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question 2
a publication that reported on the most effective ways to solve a city’s homelessness problem claimed that if its suggestions are followed and implemented, a city with ( a ) individuals experiencing homelessness this year can expect to have ( h ) homeless people ( t ) years later, according to the model
( h(t) = ae^{-0.16t} )
given this model, in how many years can the city expect a reduction of 90% of its population experiencing homelessness?
enter your answer as a number (no labels) and round to the nearest tenth of a year.
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Explanation:

Step1: Understand the problem

We need to find the time \( t \) when the homeless population is reduced by 90%, which means the remaining population is 10% of the initial population \( A \). So we set \( H(t) = 0.1A \) in the model \( H(t)=Ae^{-0.16t} \).

Step2: Set up the equation

Substitute \( H(t) = 0.1A \) into the model:
\[
0.1A = Ae^{-0.16t}
\]
Since \( A
eq 0 \) (because there is an initial homeless population), we can divide both sides by \( A \):
\[
0.1 = e^{-0.16t}
\]

Step3: Take the natural logarithm of both sides

To solve for \( t \), we take the natural logarithm of both sides:
\[
\ln(0.1) = \ln(e^{-0.16t})
\]
Using the property of logarithms \( \ln(e^x)=x \), the right side simplifies to \( -0.16t \):
\[
\ln(0.1) = -0.16t
\]

Step4: Solve for \( t \)

We can solve for \( t \) by dividing both sides by \( -0.16 \):
\[
t = \frac{\ln(0.1)}{-0.16}
\]
Calculate \( \ln(0.1) \approx -2.302585 \), then:
\[
t = \frac{-2.302585}{-0.16} \approx 14.4
\]

Answer:

14.4