elias purchases a home for $38,900. the value of the home, in thousands of dollars, since his purchase is shown in the table. find an exponential function that models the data. round numerical values to the nearest hundredth. let x be the number of years since the purchase. the function ( f(x)=39.59(1.09)^x ) models the data. use the model to predict the homes value. after 12 years, the homes value will be about $ thousand. after 35 years, the homes value will be about $ thousand.

Question

Accepted Answer

# Explanation:
## Step1: Substitute $x = 12$ into the function $f(x)=39.59\times(1.09)^{x}$
$$
\begin{align*}
f(12)&=39.59\times(1.09)^{12}\
\end{align*}
$$
First, calculate $(1.09)^{12}$. Using the formula $a^{n}=e^{n\ln(a)}$, $\ln(1.09)\approx0.0862$, $n = 12$, so $n\ln(a)=12\times0.0862 = 1.0344$, $e^{1.0344}\approx2.8127$ (alternatively, using a calculator $(1.09)^{12}\approx2.8127$)
$$
\begin{align*}
f(12)&=39.59\times2.8127\
&\approx39.59\times2.81\
&=(40 - 0.41)\times2.81\
&=40\times2.81-0.41\times2.81\
&=112.4-1.1521\
&\approx111.25
\end{align*}
$$
## Step2: Substitute $x = 35$ into the function $f(x)=39.59\times(1.09)^{x}$
$$
\begin{align*}
f(35)&=39.59\times(1.09)^{35}\
\end{align*}
$$
Calculate $(1.09)^{35}$. Using $a^{n}=e^{n\ln(a)}$, $\ln(1.09)\approx0.0862$, $n = 35$, $n\ln(a)=35\times0.0862=3.017$, $e^{3.017}\approx20.49$ (alternatively, using a calculator $(1.09)^{35}\approx20.49$)
$$
\begin{align*}
f(35)&=39.59\times20.49\
&\approx39.59\times20.5\
&=(40 - 0.41)\times20.5\
&=40\times20.5-0.41\times20.5\
&=820 - 8.405\
&\approx811.6
\end{align*}
$$
Since $811.6$ is closest to $808$ among the given options for $x = 35$

# Answer:
After 12 years: $111.25$ (not in the dropdown, but if following the model calculation). After 35 years: $808$

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QUESTION IMAGE

Question

Explanation:

Step1: Substitute \(x = 12\) into the function \(f(x)=39.59\times(1.09)^{x}\)

Step2: Substitute \(x = 35\) into the function \(f(x)=39.59\times(1.09)^{x}\)

Answer: