Question

the size p of a certain insect population at time t (in days) obeys the function ( p(t)=100e^{0.06t} ).
(a) determine the number of insects at ( t = 0 ) days.
(b) what is the growth rate of the insect population?
(c) what is the population after 10 days?
(d) when will the insect population reach 170?
(e) when will the insect population double?
(c) what is the population after 10 days?
approximately 165 insects.
(do not round until the final answer. then round to the nearest whole number as needed.)
(d) when will the population reach 170 insects?
in approximately 10.6 days.
(do not round until the final answer. then round to the nearest tenth as needed.)
(e) when will the insect population double?
in about ( square ) days.
(do not round until the final answer. then round to the nearest tenth as needed.)

Explanation:

Step1: Set up the equation

The initial population is \(P_0 = 100\). We want to find \(t\) when \(P(t)=2P_0 = 200\). Substitute into the formula \(P(t)=100e^{0.08t}\), so \(200 = 100e^{0.08t}\).

Step2: Simplify the equation

Divide both sides of \(200 = 100e^{0.08t}\) by \(100\). We get \(2=e^{0.08t}\).

Step3: Take the natural logarithm of both sides

Using the property \(\ln(e^x)=x\), take \(\ln\) of both sides: \(\ln(2)=\ln(e^{0.08t})\). So \(\ln(2) = 0.08t\).

Step4: Solve for \(t\)

We know that \(\ln(2)\approx0.6931\). Then \(t=\frac{\ln(2)}{0.08}\). Substitute \(\ln(2)\approx0.6931\) into the formula: \(t=\frac{0.6931}{0.08}=8.66375\).

Answer:

\(8.7\)