1. (1.5 points) In a suburb, housing costs have been increasing at 5.2% per year compounded annually for the past 8 years. A house worth $260,000 now would have had what value 8 years ago?

To solve this problem, we will use the formula for compound interest in reverse, as we're looking for the value of the house 8 years ago. The compound interest formula is:

\[
A = P(1 + r)^t
\]

Where:
- $ A $ is the amount (future value) after $ t $ years,
- $ P $ is the initial amount (the value we want to find),
- $ r $ is the annual growth rate (5.2% or 0.052),
- $ t $ is the time in years (8 years).

We need to find the initial amount $ P $. Rearranging the formula:

\[
P = \frac{A}{(1 + r)^t}
\]

Now, substitute the values into the formula:

\[
P = \frac{260,000}{(1 + 0.052)^8}
\]

First, calculate $ (1 + 0.052)^8 $:

\[
(1 + 0.052)^8 = 1.052^8
\]

Now let's compute the value of $ P $.The value of the house 8 years ago was approximately **$173,319.50**.

Would you like more details or have any questions?

Here are five related questions for further practice:
1. What would be the value of a house in 5 years if it costs $260,000 now and continues to appreciate at 5.2% per year?
2. If a house appreciated at 6% per year instead of 5.2%, what would have been its value 8 years ago?
3. How long would it take for a house's value to double at an annual growth rate of 5.2%?
4. What would the house's value be 8 years from now at the same growth rate of 5.2%?
5. How would the house's initial value change if the growth rate was 4.5% instead of 5.2% over the past 8 years?

**Tip**: When dealing with compound interest, knowing how to rearrange the formula helps in both forward and reverse calculations.

In a suburb, housing costs have been increasing at 5.2% per year compounded annually for the past 8 years. A house worth $260,000 now would have had what value 8 years ago?

Learn how to calculate the value of a house 8 years ago given a current value of $260,000 and an annual appreciation rate of 5.2% compounded annually. This solution uses the compound interest formula and is suitable for high school students in Grades 9-12.

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