a population of bacteria is growing according to the equation ( p(t)=1000e^{0.18t} ). use a graphing…

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.

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.

Answer

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)