Suppose that $$ \$ 12,000 $$ is invested at an interest rate of $$ 5.7 \% $$ per year, compounded continuously. What is the doubling time? A. 13.2 yr B. 2 yr C. 11.2 yr D. 12.2 yr

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To find the doubling time for an investment compounded continuously, we can use the formula for continuous compounding:

$$
A = Pe^{rt}
$$

where:
- $$ A $$ is the amount of money accumulated after time $$ t $$,
- $$ P $$ is the principal amount (the initial amount of money),
- $$ r $$ is the annual interest rate (as a decimal),
- $$ t $$ is the time in years,
- $$ e $$ is the base of the natural logarithm (approximately equal to 2.71828).

In this case, we want to find the time $$ t $$ when the amount $$ A $$ is double the principal $$ P $$. Thus, we set $$ A = 2P $$.

Substituting into the formula gives:

$$
2P = Pe^{rt}
$$

We can divide both sides by $$ P $$ (assuming $$ P \neq 0 $$):

$$
2 = e^{rt}
$$

Taking the natural logarithm of both sides:

$$
\ln(2) = rt
$$

Now, we can solve for $$ t $$:

$$
t = \frac{\ln(2)}{r}
$$

Given that the interest rate $$ r = 5.7\% = 0.057 $$, we can substitute this value into the equation:

$$
t = \frac{\ln(2)}{0.057}
$$

Calculating $$ \ln(2) $$:

$$
\ln(2) \approx 0.693147
$$

Now substituting this value into the equation for $$ t $$:

$$
t \approx \frac{0.693147}{0.057} \approx 12.16
$$

Rounding this to one decimal place gives approximately $$ 12.2 $$ years.

Thus, the doubling time is:

$$
\boxed{12.2 \text{ yr}}
$$
The doubling time is approximately 12.2 years.

Suppose that \( \$ 12,000 \) is invested at an interest rate of \( 5.7 \% \) per year, compounded continuously. What is the doubling time? A. 13.2 yr B. 2 yr C. 11.2 yr D. 12.2 yr

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