Question

The population P of a bacteria culture is modeled by P = 4100e^kt where t is the time in

hours. If the population of the culture was 5800 after 40 hours, how long does it take for
the population to double? Round to the nearest tenth of an hour.

Show work please
A LOT OF POINTS

Answer 1

It would take around 122 hours to double the population.

Explanation:

To answer the question, we first need to find the constant k, using the given information and the expression.

`P=4100e^(kt) \\5800=4100e^(k(40)) \\(5800)/(4100)=e^(40k)\\e^(40k)=1.41\\lne^(40k)=ln1.41\\40k=0.34\\k=(0.34)/(40)\approx 0.0085`

Now that we have the constant. We can find the time it would take to double the population which would be 11600:

`P=4100e^(kt)\\11600=4100e^(0.0085t)\\(11600)/(4100)= e^(0.0085t)\\e^(0.0085t)=2.83\\lne^(0.0085t)=ln2.83\\0.0085t=1.04\\t=(1.04)/(0.0085)\approx 122.35`

Therefore, it would take around 122 hours to double the population.

Answer 2

Enter the given values into the equation and solve.

5800 = 4100e^(k*40)

Divide both sides by 4100 and simplify:

58 / 41 = e^(k*40)

Remove e by taking the logarithm of both sides:

ln(58/41) = k *40

Divide both sides by 40:

k = ln(58/41)/40

k = 0.00867

Now for the population to double set up the equation:

2*4100 = 4100e^kt

The 4100 cancels out on both sides:

2 = e^kt

Take the logarithm of both sides:

ln(2) = k*t

Divide both sides by k

t = ln(2) /k

replace k with the value from above:

t = ln(2) / 0.00867

t = 79.95

Rounded to the nearest tenth = 80.0 hours to double.