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Drag each value and expression to the correct location on the equation. Not all values and expressions will be used.

A patch of farmland is currently worth [tex]$\$[/tex]78,125[tex]$. The expected increase in its market value can be modeled by the function below, where $[/tex]t[tex]$ is the time in years.

\[ p(t) = 78,125 e^{0.025 t} \]

How many years will it take for the farmland's market value to reach $[/tex]\[tex]$125,000$[/tex]?

[tex]\[
\ln \left(\frac{125,000}{78,125}\right)
\][/tex]

[tex]\[
6.4
\][/tex]

[tex]\[
\ln \left(\frac{1.6}{0.025}\right) \quad \ln \left(\frac{0.025}{1.6}\right)
\][/tex]

[tex]\[
13.6
\][/tex]

[tex]\[
t \approx \frac{\ln (1.6)}{0.025} \quad \text{years}
\][/tex]

[tex]\[
18.8
\][/tex]



Answer :

GinnyAnswer
Let's go through the solution step by step:

1. Given Function:
[tex]\[ p(t) = 78,125 \cdot e^{0.025t} \][/tex]

We need to find the time [tex]\(t\)[/tex] when the market value [tex]\(p(t)\)[/tex] will be $125,000.

2. Formulate the Equation:
[tex]\[ 125,000 = 78,125 \cdot e^{0.025t} \][/tex]

3. Isolate the Exponential Term:
[tex]\[ \frac{125,000}{78,125} = e^{0.025t} \][/tex]

Calculate the fraction on the left side:
[tex]\[ \frac{125,000}{78,125} \approx 1.6 \][/tex]

So the equation becomes:
[tex]\[ 1.6 = e^{0.025t} \][/tex]

4. Take the Natural Logarithm of Both Sides:
[tex]\[ \ln(1.6) = \ln(e^{0.025t}) \][/tex]

Using the property of logarithms [tex]\(\ln(e^x) = x\)[/tex]:
[tex]\[ \ln(1.6) = 0.025t \][/tex]

5. Solve for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{\ln(1.6)}{0.025} \][/tex]

6. Calculate the Final Value:
[tex]\[ t \approx 18.8 \text{ years} \][/tex]

So, we have:
[tex]\[ t = \vdots 18.8 \vdots \quad \text{years} \approx \frac{\ln (1.6)}{0.025} \vdots \quad \text{years} \][/tex]

Therefore, the correct values and expressions should be placed as follows:
- [tex]\(t = \vdots 18.8 \vdots\)[/tex] years
- [tex]\(\approx \frac{\ln(1.6)}{0.025} \vdots\)[/tex] years

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