A= 4000 P=2000 r=0.035 t=? Solve this in compounded continuously 

The formula for continuously compounded interest is given by:

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

Where:
- \( A \) is the final amount (4000 in this case),
- \( P \) is the initial principal (2000 here),
- \( r \) is the annual interest rate (0.035 here),
- \( t \) is the time in years (which we are solving for),
- \( e \) is the mathematical constant (approximately 2.71828).

### Step-by-step solution:
1. Substitute the values into the formula:

\[
4000 = 2000 \cdot e^{0.035t}
\]

2. Divide both sides by 2000 to isolate the exponential term:

\[
2 = e^{0.035t}
\]

3. Take the natural logarithm (ln) of both sides:

\[
\ln(2) = 0.035t
\]

4. Solve for \( t \):

\[
t = \frac{\ln(2)}{0.035}
\]

Let me calculate the value of \( t \).The time \( t \) is approximately 19.80 years.

Would you like further details on this calculation or have any other questions?

Here are 5 related questions you might find helpful:
1. What is the difference between simple and compound interest?
2. How does continuous compounding differ from standard compounding?
3. What would happen to the time if the interest rate increased to 5%?
4. Can the formula for continuous compounding be applied to non-monetary contexts?
5. How does the value of \( e \) affect continuous compounding?

**Tip:** Continuous compounding allows interest to be added to the principal an infinite number of times per year, which results in slightly more interest than traditional compounding.

A= 4000 P=2000 r=0.035 t=? Solve this in compounded continuously

Solve a compound interest problem where A=4000, P=2000, r=0.035, and t is unknown. Learn how to apply the continuous compounding formula A = P * e^(rt) and use logarithms to calculate the time needed to double the investment. Suitable for high school students in Grades 10-12.

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