We can solve this problem using the compound interest formula:

$A = P \left(1 + r\right)^t$

Where:

• $A$ is the future value of the investment ($11,100).
• $P$ is the initial principal ($5,500).
• $r$ is the annual interest rate (5.25% or 0.0525 as a decimal).
• $t$ is the time in years.

Plugging in the values:

$11100 = 5500 \left(1 + 0.0525\right)^t$

First, divide both sides by 5500 to isolate the growth factor:

$\frac{11100}{5500} = \left(1.0525\right)^t$

$2.01818 = \left(1.0525\right)^t$

Next, take the natural logarithm of both sides to solve for $t$:

$\ln(2.01818) = t \cdot \ln(1.0525)$

$t = \frac{\ln(2.01818)}{\ln(1.0525)}$

Now, calculate $t$:

$t \approx \frac{0.70113}{0.05117} \approx 13.7$

So, it will take approximately 14 years for the account value to reach $11,100.

Would you like more details or have any questions?

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Here are 8 related questions:

1. How would the time change if the interest rate were increased to 6%?
2. What if the principal amount were $7,000 instead of $5,500?
3. How long would it take to triple the initial investment at 5.25% interest?
4. What would be the final amount after 14 years if the principal was $5,500 but the interest rate was 4%?
5. How does the compounding frequency (monthly vs. annually) affect the time to double the investment?
6. What would be the time to reach $15,000 with the same initial investment and interest rate?
7. If you need the money in 10 years, what interest rate would be required to reach $11,100?
8. How does the concept of present value relate to this calculation?

Tip: Keep in mind that small differences in interest rates can significantly affect the time required for an investment to grow, especially over longer periods.