Questions: A famous painting was sold in 1946 for 21,580. In 1988 the painting was sold for 30.8 million. What rate of interest compounded continuously did this investment earn? As an investment, the painting earned an interest rate of %. (Round to one decimal place as needed.)

Solution Steps

To find the rate of interest compounded continuously, we can use the formula for continuous compounding: \( A = Pe^{rt} \), where \( A \) is the amount of money accumulated after time \( t \), \( P \) is the principal amount (initial investment), \( r \) is the rate of interest, and \( t \) is the time in years. We need to solve for \( r \) given \( A = 30.8 \) million, \( P = 21,580 \), and \( t = 1988 - 1946 \).

We are given the following values:

• Initial amount \( P = 21580 \)
• Final amount \( A = 30.8 \times 10^6 = 30800000 \)
• Time period \( t = 1988 - 1946 = 42 \) years

The formula for continuous compounding is given by: \[ A = Pe^{rt} \] We need to solve for the rate \( r \).

Rearranging the formula to solve for \( r \): \[ r = \frac{\ln\left(\frac{A}{P}\right)}{t} \]

Substituting the known values into the equation: \[ r = \frac{\ln\left(\frac{30800000}{21580}\right)}{42} \]

Calculating the value of \( r \): \[ r \approx 0.1729 \]

To express \( r \) as a percentage: \[ \text{Interest Rate} = r \times 100 \approx 17.29\% \] Rounding to one decimal place gives us \( 17.3\% \).

Final Answer