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
