Question

a biologist monitors the fish population in an aquarium. there are 400 fish when she begins her study. the number of fish after t days can be modeled by the equation ( p = 400e^{0.014t} ).
how many days will it take for the fish population to double? round your answer to the nearest whole number.
( t = )

Explanation:

Step1: Set up the equation

When the population doubles, \(P = 2\times400=800\). Substitute \(P = 800\) into the equation \(P = 400e^{0.014t}\), we get \(800=400e^{0.014t}\).

Step2: Simplify the equation

Divide both sides of the equation \(800 = 400e^{0.014t}\) by \(400\), we have \(\frac{800}{400}=e^{0.014t}\), which simplifies to \(2=e^{0.014t}\).

Step3: Take the natural logarithm of both sides

Since \(y = e^{x}\) and \(x=\ln(y)\) are inverse - functions, taking the natural logarithm of both sides of \(2 = e^{0.014t}\), we get \(\ln(2)=\ln(e^{0.014t})\). Using the property \(\ln(e^{a})=a\), the right - hand side simplifies to \(0.014t\). So, \(\ln(2)=0.014t\).

Step4: Solve for \(t\)

We know that \(\ln(2)\approx0.693\). Then \(t=\frac{\ln(2)}{0.014}\). Substitute \(\ln(2)\approx0.693\) into the formula, \(t=\frac{0.693}{0.014}\).

Answer:

\(t = 49.5\approx50\)