DETERMINE THE NUMBER OF YEARS IT WILL TAKE FOR 25,OOO TO GROW TO 29,000 AT AN ANNUAL INTEREST OF 8%, USING ECXACT TIME IN THE YEAR 2019.

To determine the number of years it will take for an investment of $25,000 to grow to $29,000 at an annual interest rate of 8%, we'll assume the interest is compounded annually. The formula for compound interest is:

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

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 in years.

Given:
- $ A = 29,000 $
- $ P = 25,000 $
- $ r = 0.08 $ (8% as a decimal)
- $ n = 1 $ (since it’s compounded annually)

We need to find $ t $, the number of years.

The formula simplifies to:

\[
29,000 = 25,000 \times \left(1 + 0.08\right)^t
\]

\[
\frac{29,000}{25,000} = \left(1.08\right)^t
\]

\[
1.16 = (1.08)^t
\]

To solve for $ t $, take the natural logarithm (ln) of both sides:

\[
\ln(1.16) = t \cdot \ln(1.08)
\]

\[
t = \frac{\ln(1.16)}{\ln(1.08)}
\]

Now, let's calculate the value of $ t $.

It will take approximately 1.93 years for $25,000 to grow to $29,000 at an annual interest rate of 8%.

Would you like more details, or do you have any other questions?

Here are 5 related questions:
1. How does the time change if the interest rate increases to 10%?
2. What if the interest is compounded quarterly instead of annually?
3. How long would it take for $25,000 to double at an 8% annual interest rate?
4. What would the final amount be after 5 years at an 8% interest rate?
5. How does continuous compounding affect the time required to reach $29,000?

**Tip:** Remember that compound interest grows faster over time compared to simple interest due to the effect of earning interest on previously accumulated interest.

Learn how long it takes for $25,000 to grow to $29,000 at an annual interest rate of 8% using compound interest calculations. Follow a step-by-step solution with formulas and calculations. Suitable for advanced high school students.

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