Calculate the interest rate, compounded monthly, if an investment of R40 000 is to double in 6 years.

To calculate the interest rate needed for an investment to double in a given time, we use the formula for compound interest:

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- \( A \) is the final amount (R80,000, since the investment doubles).
- \( P \) is the initial principal (R40,000).
- \( r \) is the annual interest rate (which we want to find).
- \( n \) is the number of times the interest is compounded per year (12 times for monthly compounding).
- \( t \) is the time in years (6 years).

Given that the investment doubles, \( A = 2P \), so:

\[
2P = P \left(1 + \frac{r}{12}\right)^{12 \times 6}
\]

Dividing both sides by \( P \) (which is R40,000):

\[
2 = \left(1 + \frac{r}{12}\right)^{72}
\]

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

\[
\ln(2) = \ln\left(\left(1 + \frac{r}{12}\right)^{72}\right)
\]

Using the logarithm power rule \( \ln(a^b) = b \cdot \ln(a) \):

\[
\ln(2) = 72 \cdot \ln\left(1 + \frac{r}{12}\right)
\]

Now, solve for \( \ln\left(1 + \frac{r}{12}\right) \):

\[
\ln\left(1 + \frac{r}{12}\right) = \frac{\ln(2)}{72}
\]

Now, exponentiate both sides to get rid of the logarithm:

\[
1 + \frac{r}{12} = e^{\frac{\ln(2)}{72}}
\]

Subtract 1 from both sides:

\[
\frac{r}{12} = e^{\frac{\ln(2)}{72}} - 1
\]

Finally, solve for \( r \):

\[
r = 12 \times \left(e^{\frac{\ln(2)}{72}} - 1\right)
\]

Let's compute this value:

\[
r \approx 12 \times \left(e^{\frac{\ln(2)}{72}} - 1\right)
\]

Now, I'll calculate this value.

Learn how to calculate the interest rate needed for an investment of R40,000 to double in 6 years with monthly compounding. Follow a step-by-step solution using compound interest formulas and logarithmic functions.

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