Question

sections 4.1 - 4.2
score: 11.8/20 answered: 12/20
question 13
a population of bacteria is growing according to the equation ( p(t)=1300 e^{0.23 t} ). estimate when the
population will exceed 3045.
( t= )
give your answer accurate to one decimal place.
question help: video 1 video 2
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Explanation:

Step1: Set up the inequality

We want to find \(t\) when \(P(t)=1300e^{0.23t}>3045\). First, divide both sides of the inequality by \(1300\):
\(\frac{1300e^{0.23t}}{1300}>\frac{3045}{1300}\), which simplifies to \(e^{0.23t}>2.3423\).

Step2: Take the natural logarithm of both sides

Using the property \(\ln(e^{x}) = x\), if \(e^{0.23t}>2.3423\), then \(\ln(e^{0.23t})>\ln(2.3423)\). So \(0.23t>\ln(2.3423)\).

Step3: Solve for \(t\)

We know that \(\ln(2.3423)\approx0.851\). Then \(t >\frac{\ln(2.3423)}{0.23}\). Substitute \(\ln(2.3423)\approx0.851\) into the formula: \(t>\frac{0.851}{0.23}\).
\(\frac{0.851}{0.23}\approx3.7\)

Answer:

\(t = 3.7\)