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]\[
\begin{array}{c}
\ln \left(\frac{1.6}{0.025}\right) \quad 13.6 \quad \ln \left(\frac{1.6}{0.078}\right) \quad \frac{\ln (1.6)}{0.025} \quad \ln \left(\frac{0.025}{1.6}\right) \quad 6.4 \quad 18.8 \\
t = \vdots
\end{array}
\][/tex]

[tex]\(\approx\)[/tex] years



Answer :

To determine how many years it will take for the farmland's market value to reach [tex]$125,000$[/tex] dollars, we start with the given exponential growth function:

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

We need to find [tex]\( t \)[/tex] when the market value [tex]\( p(t) \)[/tex] is [tex]$125,000. First, set the function equal to the target value: \[ 125,000 = 78,125 \cdot e^{0.025t} \] To isolate the exponential term, divide both sides by 78,125: \[ \frac{125,000}{78,125} = e^{0.025t} \] Simplify the left-hand side: \[ \frac{125,000}{78,125} = 1.6 \] So, we have: \[ 1.6 = e^{0.025t} \] Next, take the natural logarithm of both sides to solve for \( t \): \[ \ln(1.6) = \ln(e^{0.025t}) \] Since \( \ln(e^x) = x \), this simplifies to: \[ \ln(1.6) = 0.025t \] Finally, solve for \( t \) by dividing both sides by 0.025: \[ t = \frac{\ln(1.6)}{0.025} \] Using the calculated result: \[ t \approx 18.8 \] Therefore, the number of years it will take for the farmland's market value to reach $[/tex]125,000 is approximately:

[tex]\[ t \approx 18.8 \, \text{years} \][/tex]