# If there are 500 present at a given time and 1000 present 2 hours later, how many hours will it...

## Question:

If there are 500 present at a given time and 1000 present 2 hours later, how many hours will it take for the number of bacteria to be 2500?

## Population Growth:

Populations like bacteria tend to grow at a rate proportional to their size until they start to bump up against the carrying capacity of they environment. So if there are {eq}P(t) {/eq} bacteria present at time {eq}t {/eq}, then we have {eq}\dfrac{dP}{dt}=kP {/eq} for some constant {eq}k {/eq}. This gives us a separable differential equation and we have

{eq}\dfrac{1}{P}\, dP=k\, dt {/eq}

Integrating gives us

{eq}\ln |P|=kt+C \\ |P|=e^{kt+C}\\ |P|=e^Ce^{kt} {/eq}

Since {eq}P>0 {/eq} we can drop the absolute value and we have

{eq}P=Ce^{kt} {/eq}.

Let {eq}P(t) {/eq} be the population at time {eq}t {/eq}. Assuming the growth is exponential, then {eq}P=Ce^{kt} {/eq}. We then have {eq}P(0)=500 {/eq}. This gives us

{eq}500=P(0)=Ce^0\\ C=500 {/eq}

This gives us

{eq}P=500e^{kt} {/eq}

Next we note that {eq}P(2)=1000 {/eq} so that

{eq}1000=P(2)=500e^{2k}\\ e^{2k}=2\\ 2k=\ln 2\\ k=\dfrac{\ln 2}{2} {/eq}

This gives us

{eq}P=500e^{\frac{\ln 2}{2}t} {/eq}

We need to find {eq}t {/eq} such that {eq}P(t)=2500 {/eq}. This gives us

{eq}2500=P(t)\\ 2500=500e^{\frac{\ln 2}{2}t}\\ e^{\frac{\ln 2}{2}t}=5\\ \frac{\ln 2}{2}t=\ln 5\\ t=\dfrac{2\ln 5}{\ln 2}=4.644 \text{ hours} {/eq} 