Determine how much time is required for an investment to double in value if interest is earned at the rate of 6.75​% compounded annually.
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Part 1
The investment will double in value in how many years

To determine the time required for an investment to double in value with an interest rate of 6.75% compounded annually, we can use the formula for compound interest:

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

Where:
- \( A \) is the amount of money accumulated after \( n \) years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time the money is invested for, in years.

Since the problem states that the interest is compounded annually, \( n = 1 \). Also, for the investment to double, \( A = 2P \).

Substituting these values into the formula:

\[
2P = P \left(1 + \frac{r}{1}\right)^{1 \cdot t}
\]

Simplifying:

\[
2 = \left(1 + r\right)^t
\]

Taking the natural logarithm (ln) of both sides:

\[
\ln(2) = t \cdot \ln(1 + r)
\]

Now, solve for \( t \):

\[
t = \frac{\ln(2)}{\ln(1 + r)}
\]

Given \( r = 6.75\% = 0.0675 \):

\[
t = \frac{\ln(2)}{\ln(1 + 0.0675)}
\]

Let's calculate this:

\[
t \approx \frac{0.6931}{\ln(1.0675)}
\]
\[
t \approx \frac{0.6931}{0.0654}
\]
\[
t \approx 10.6 \text{ years}
\]

Thus, the investment will double in value in approximately **10.6 years**.

Would you like more details or have any questions?

Here are some related questions you might find interesting:
1. What is the rule of 72, and how can it be applied here?
2. How would the doubling time change if the interest rate was 5% instead of 6.75%?
3. What if the interest was compounded semi-annually instead of annually?
4. How does the doubling time change if continuous compounding is used?
5. What is the relationship between the interest rate and the doubling time?

**Tip:** The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \) (where \( e \) is approximately 2.718). It is commonly used in calculations involving growth and decay, such as in compound interest.

Learn how to calculate the time required for an investment to double in value with a 6.75% annual interest rate using compound interest formulas and natural logarithms. Get a detailed solution with step-by-step calculations.

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