Question

a population of bacteria is growing according to the equation ( p(t)=650 e^{0.06 t} ). use a graphing calculator to estimate when the population will exceed 763.
( t= )
give your answer accurate to one decimal place.
question help: video 1 read 1 video 2 written example 1

Explanation:

Step1: Set up the inequality

We want to find \(t\) when \(P(t)=650e^{0.06t}>763\). First, divide both sides of the inequality by \(650\):
\(e^{0.06t}>\frac{763}{650}\)
\(e^{0.06t}>1.173846\)

Step2: Take the natural - logarithm of both sides

Using the property \(\ln(e^{x}) = x\), we have \(\ln(e^{0.06t})>\ln(1.173846)\)
\(0.06t>\ln(1.173846)\)

Step3: Solve for \(t\)

We know that \(\ln(1.173846)\approx0.1605\). Then \(t>\frac{\ln(1.173846)}{0.06}\)
\(t>\frac{0.1605}{0.06}\)
\(t > 2.7\)

Answer:

\(t = 2.7\)