Question

a population of bacteria is growing according to the exponential model $p = 100e^{0.70t}$, where $p$ is the number of colonies and $t$ is measured in hours. after how many hours will 300 colonies be present? round answer to the nearest tenth.
a 0.7
b 1.6
c 6.7
d 7.2

Explanation:

Step1: Set up the equation

We know the model is \( P = 100e^{0.70t} \) and we want to find \( t \) when \( P = 300 \). So we set up the equation:
\( 300 = 100e^{0.70t} \)

Step2: Divide both sides by 100

Divide both sides of the equation by 100 to simplify:
\( \frac{300}{100} = e^{0.70t} \)
\( 3 = e^{0.70t} \)

Step3: Take the natural logarithm of both sides

Take the natural log (\( \ln \)) of both sides to solve for the exponent:
\( \ln(3) = \ln(e^{0.70t}) \)
Using the property \( \ln(e^x)=x \), we get:
\( \ln(3) = 0.70t \)

Step4: Solve for t

Divide both sides by 0.70 to solve for \( t \):
\( t = \frac{\ln(3)}{0.70} \)
Calculate \( \ln(3) \approx 1.0986 \), then:
\( t \approx \frac{1.0986}{0.70} \approx 1.569 \)
Round to the nearest tenth, \( t \approx 1.6 \)

Answer:

B. 1.6