \\ 2=(1.16)^t \end{gathered}$

Apply the laws of exponents:

$ln(2)=tln(1.16)$

Solve for t:

$t=\frac{ln(2)}{ln(1.16)}=4.7$

Answer: B. 4.7 hours