Cameron invested $690 in an account paying an interest rate of 9, start fraction, 5, divided by, 8, end fraction9 
8
5
​
 % compounded monthly. Alonso invested $690 in an account paying an interest rate of 10, one eigth10 
8
1
​
 % compounded continuously. To the nearest dollar, how much money would Alonso have in his account when Cameron's money has tripled in value?

To determine how much money Alonso will have in his account when Cameron's investment triples, we need to calculate the amount of time it takes for Cameron's investment to triple, and then use that same amount of time to calculate Alonso's investment value.

### Step 1: Calculate the time it takes for Cameron's investment to triple
Cameron's investment is compounded monthly, and the formula for compound interest is:

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

Where:
- $A$ is the final amount,
- $P$ is the initial principal (in this case, $P = 690$),
- $r$ is the annual interest rate (expressed as a decimal),
- $n$ is the number of times the interest is compounded per year (monthly, so $n = 12$),
- $t$ is the time in years.

We know Cameron's investment will triple, so $A = 3P = 3 \times 690 = 2070$. The interest rate is given as $9 \frac{5}{8} \%$, which is $r = 9.625\%$ or $r = 0.09625$.

Using the formula:

\[
2070 = 690 \left(1 + \frac{0.09625}{12}\right)^{12t}
\]

Simplifying:

\[
2070 = 690 \left(1 + 0.00802083\right)^{12t}
\]

\[
2070 = 690 \left(1.00802083\right)^{12t}
\]

Divide both sides by 690:

\[
3 = \left(1.00802083\right)^{12t}
\]

Now take the natural logarithm (ln) of both sides:

\[
\ln(3) = 12t \cdot \ln(1.00802083)
\]

\[
\ln(3) \approx 1.0986, \quad \ln(1.00802083) \approx 0.007987
\]

\[
1.0986 = 12t \cdot 0.007987
\]

Now solve for $t$:

\[
t = \frac{1.0986}{12 \times 0.007987} \approx \frac{1.0986}{0.095844} \approx 11.46 \text{ years}
\]

So it takes about 11.46 years for Cameron's investment to triple.

### Step 2: Calculate Alonso's investment after 11.46 years
Alonso's investment is compounded continuously. The formula for continuously compounded interest is:

\[
A = P e^{rt}
\]

Where:
- $A$ is the final amount,
- $P = 690$,
- $r = 10 \frac{1}{8}\%$ or $r = 0.10125$,
- $t = 11.46$ years,
- $e$ is Euler's number (approximately 2.718).

Using the formula:

\[
A = 690 e^{0.10125 \times 11.46}
\]

First, calculate the exponent:

\[
0.10125 \times 11.46 \approx 1.1609
\]

Now calculate $e^{1.1609}$:

\[
e^{1.1609} \approx 3.192
\]

Now calculate Alonso's final amount:

\[
A = 690 \times 3.192 \approx 2202.48
\]

To the nearest dollar, Alonso will have **$2202** in his account when Cameron's money has tripled.

---

Do you want more details or have any questions?

Here are 5 related questions:
1. What if Cameron’s interest was compounded quarterly instead of monthly?
2. How much longer would it take Cameron's investment to quadruple?
3. What would happen if Alonso's account was compounded monthly instead of continuously?
4. How does the difference in compounding frequency affect long-term investment returns?
5. How would a higher interest rate affect Alonso's total in the same timeframe?

**Tip**: Continuous compounding generally leads to a higher return compared to monthly or yearly compounding, as interest is applied constantly over time.

Cameron invested $690 in an account paying an interest rate of 9 5/8% compounded monthly. Alonso invested $690 in an account paying an interest rate of 10 1/8% compounded continuously. To the nearest dollar, how much money would Alonso have in his account when Cameron's money has tripled in value?

This math problem explores the concepts of compound interest and continuous compounding. It involves calculating how long it takes for Cameron's $690 investment to triple with 9 5/8% monthly compounded interest, and determining the value of Alonso's investment, compounded continuously at 10 1/8%, after the same time period. Perfect for students in Grades 11-12.

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