Question

a population of bacteria is growing according to the equation ( p(t)=1000e^{0.18t} ). use a graphing calculator to estimate when the population will exceed 3281. ( t = ) give your answer accurate to one decimal place.

Explanation:

Step1: Set up the inequality

We want to find \(t\) when \(P(t)=1000e^{0.18t}>3281\). So, first, divide both sides of the inequality by \(1000\):
\(e^{0.18t}>\frac{3281}{1000} = 3.281\)

Step2: Take the natural logarithm of both sides

Using the property \(\ln(e^{x})=x\), if \(e^{0.18t}>3.281\), then \(\ln(e^{0.18t})>\ln(3.281)\). So, \(0.18t>\ln(3.281)\)

Step3: Solve for \(t\)

We know that \(\ln(3.281)\approx1.189\). Then \(t >\frac{\ln(3.281)}{0.18}\). Substitute \(\ln(3.281)\approx1.189\) into the formula: \(t>\frac{1.189}{0.18}\approx6.6\)

Answer:

\(t = 6.6\)