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This problem involves continuous compounding interest. The formula for continuous compounding is:

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

Where:
- \( A \) is the final amount (in this case, double the principal \( P \)),
- \( P \) is the initial principal,
- \( r \) is the annual interest rate (as a decimal),
- \( t \) is the time in years,
- \( e \) is the base of the natural logarithm.

In this case, we want to find the time \( t \) it takes for the balance to double. So, \( A = 2P \), and the interest rate is given as 1.84% per year, or \( r = 0.0184 \).

The equation becomes:

\[
2P = P e^{0.0184t}
\]

Dividing both sides by \( P \):

\[
2 = e^{0.0184t}
\]

Now, take the natural logarithm (ln) of both sides to solve for \( t \):

\[
\ln(2) = 0.0184t
\]

\[
t = \frac{\ln(2)}{0.0184}
\]

I'll calculate the value of \( t \).The time it will take for the balance to double with a continuous interest rate of 1.84% per year is approximately **37.67 years**.

Would you like further clarification on any steps? Here are some related questions:

1. What is the formula for compound interest when it is compounded at regular intervals?
2. How would the doubling time change if the interest rate were higher?
3. How can the natural logarithm be used in other financial calculations?
4. What are some common applications of continuous compounding in finance?
5. How does continuous compounding differ from discrete compounding?

**Tip:** Continuous compounding grows an amount faster than regular compounding because interest is added at every possible instant, leading to a higher total over time.

If a bank pays 1.84% per year interest compounded continuously, how long does it take for the balance in an account to double?

Explore how to calculate the doubling time for an investment with a continuous compounding interest rate of 1.84%. This detailed guide uses the formula A = P e^{rt} and involves solving using natural logarithms.

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