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Suppose that $10,000 is invested and at the end of 5 years, the value of the account is $13,771.28. Use the model A=Pe^(rt) to determine the average rate of return r under continuous compounding.

Suppose that $10,000 \$ 10,000 is invested and at the end of 55 years, the value of the account is $13,771.28 \$ 13,771.28 . Use the model A=Pert A=P e^{r t} to determine the average rate of return r r under continuous compounding.

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Full solution

Q. Suppose that $10,000 \$ 10,000 is invested and at the end of 55 years, the value of the account is $13,771.28 \$ 13,771.28 . Use the model A=Pert A=P e^{r t} to determine the average rate of return r r under continuous compounding.
  1. Isolate erte^{rt}: We have the formula for continuous compounding: A=PertA = Pe^{rt}. Here, A=$13,771.28A = \$13,771.28, P=$10,000P = \$10,000, and t=5t = 5 years. We need to find rr.
  2. Calculate ln⁡(1.377128)\ln(1.377128): First, divide both sides of the equation by PP to isolate erte^{rt}. So, ert=AP=$(13,771.28)$(10,000)=1.377128e^{rt} = \frac{A}{P} = \frac{\$(13,771.28)}{\$(10,000)} = 1.377128.
  3. Simplify to rtrt: Now, take the natural logarithm (ln⁡\ln) of both sides to get rid of the exponent.ln⁡(ert)=ln⁡(1.377128)\ln(e^{rt}) = \ln(1.377128).
  4. Solve for rr: Since ln⁡(e(rt))\ln(e^{(rt)}) simplifies to rtrt, we have rt=ln⁡(1.377128)rt = \ln(1.377128).\newlineCalculate ln⁡(1.377128)\ln(1.377128) using a calculator.\newlinert=ln⁡(1.377128)≈0.319rt = \ln(1.377128) \approx 0.319.
  5. Calculate rr: Finally, divide both sides by tt to solve for rr.r=rtt=0.3195r = \frac{rt}{t} = \frac{0.319}{5}.
  6. Calculate r: Finally, divide both sides by t to solve for r.\newliner=rtt=0.3195r = \frac{rt}{t} = \frac{0.319}{5}.Calculate r.\newliner≈0.3195≈0.0638r \approx \frac{0.319}{5} \approx 0.0638 or 6.38%6.38\%.

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