Math Problem Statement

Determine how long it takes for 40,000 dollars to grow to 280,000 dollars at 7% compounded continuously. Round your answer to the nearest integer.

Solution

To solve this problem, we can use the formula for continuous compounding:

A=PertA = P e^{rt}

where:

  • AA is the final amount (280,000 dollars),
  • PP is the initial principal (40,000 dollars),
  • rr is the interest rate (0.07 or 7%),
  • tt is the time in years (the unknown we need to solve for).

Rearrange the formula to solve for tt:

t=ln(AP)rt = \frac{\ln \left(\frac{A}{P}\right)}{r}

Step-by-Step Solution:

  1. Substitute the values into the equation: t=ln(280,00040,000)0.07t = \frac{\ln \left(\frac{280,000}{40,000}\right)}{0.07}

  2. Calculate inside the logarithm: 280,00040,000=7\frac{280,000}{40,000} = 7

  3. Take the natural logarithm: ln(7)1.9459\ln(7) \approx 1.9459

  4. Divide by the interest rate: t=1.94590.0727.8t = \frac{1.9459}{0.07} \approx 27.8

  5. Round to the nearest integer: t28 yearst \approx 28 \text{ years}

Answer:

It will take approximately 28 years for 40,000 dollars to grow to 280,000 dollars at 7% interest compounded continuously.

Would you like more details, or do you have any questions?


Related Questions

  1. How would the time change if the interest rate was 5% instead of 7%?
  2. What is the formula for calculating compound interest if compounding is done annually instead of continuously?
  3. How does continuous compounding differ from periodic compounding?
  4. If the final amount were 500,000 dollars, how much time would it take with the same rate?
  5. What would be the value of 40,000 dollars after 10 years at 7% compounded continuously?

Tip:

For continuous compounding problems, remember that the exponential function grows faster than periodic compounding due to its constant growth rate effect.

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Continuous Compounding

Formulas

A = Pe^(rt)
t = ln(A / P) / r

Theorems

Properties of Natural Logarithms

Suitable Grade Level

Grades 11-12