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 the population from ( a_0 ) to ( 2a_0 ), substitute ( a = 2a_0 ).
substitute.
( 2a_0 = a_0 e^{kt} )
divide both sides by ( a_0 ).
( square = e^{kt} )

Explanation:

Step1: Start with the exponential growth model

We know the exponential growth model is \( A = A_0 e^{kt} \), where \( A \) is the final amount, \( A_0 \) is the initial amount, \( k \) is the growth rate, and \( t \) is the time.

Step2: Substitute \( A = 2A_0 \) into the model

Substitute \( A = 2A_0 \) into \( A = A_0 e^{kt} \), we get \( 2A_0 = A_0 e^{kt} \).

Step3: Divide both sides by \( A_0 \)

Divide both sides of the equation \( 2A_0 = A_0 e^{kt} \) by \( A_0 \) (assuming \( A_0
eq0 \)). Since \( A_0\) is a non - zero initial population, we have \( \frac{2A_0}{A_0}=\frac{A_0 e^{kt}}{A_0} \), which simplifies to \( 2 = e^{kt} \).

Step4: Take the natural logarithm of both sides

Take the natural logarithm of both sides of the equation \( 2 = e^{kt} \). We know that \( \ln(e^{x})=x \), so \( \ln(2)=\ln(e^{kt}) \), which gives \( \ln(2)=kt \).

Step5: Solve for \( t \)

Divide both sides of the equation \( \ln(2)=kt \) by \( k \) (assuming \( k
eq0 \)) to solve for \( t \). We get \( t=\frac{\ln(2)}{k} \).

Answer:

When we divide both sides of \( 2A_0 = A_0 e^{kt} \) by \( A_0 \), the result is \( 2 = e^{kt} \), so the box should be filled with \( 2 \).