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How much should be invested now at an interest rate of 6%6\% per year, compounded continuously, to have $2000\$2000 in five years?\newlineDo not round any intermediate computations, and round your answer to the nearest cent.\newlines‾1\text{s}\underline{\hspace{1em}}1

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Q. How much should be invested now at an interest rate of 6%6\% per year, compounded continuously, to have $2000\$2000 in five years?\newlineDo not round any intermediate computations, and round your answer to the nearest cent.\newlines‾1\text{s}\underline{\hspace{1em}}1
  1. Understand the Formula: To solve this problem, we will use the formula for continuous compounding, which is A=PertA = Pe^{rt}, where AA is the amount of money accumulated after nn years, including interest, PP is the principal amount (the initial amount of money), rr is the annual interest rate (decimal), tt is the time the money is invested for, and ee is the base of the natural logarithm. We are given A=$2000A = \$2000, r=6%r = 6\% or 0.060.06, and AA00 years. We need to solve for PP.
  2. Rearrange the Formula: First, we need to rearrange the formula to solve for PP. The rearranged formula is P=AertP = \frac{A}{e^{rt}}.
  3. Plug in Values: Now we can plug in the values we have into the rearranged formula. So we get P=2000e0.06×5P = \frac{2000}{e^{0.06 \times 5}}.
  4. Calculate Exponent: Calculating the exponent part, we have e(0.06×5)=e0.3e^{(0.06\times 5)} = e^{0.3}.
  5. Find Value: Using a calculator to find the value of e0.3e^{0.3}, we get approximately 1.34985880757600321.3498588075760032.
  6. Divide Amount: Now we divide $2000\$2000 by this value to find PP. So P=20001.3498588075760032P = \frac{2000}{1.3498588075760032}.
  7. Perform Division: Performing the division, we get P≈1481.4722584856353P \approx 1481.4722584856353.
  8. Round to Nearest Cent: Rounding to the nearest cent, we get P≈$1481.47P \approx \$1481.47.

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