Question

homework 6.1 exponential functions
score: 20/140 answered: 2/14
question 3
a population of bacteria is growing according to the equation ( p(t)=350 e^{0.23 t} ).
use a graphing calculator to estimate when the population will exceed 1491.
( t= )
give your answer accurate to one decimal place.
question help: video

Explanation:

Step1: Set up the inequality

Set \(P(t)>1491\), so \(350e^{0.23t}>1491\).

Step2: Solve for \(e^{0.23t}\)

Divide both sides by \(350\): \(e^{0.23t}>\frac{1491}{350}\approx4.26\).

Step3: Take the natural logarithm of both sides

\(\ln(e^{0.23t})>\ln(4.26)\). Since \(\ln(e^{x}) = x\), we have \(0.23t>\ln(4.26)\).

Step4: Solve for \(t\)

\(t>\frac{\ln(4.26)}{0.23}\). Calculate \(\ln(4.26)\approx1.45\), then \(t>\frac{1.45}{0.23}\approx6.3\).

Answer:

\(t = 6.3\)