Chase invested $39,000 in an account paying an interest rate of 3.2% compounded daily. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach $66,100?

The formula to calculate compound interest is:

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

Where:
- $ A $ is the final amount ($66,100)
- $ P $ is the principal or initial investment ($39,000)
- $ r $ is the annual interest rate (3.2% or 0.032)
- $ n $ is the number of times the interest is compounded per year (daily compounding means $ n = 365 $)
- $ t $ is the time in years, which we are solving for.

Rearranging the formula to solve for $ t $:

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

Now let's calculate $ t $.It will take approximately **16.5 years** for Chase's investment to grow from $39,000 to $66,100, assuming a 3.2% interest rate compounded daily.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How does increasing the interest rate affect the time needed to reach the target amount?
2. What would happen if the interest were compounded monthly instead of daily?
3. How long would it take for Chase's account to double in value at the same interest rate?
4. What is the difference between simple interest and compound interest in this context?
5. How much interest is earned after 10 years?

**Tip:** The more frequently interest is compounded, the faster the investment grows due to interest being calculated on previously earned interest.

Find out how long it will take for a $39,000 investment to grow to $66,100 at an interest rate of 3.2% compounded daily. This problem involves the compound interest formula, logarithms, and exponential growth, suitable for high school students in Grades 10-12.

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