Question

A population of bacteria is growing according to the equation P(t)=1600e ^.05t. Estimate when the population will exceed 2282.

Answer 1

To solve the problem we need to use the next given equation:

$P\left(t\right)=1600e^{0.05t}$

If the population will exceed 2282. Then p(x)= 2282.

$2282=1,600e^{0.05t}$

Solve for t:

$\frac{2282}{1600}=e^{0.05t}$

Now, we need to take logarithms:

$\begin{gathered} \ln(\frac{2282}{1600})=\ln e^{0.05t} \\ \ln(\frac{2282}{1600})=0.05t\ast\ln e \end{gathered}$

Where ln*e = 1.Then:

$\begin{gathered} \ln(\frac{2282}{1600})=0.05t \\ Where \\ t=\frac{\ln(\frac{2282}{1600})}{0.05} \\ t=7.100 \end{gathered}$

Hence, the population will exceed 2282 when t=7.100