Question

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?

Answer 1

I tried this:

Explanation:

I would write the growth function as:
`y=100e^(kt)`
So that at `t=0` we get:
`y=100e^(0.02*0)=100` bacteria.
At `t=5` hours we will get:
`y=100e^(5*0.02)=110` bacteria.
To double the initial number we need to solve:
`200=100e^(0.02t)`
`e^(0.02t)=2`
Apply natural logs to both sides:
`ln(e^(0.02t))=ln(2)`
`0.02t=ln(2)`
`t=(ln(2))/0.02=34.65~~34.6`
`34` hours and 36 minutes approx.