Question

What constant interest rate is required if an initial deposit placed into an account accrues interest compounded continuously is to double its value in six years?

• A. $\frac{1}{6}ln4$
• B. $\frac{1}{6}ln2$
• C. $\frac{1}{6}ln3$
• D. $\frac{1}{2}ln2$

Answer 1

Answer: (B)

$\frac{1}{6}ln2$

Answer 2

Concept: The formula for continuous compounding is $A = Pe^{rt}$, where:

• $A$ is the final amount
• $P$ is the principal (initial deposit)
• $r$ is the annual interest rate
• $t$ is the time in years.

Given that the initial deposit is to double its value, we have $A = 2P$. The time given is $t = 6$ years.

Substitute these values into the continuous compounding formula:

$2P = Pe^{r \times 6}$

Divide both sides by $P$:

$2 = e^{6r}$

To solve for $r$, take the natural logarithm ($\ln$) of both sides:

$\ln(2) = \ln(e^{6r})$

Using the logarithm property $\ln(e^x) = x$:

$\ln(2) = 6r$

Now, solve for $r$:

$r = \frac{\ln(2)}{6}$

This can also be written as $r = \frac{1}{6}\ln2$.