Question

use the exponential growth model, ( a = a_0 e^{kt} ), to show that the time it takes a population to double (to grow from ( a_0 ) to ( 2a_0 )) is given by ( t = \frac{ln 2}{k} ). to find the time it takes to grow the population from ( a_0 ) to ( 2a_0 ), substitute ( a = ) options: ( 2a_0 ), ( \frac{a_0}{2} ), ( a_0 )

Explanation:

Step1: Identify the target population

We want the population to grow from \( A_0 \) to \( 2A_0 \), so the final population \( A \) should be \( 2A_0 \).

Step2: Substitute into the formula

The exponential growth model is \( A = A_0 e^{kt} \). When the population doubles, \( A = 2A_0 \), so we substitute \( A = 2A_0 \) into the model.

Step3: Solve for \( t \)

Starting with \( 2A_0 = A_0 e^{kt} \), divide both sides by \( A_0 \) (assuming \( A_0
eq 0 \)) to get \( 2 = e^{kt} \). Then take the natural logarithm of both sides: \( \ln(2) = \ln(e^{kt}) \). Since \( \ln(e^{kt}) = kt \), we have \( \ln(2) = kt \). Solving for \( t \) gives \( t=\frac{\ln(2)}{k} \), which matches the given formula.

Answer:

To find the time to grow from \( A_0 \) to \( 2A_0 \), substitute \( A = 2A_0 \) (the second option among the given choices for \( A \)). The time \( t \) is derived as \( t = \frac{\ln(2)}{k} \) by substituting \( A = 2A_0 \) into \( A = A_0 e^{kt} \), dividing by \( A_0 \), taking the natural log, and solving for \( t \).