To determine the year when the price of 9-volt batteries will reach [tex]$3 given the price function \( P(t) = 1.1 \cdot e^{0.047t} \), we follow these steps:
1. Set up the equation: We need to find \( t \) when \( P(t) = 3 \) dollars. Therefore, we set up the equation:
\[
1.1 \cdot e^{0.047t} = 3
\]
2. Isolate the exponential term: Divide both sides by 1.1 to isolate the exponential term:
\[
e^{0.047t} = \frac{3}{1.1}
\]
3. Simplify the right-hand side: Calculate the value of \(\frac{3}{1.1}\):
\[
e^{0.047t} = 2.7272\overline{72}
\]
4. Take the natural logarithm of both sides: To solve for \( t \), take the natural logarithm (\(\ln\)) of both sides:
\[
0.047t = \ln(2.7272\overline{72})
\]
5. Calculate the natural logarithm: Compute the natural logarithm of 2.7272\overline{72}:
\[
\ln(2.7272\overline{72}) \approx 1.0031
\]
6. Solve for \( t \): Divide both sides by 0.047 to solve for \( t \):
\[
t = \frac{1.0031}{0.047} \approx 21.3469
\]
7. Determine the year: Add the solution \( t \) to the base year 1980 to determine the actual year:
\[
\text{Year} = 1980 + 21.3469 \approx 2001.3469
\]
Since we need a whole year, we round to the nearest integer, which gives 2001.
Therefore, the price of 9-volt batteries will reach \$[/tex]3 in the year 2001.
Answer: B. 2001
