Questions: Find the time required for an investment of 8600 dollars to grow to 11000 dollars at an interest rate of 4.5% per year, compounded continuously. Round to the nearest hundredth of year. t= years.

Find the time required for an investment of 8600 dollars to grow to 11000 dollars at an interest rate of 4.5% per year, compounded continuously. Round to the nearest hundredth of year.
t= years.
Transcript text: Find the time required for an investment of 8600 dollars to grow to 11000 dollars at an interest rate of $4.5 \%$ per year, compounded continuously. Round to the nearest hundredth of year. $t=$ $\square$ years.
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Solution

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Solution Steps

To find the time required for an investment to grow to a certain amount with continuous compounding, we use the formula for continuous compound interest: \( A = Pe^{rt} \), where \( A \) is the amount of money accumulated after time \( t \), \( P \) is the principal amount (initial investment), \( r \) is the annual interest rate (as a decimal), and \( t \) is the time in years. We need to solve for \( t \).

  1. Substitute the given values into the formula: \( A = 11000 \), \( P = 8600 \), and \( r = 0.045 \).
  2. Rearrange the formula to solve for \( t \): \( t = \frac{\ln(A/P)}{r} \).
  3. Calculate \( t \) using Python.
Step 1: Identify the Variables

We are given the following values:

  • Principal amount \( P = 8600 \)
  • Future amount \( A = 11000 \)
  • Annual interest rate \( r = 0.045 \)
Step 2: Use the Continuous Compounding Formula

The formula for continuous compounding is given by: \[ A = Pe^{rt} \] To find the time \( t \), we rearrange the formula: \[ t = \frac{\ln\left(\frac{A}{P}\right)}{r} \]

Step 3: Substitute the Values

Substituting the known values into the rearranged formula: \[ t = \frac{\ln\left(\frac{11000}{8600}\right)}{0.045} \]

Step 4: Calculate \( t \)

Calculating the value: \[ t \approx 5.469623767531301 \] Rounding to the nearest hundredth: \[ t \approx 5.47 \]

Final Answer

The time required for the investment to grow is approximately \\(\boxed{5.47}\\) years.

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