# At the start of an experiment, there are 100 bacteria. If the bacteria follow an exponential growth pattern with rate k = 0.02, what will be the population after 5 hours? How long will it take for the population to double?

Jan 24, 2017

I tried this:

#### Explanation:

I would write the growth function as:
$y = 100 {e}^{k t}$
So that at $t = 0$ we get:
$y = 100 {e}^{0.02 \cdot 0} = 100$ bacteria.
At $t = 5$ hours we will get:
$y = 100 {e}^{5 \cdot 0.02} = 110$ bacteria.
To double the initial number we need to solve:
$200 = 100 {e}^{0.02 t}$
${e}^{0.02 t} = 2$
Apply natural logs to both sides:
$\ln \left({e}^{0.02 t}\right) = \ln \left(2\right)$
$0.02 t = \ln \left(2\right)$
$t = \frac{\ln \left(2\right)}{0.02} = 34.65 \approx 34.6$
$34$ hours and 36 minutes approx.