Question: A painting sold for $220 in 1976 and was sold again in 1986 for $466. Assume that the growth in the…

A painting sold for $220 in 1976 and was sold again in 1986 for $466. Assume that the growth in the value V of the collector’s item was exponential.

a) Find the value k of the exponential growth rate. Assume V_0 = 220. k = 0.075 (Round to the nearest thousandth.) b) Find the exponential growth function in terms of t, where t is the number of years since 1976. V(t) = 220 * e^(0.075t) c) Estimate the value of the painting in 2005. $ (Round to the nearest dollar.)

Solution

a) To find the value \( k \) of the exponential growth rate, we use the formula for exponential growth: \[ V(t) = V_0 \cdot e^{kt} \] We know \( V_0 = 220 \), \( V(t) = 466 \), and \( t = 10 \) years (from 1976 to 1986). Plug these values into the equation: \[ 466 = 220 \cdot e^{10k} \] Divide both sides by 220: \[ e^{10k} = \frac{466}{220} \] Take the natural logarithm on both sides: \[ 10k = \ln \left(\frac{466}{220}\right) \] Solve for \( k \): \[ k = \frac{\ln \left(\frac{466}{220}\right)}{10} \approx 0.075 \] b) The exponential growth function in terms of \( t \) is: \[ V(t) = 220 \cdot e^{0.075t} \] c) To estimate the value of the painting in 2005, calculate \( V(29) \) since 2005 is 29 years after 1976: \[ V(29) = 220 \cdot e^{0.075 \times 29} \] Calculate the exponent: \[ V(29) \approx 220 \cdot e^{2.175} \] Calculate the value: \[ V(29) \approx 220 \cdot 8.814 \] \[ V(29) \approx 1939 \] Therefore, the estimated value of the painting in 2005 is \$1939.