Question

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

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:

\(t = 7.3\)