To determine the amount of money [tex]\( P \)[/tex] that should be invested at a rate [tex]\( r = 5\% \)[/tex] (or 0.05 as a decimal) to produce a final balance of [tex]$140,000 in \( t \) years with continuous compounding, we use the formula for continuous compounding:
\[ P = \frac{A}{e^{rt}} \]
where:
- \( A \) is the final balance, which is $[/tex]140,000.
- [tex]\( r \)[/tex] is the annual interest rate, expressed as a decimal (0.05).
- [tex]\( t \)[/tex] is the number of years.
- [tex]\( e \)[/tex] is the base of the natural logarithm, approximately equal to 2.71828.
We'll calculate the initial investment [tex]\( P \)[/tex] for each given [tex]\( t \)[/tex] using the formula above.
1. For [tex]\( t = 1 \)[/tex]:
[tex]\[ P = \frac{140,000}{e^{0.05 \cdot 1}} \][/tex]
[tex]\[ P = \frac{140,000}{e^{0.05}} \][/tex]
[tex]\[ P \approx 133,172.12 \][/tex]
2. For [tex]\( t = 10 \)[/tex]:
[tex]\[ P = \frac{140,000}{e^{0.05 \cdot 10}} \][/tex]
[tex]\[ P = \frac{140,000}{e^{0.5}} \][/tex]
[tex]\[ P \approx 84,914.29 \][/tex]
3. For [tex]\( t = 20 \)[/tex]:
[tex]\[ P = \frac{140,000}{e^{0.05 \cdot 20}} \][/tex]
[tex]\[ P = \frac{140,000}{e^{1}} \][/tex]
[tex]\[ P \approx 51,503.12 \][/tex]
4. For [tex]\( t = 30 \)[/tex]:
[tex]\[ P = \frac{140,000}{e^{0.05 \cdot 30}} \][/tex]
[tex]\[ P = \frac{140,000}{e^{1.5}} \][/tex]
[tex]\[ P \approx 31,238.22 \][/tex]
5. For [tex]\( t = 40 \)[/tex]:
[tex]\[ P = \frac{140,000}{e^{0.05 \cdot 40}} \][/tex]
[tex]\[ P = \frac{140,000}{e^{2}} \][/tex]
[tex]\[ P \approx 18,946.94 \][/tex]
6. For [tex]\( t = 50 \)[/tex]:
[tex]\[ P = \frac{140,000}{e^{0.05 \cdot 50}} \][/tex]
[tex]\[ P = \frac{140,000}{e^{2.5}} \][/tex]
[tex]\[ P \approx 11,491.90 \][/tex]
So, the completed table with the computed amounts of [tex]\( P \)[/tex] is:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$t$ & $P$ \\
\hline
1 & 133,172.12 \\
\hline
10 & 84,914.29 \\
\hline
20 & 51,503.12 \\
\hline
30 & 31,238.22 \\
\hline
40 & 18,946.94 \\
\hline
50 & 11,491.90 \\
\hline
\end{tabular}
\][/tex]
