Substitute the given values in the formula A = P to find the missing quantity.
1. P = 750,000, r = 18% per year, t = 8 years
2. A = 1,740,000, r = 7% per year, t = 5 years
3. A = 675,000, P = 425,000, r = 4% per year

To solve these problems, we use the formula for compound interest, assuming annual compounding. The formula is:

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

Where:
- \( A \) is the amount after interest.
- \( P \) is the principal (initial amount).
- \( r \) is the interest rate per period (expressed as a decimal).
- \( t \) is the time the money is invested or borrowed for, in years.

Let's substitute the given values into the formula.

### 1. \( P = 750,000 \), \( r = 18\% \) (or 0.18), \( t = 8 \) years.
We are looking for \( A \).

\[
A = 750,000(1 + 0.18)^8
\]
\[
A = 750,000(1.18)^8
\]
\[
A \approx 750,000(3.948717)
\]
\[
A \approx 2,961,537.75
\]

So, \( A \approx 2,961,537.75 \).

### 2. \( A = 1,740,000 \), \( r = 7\% \) (or 0.07), \( t = 5 \) years.
We are looking for \( P \).

\[
1,740,000 = P(1 + 0.07)^5
\]
\[
1,740,000 = P(1.07)^5
\]
\[
1,740,000 = P(1.402552)
\]
\[
P = \frac{1,740,000}{1.402552}
\]
\[
P \approx 1,240,596.94
\]

So, \( P \approx 1,240,596.94 \).

### 3. \( A = 675,000 \), \( P = 425,000 \), \( r = 4\% \) (or 0.04).
We are looking for \( t \).

\[
675,000 = 425,000(1 + 0.04)^t
\]
\[
675,000 = 425,000(1.04)^t
\]
\[
\frac{675,000}{425,000} = (1.04)^t
\]
\[
1.588235 = (1.04)^t
\]

Now, we solve for \( t \) using logarithms:

\[
t = \frac{\log(1.588235)}{\log(1.04)}
\]
\[
t \approx \frac{0.2005}{0.017033}
\]
\[
t \approx 11.77
\]

So, \( t \approx 11.77 \) years.

---

Would you like more details or explanations?

Here are 5 follow-up questions:
1. What happens if the interest is compounded more frequently, like monthly or quarterly?
2. How does changing the interest rate affect the growth over time?
3. Can the formula be modified for continuous compounding?
4. What would happen if \( t \) were a fraction, like investing for half a year?
5. How would the solution change if the interest rate was variable?

**Tip:** When solving for \( t \), logarithms are essential to undo exponential functions. Familiarity with logarithmic properties will speed up these calculations.

Substitute the given values in the formula A = P(1 + r)^t to find the missing quantity.
1. P = 750,000, r = 18% per year, t = 8 years
2. A = 1,740,000, r = 7% per year, t = 5 years
3. A = 675,000, P = 425,000, r = 4% per year

Learn how to solve compound interest problems using the formula A = P(1 + r)^t. This includes finding the future value, principal, and time in investment scenarios. The examples include P = 750,000, r = 18%, t = 8 years, and more, using key algebraic and logarithmic concepts. Suitable for Grades 9-12.

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