To solve this problem, we will use the compound interest formula:

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

Where:

• $A$ is the amount of money accumulated after interest (in this case, $14,200),
• $P$ is the principal amount (initial investment, in this case, $9,000),
• $r$ is the annual interest rate (4.25% or 0.0425),
• $n$ is the number of times the interest is compounded per year (semi-annually means $n = 2$),
• $t$ is the time in years, which we need to find.

Step-by-step solution:

1. Substitute the given values into the formula: $14,200 = 9,000 \left(1 + \frac{0.0425}{2}\right)^{2t}$
2. Simplify the equation: $14,200 = 9,000 \left(1 + 0.02125\right)^{2t}$ $14,200 = 9,000 \left(1.02125\right)^{2t}$
3. Divide both sides by 9,000: $\frac{14,200}{9,000} = \left(1.02125\right)^{2t}$ $1.5778 \approx \left(1.02125\right)^{2t}$
4. Take the natural logarithm (ln) of both sides: $\ln(1.5778) = \ln\left(1.02125^{2t}\right)$
5. Simplify using the logarithmic identity $\ln(a^b) = b \ln(a)$: $\ln(1.5778) = 2t \ln(1.02125)$
6. Solve for $t$: $t = \frac{\ln(1.5778)}{2 \ln(1.02125)}$

Now, I'll compute the value of $t$. math

Given values

A = 14200 P = 9000 r = 0.0425 n = 2

Calculating t using the formula derived

left_side = math.log(A / P) right_side = 2 * math.log(1 + r / n)

Solving for t

t = left_side / right_side tThe time required for the investment to grow to $14,200 is approximately 10.8 years, to the nearest tenth of a year.

Would you like any further details or clarification on this? Here are some related questions that might help deepen your understanding:

1. How does the compound interest formula change when interest is compounded annually, quarterly, or monthly?
2. What is the difference between simple interest and compound interest in terms of investment growth?
3. How does the principal amount $P$ influence the time it takes to reach a specific amount of money with compound interest?
4. How does the interest rate $r$ impact the time required for the investment to grow?
5. How do you calculate compound interest for multiple periods of time, like every month or quarter?

Tip: For quicker approximations, remember that compound interest increases faster as the compounding frequency (n) increases.