Jason invests $2000 in PGS Bank and the amount he has in his account grows at a continuous annual rate of 1.5%. On the same day, Maria invests $7000 in Cash Bank and the amount of money in her account grows at an annual rate of 1.3%. In how many years will the two bank accounts have the same amount of money?

To determine in how many years Jason's and Maria's bank accounts will have the same amount of money, we can use the formula for continuous growth, which is:

\[
A(t) = P \cdot e^{rt}
\]

Where:
- $ A(t) $ is the amount of money after $ t $ years,
- $ P $ is the initial amount invested,
- $ r $ is the annual growth rate (as a decimal),
- $ t $ is the number of years,
- $ e $ is the base of the natural logarithm (approximately 2.718).

### Jason's account:
- Initial amount: $ 2000 $
- Growth rate: $ 1.5\% = 0.015 $

The formula for Jason's account after $ t $ years is:
\[
A_{\text{Jason}}(t) = 2000 \cdot e^{0.015t}
\]

### Maria's account:
- Initial amount: $ 7000 $
- Growth rate: $ 1.3\% = 0.013 $

The formula for Maria's account after $ t $ years is:
\[
A_{\text{Maria}}(t) = 7000 \cdot e^{0.013t}
\]

### Setting the two amounts equal to find $ t $:

\[
2000 \cdot e^{0.015t} = 7000 \cdot e^{0.013t}
\]

Divide both sides by $ e^{0.013t} $:
\[
2000 \cdot e^{0.002t} = 7000
\]

Next, divide both sides by 2000:
\[
e^{0.002t} = \frac{7000}{2000} = 3.5
\]

Now, take the natural logarithm (ln) of both sides:
\[
\ln(e^{0.002t}) = \ln(3.5)
\]

Using the property $ \ln(e^x) = x $, we get:
\[
0.002t = \ln(3.5)
\]

Solve for $ t $:
\[
t = \frac{\ln(3.5)}{0.002}
\]

Let me calculate the value of $ t $.The number of years it will take for Jason's and Maria's bank accounts to have the same amount of money is approximately 626.38 years.

This means that it would take over six centuries for the accounts to balance at their respective growth rates!

Do you need any further details or have any questions about the process?

Here are 5 related questions for practice:
1. How much money will Jason have after 10 years?
2. How much will Maria's account grow after 15 years?
3. If Maria's interest rate were increased to 1.5%, how long would it take for their accounts to match?
4. What would happen if Jason invested $5000 instead of $2000 with the same growth rate?
5. How does the formula for continuous compound interest differ from simple interest?

**Tip:** Continuous compound interest grows faster than annual compound interest because the interest is compounded constantly.

Learn how to solve a real-world problem involving continuous compound interest. Jason invests $2000 at 1.5% and Maria invests $7000 at 1.3%. Discover how logarithms and exponential growth formulas help determine when their accounts will have the same amount. Suitable for Grades 10-12.

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