An investment of $550 compounded daily at 7.3% per year to produce a final amount of $675. (Answer to the nearest day.)

We can solve this using the formula for compound interest:

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

Where:
- $ A $ is the final amount ($675),
- $ P $ is the principal amount ($550),
- $ r $ is the annual interest rate (7.3% or 0.073),
- $ n $ is the number of times the interest is compounded per year (daily means $ n = 365 $),
- $ t $ is the time in years.

We are solving for $ t $, the time in years. Rearranging the formula to isolate $ t $:

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

Let’s calculate $ t $ and then convert it to days.The investment will take approximately 1,024 days to grow from $550 to $675, compounded daily at a rate of 7.3%.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How would the result change if the interest were compounded monthly instead of daily?
2. What if the interest rate was 6.5% instead of 7.3%?
3. How would you calculate the final amount after a different period of time, say 800 days?
4. What would happen if the principal amount was increased to $1,000?
5. How does the formula change if the interest is compounded continuously?

**Tip:** Always keep track of how frequently interest is compounded—it significantly affects the final amount.

Solve how long it takes for $550 to grow to $675 with daily compounding at 7.3% per year using compound interest formulas and logarithms. Perfect for high school students learning about exponential growth and interest calculations.

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