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The population of a town can be modeled using the formula P=25000e^0.03t , where t is the number of years after 2012 and P is t...Question
the population of a town can be modeled using hte formula p=20,000 e^0.02t , where t is the number of years after 2012 and p is the town's population . which of the following equations can be used to find the number of years after 2012 hat the population will doubke to 40,000 ?
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Answered by
GPT-4o mini
To find the number of years after 2012 that the town's population will double to 40,000, we can set up the equation based on the population model provided:
Given:
\[
p = 20,000 e^{0.02t}
\]
We want to find \( t \) when \( p = 40,000 \). So, we set up the equation:
\[
40,000 = 20,000 e^{0.02t}
\]
Now, divide both sides by 20,000:
\[
\frac{40,000}{20,000} = e^{0.02t}
\]
This simplifies to:
\[
2 = e^{0.02t}
\]
To solve for \( t \), we take the natural logarithm of both sides:
\[
\ln(2) = \ln(e^{0.02t})
\]
Using the property of logarithms that \( \ln(e^x) = x \), we have:
\[
\ln(2) = 0.02t
\]
Now we can solve for \( t \):
\[
t = \frac{\ln(2)}{0.02}
\]
This equation can be used to find the number of years after 2012 that the population will double to 40,000.
Given:
\[
p = 20,000 e^{0.02t}
\]
We want to find \( t \) when \( p = 40,000 \). So, we set up the equation:
\[
40,000 = 20,000 e^{0.02t}
\]
Now, divide both sides by 20,000:
\[
\frac{40,000}{20,000} = e^{0.02t}
\]
This simplifies to:
\[
2 = e^{0.02t}
\]
To solve for \( t \), we take the natural logarithm of both sides:
\[
\ln(2) = \ln(e^{0.02t})
\]
Using the property of logarithms that \( \ln(e^x) = x \), we have:
\[
\ln(2) = 0.02t
\]
Now we can solve for \( t \):
\[
t = \frac{\ln(2)}{0.02}
\]
This equation can be used to find the number of years after 2012 that the population will double to 40,000.
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