To determine the time it takes for a given investment to double when compounded continuously, we use the formula for continuous compounding:
[tex]\[ A = P \cdot e^{rt} \][/tex]
Here, [tex]\( A \)[/tex] is the amount of money accumulated after time [tex]\( t \)[/tex], [tex]\( P \)[/tex] is the principal amount (the initial amount of money), [tex]\( r \)[/tex] is the annual interest rate, and [tex]\( t \)[/tex] is the time in years. For the case where the amount doubles, [tex]\( A = 2P \)[/tex].
So, our equation becomes:
[tex]\[ 2P = P \cdot e^{rt} \][/tex]
We can simplify this equation by dividing both sides by [tex]\( P \)[/tex]:
[tex]\[ 2 = e^{rt} \][/tex]
Next, we take the natural logarithm (ln) of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(2) = rt \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(2)}{r} \][/tex]
Given the annual interest rate [tex]\( r = 4.3\% \)[/tex] or [tex]\( 0.043 \)[/tex] in decimal form, we can substitute this value into our formula:
[tex]\[ t = \frac{\ln(2)}{0.043} \][/tex]
We can now calculate [tex]\( \ln(2) \)[/tex]:
[tex]\[ \ln(2) \approx 0.693147 \][/tex]
Thus:
[tex]\[ t = \frac{0.693147}{0.043} \][/tex]
[tex]\[ t \approx 16.12 \][/tex]
So, the time it takes for [tex]$8,500 to double when invested at an annual interest rate of 4.3% compounded continuously is approximately 16.12 years.
Similarly, for $[/tex]629,000 to double under the same conditions, the doubling time will be the same because the formula does not depend on the principal amount, only on the rate of interest. Therefore, the time for [tex]$629,000 to double is also:
\[ t \approx 16.12 \]
To summarize:
- The time it takes for $[/tex]8,500 to double is approximately [tex]\( 16.12 \)[/tex] years.
- The time it takes for $629,000 to double is approximately [tex]\( 16.12 \)[/tex] years.
