# A bacteria population is 3000 at time t = 0 and its rate of growth is 1000-6^t bacteria per hour...

## Question:

A bacteria population is 3000 at time t = 0 and its rate of growth is {eq}1000-6^t {/eq} bacteria per hour after t hours. What is the population after one hour? (Round your answer to the nearest whole number.)

## Exponential Growth Function

This problem presents us with the rate of exponential growth. We need to find a function that tells us the growth. Once we have the growth function we will use it to find the population after the given time period.

• I wonder if the rate of growth function is properly written but I shall consider the one mentioned in the question.

Growth rate of the population is given by:

{eq}\displaystyle \frac{dp}{dt}=1000-6^t {/eq}

It can be written as,

{eq}\displaystyle dp=(1000-6^t)dt {/eq}

Integrating both the sides, we get,

{eq}\displaystyle \int_{p(0)}^{p(t)}dp=\int_{0}^{t}(1000-6^t)dt {/eq}

{eq}\displaystyle p(t)-p(0)=1000t-\frac{6^t}{\ln 6}+\frac{1}{\ln 6} {/eq}

{eq}\displaystyle p(t)=p(0)+1000t-\frac{6^t}{\ln 6}+\frac{1}{\ln 6} {/eq}

Plugging in the value of initial population, we get,

{eq}\displaystyle p(t)=3000+1000t-\frac{6^t}{\ln 6}+\frac{1}{\ln 6} {/eq}

Therefore, the population after {eq}1 {/eq} hour is:

{eq}\displaystyle p(1)=3000+1000(1)-\frac{6^{1}}{\ln 6}+\frac{1}{\ln 6} {/eq}

{eq}\displaystyle p(1)=3000+1000-\frac{6}{\ln 6}+\frac{1}{\ln 6} {/eq}

{eq}\displaystyle p(1)=4000-\frac{6}{1.79}+\frac{1}{1.79} {/eq}

{eq}\displaystyle \boxed{\displaystyle p(1)=3997.208\approx 3997} {/eq}

So, the population of bacteria after 1 hour is 3997.