a population of bacteria is growing according to the equation ( p(t) = 1700e^{0.13t} ).\nuse a graphing…

a population of bacteria is growing according to the equation ( p(t) = 1700e^{0.13t} ).\nuse a graphing calculator to estimate when the population will exceed 4391.\n( t = )\ngive your answer accurate to one decimal place.

a population of bacteria is growing according to the equation ( p(t) = 1700e^{0.13t} ).\nuse a graphing calculator to estimate when the population will exceed 4391.\n( t = )\ngive your answer accurate to one decimal place.

Answer

Explanation:

Step1: Set up the inequality

We want to find (t) when (P(t)=1700e^{0.13t}>4391). First, divide both sides of the inequality by (1700): (e^{0.13t}>\frac{4391}{1700}) (e^{0.13t}>2.582941)

Step2: Take the natural - logarithm of both sides

Using the property (\ln(e^{x}) = x), if (e^{0.13t}>2.582941), then (\ln(e^{0.13t})>\ln(2.582941)) (0.13t>\ln(2.582941)) Since (\ln(2.582941)\approx0.95) (using a calculator), we have (0.13t > 0.95)

Step3: Solve for (t)

Divide both sides of the inequality (0.13t>0.95) by (0.13): (t>\frac{0.95}{0.13}) (t>\frac{95}{13}\approx7.3)

Answer:

(t = 7.3)