Solve the following investment equation for the number of compounding periods, n. Round your answer to the nearest whole number.
350 = 300(1.04)n

To solve the equation for the number of compounding periods \( n \), follow these steps:

Given equation:
\[
350 = 300(1.04)^n
\]

1. **Divide both sides by 300** to isolate the exponential term:
\[
\frac{350}{300} = (1.04)^n
\]

\[
\frac{7}{6} \approx 1.1667 = (1.04)^n
\]

2. **Take the natural logarithm (ln) of both sides** to solve for \( n \):
\[
\ln\left(\frac{7}{6}\right) = \ln\left((1.04)^n\right)
\]

\[
\ln\left(\frac{7}{6}\right) = n \cdot \ln(1.04)
\]

3. **Solve for \( n \)** by dividing both sides by \( \ln(1.04) \):
\[
n = \frac{\ln\left(\frac{7}{6}\right)}{\ln(1.04)}
\]

4. **Calculate the value**:
\[
n \approx \frac{\ln(1.1667)}{\ln(1.04)} \approx \frac{0.15415}{0.03922} \approx 3.93
\]

Rounding to the nearest whole number, \( n \approx 4 \).

So, the number of compounding periods \( n \) is approximately **4**.

Would you like a detailed explanation, or do you have any questions?

Here are 5 related questions for further practice:

1. If the final amount is 500, and the principal is 300 with a 5% interest rate, how many periods will it take?
2. How does changing the interest rate affect the number of compounding periods needed to reach a target amount?
3. Solve for \( n \) in the equation \( 400 = 250(1.03)^n \).
4. What happens to the value of \( n \) if the principal and the final amount are the same?
5. How would the equation change if the interest were compounded continuously?

**Tip:** When solving for variables in exponents, logarithms are the key tool to simplify the equation.

Learn how to solve the exponential growth equation 350 = 300(1.04)^n for the number of compounding periods, n. Follow a step-by-step solution with detailed explanations and rounding to the nearest whole number. Suitable for Grades 11-12.

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