Question

the number of bacteria in a certain population increases according to a continuous exponential growth model, with a growth rate parameter of 1.3% per hour. how many hours does it take for the size of the sample to double?
note: this is a continuous exponential growth model.
do not round any intermediate computations, and round your answer to the nearest hundredth.

Explanation:

Step1: Recall the continuous - growth formula

The continuous - growth formula is \(A = A_0e^{rt}\), where \(A\) is the final amount, \(A_0\) is the initial amount, \(r\) is the growth rate, and \(t\) is the time. If the population doubles, then \(A = 2A_0\), and \(r=0.013\).
Substitute \(A = 2A_0\) into the formula: \(2A_0=A_0e^{0.013t}\).
Since \(A_0
eq0\) (because we are talking about a non - zero population of bacteria), we can divide both sides of the equation by \(A_0\). We get \(2 = e^{0.013t}\).

Step2: Take the natural logarithm of both sides

Using the property \(\ln(e^{x})=x\), if \(2 = e^{0.013t}\), then \(\ln(2)=\ln(e^{0.013t})\).
By the property of logarithms \(\ln(e^{x}) = x\), the right - hand side simplifies to \(0.013t\). So, \(\ln(2)=0.013t\).
We know that \(\ln(2)\approx0.6931\).

Step3: Solve for \(t\)

We have the equation \(t=\frac{\ln(2)}{0.013}\).
Substitute \(\ln(2)\approx0.6931\) into the formula: \(t=\frac{0.6931}{0.013}\).
\(t=\frac{6931}{130}\approx53.3154\).

Answer:

\(t\approx53.32\) hours