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8.) A house purchased 5 years ago for $100,000 was just sold for $135,000. Assuming exponentia growth, approximate the annual growth rate, to the nearest percent. Work/Explana

88.) A house purchased 55 years ago for $100,000 \$ 100,000 was just sold for $135,000 \$ 135,000 . Assuming exponentia growth, approximate the annual growth rate, to the nearest percent.\newlineWork/Explana

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Q. 88.) A house purchased 55 years ago for $100,000 \$ 100,000 was just sold for $135,000 \$ 135,000 . Assuming exponentia growth, approximate the annual growth rate, to the nearest percent.\newlineWork/Explana
  1. Identify values: Identify the initial value PP, final value AA, and the number of years tt for the exponential growth formula A=P(1+r)tA = P(1 + r)^t.
    Initial value PP = $100,000\$100,000
    Final value AA = $135,000\$135,000
    Number of years tt = 55
  2. Write formula and calculate: Write down the exponential growth formula and plug in the known values.\newlineA=P(1+r)tA = P(1 + r)^t\newline$135,000=$100,000(1+r)5\$135,000 = \$100,000(1 + r)^5
  3. Isolate growth factor: Divide both sides of the equation by the initial value to isolate the growth factor on one side.\newline($135,000/$100,000)=(1+r)5(\$135,000 / \$100,000) = (1 + r)^5\newline1.35=(1+r)51.35 = (1 + r)^5
  4. Solve for (1+r)(1 + r): Take the fifth root of both sides to solve for (1+r)(1 + r).(1+r)=(1.35)15(1 + r) = (1.35)^{\frac{1}{5}}
  5. Calculate fifth root: Calculate the fifth root of 1.351.35 to find (1+r)(1 + r). \newline(1+r)≈1.351/5≈1.062(1 + r) \approx 1.35^{1/5} \approx 1.062
  6. Find growth rate: Subtract 11 from (1+r)(1 + r) to find the growth rate (r)(r). \newliner≈1.062−1r \approx 1.062 - 1\newliner≈0.062r \approx 0.062
  7. Convert to percentage: Convert the decimal growth rate to a percentage and round to the nearest percent.\newliner≈0.062×100%r \approx 0.062 \times 100\%\newliner≈6.2%r \approx 6.2\%\newlineRound to the nearest percent: r≈6%r \approx 6\%

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