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An investment of $1,200\$1,200 is made in an account with continuous compounding. After 88 years, the account balance is $2,400\$2,400. Determine the annual interest rate. Round your answer to the nearest two decimal places in percentage form. Use the formula A=PertA = Pe^{rt}, where AA is the balance (final amount), PP is the principal (starting amount), ee is the base of natural logarithms (2.71828\approx 2.71828),rr is the interest rate expressed as a decimal, and tt is the time in years.

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Q. An investment of $1,200\$1,200 is made in an account with continuous compounding. After 88 years, the account balance is $2,400\$2,400. Determine the annual interest rate. Round your answer to the nearest two decimal places in percentage form. Use the formula A=PertA = Pe^{rt}, where AA is the balance (final amount), PP is the principal (starting amount), ee is the base of natural logarithms (2.71828\approx 2.71828),rr is the interest rate expressed as a decimal, and tt is the time in years.
  1. Identify Given Values: Identify the given values: P=$1,200P = \$1,200 A=$2,400A = \$2,400 t=8 yearst = 8 \text{ years} Use the formula A=PertA = Pe^{rt}.
  2. Rearrange Formula for r: Rearrange the formula to solve for rr: A=PertA = Pe^{rt} 2400=1200imese8r2400 = 1200 imes e^{8r} Divide both sides by 12001200: 2=e8r2 = e^{8r}
  3. Take Natural Logarithm: Take the natural logarithm (ln\ln) of both sides to solve for rr: ln(2)=ln(e8r)\ln(2) = \ln(e^{8r}) ln(2)=8r\ln(2) = 8r
  4. Divide to Isolate rr: Divide both sides by 88 to isolate rr:\newliner=ln(2)8r = \frac{\ln(2)}{8}
  5. Calculate ln(2) \ln(2) : Calculate ln(2) \ln(2) using a calculator: ln(2)0.69315 \ln(2) \approx 0.69315 r=0.693158 r = \frac{0.69315}{8}
  6. Perform Division: Perform the division: r0.08664r \approx 0.08664
  7. Convert to Percentage: Convert rr to a percentage by multiplying by 100100: r0.08664×100r \approx 0.08664 \times 100 r8.66%r \approx 8.66\%

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