Question: A painting sold for $272 in 1976 and was sold again in 1990 for $483. Assume that the growth in the…

A painting sold for $272 in 1976 and was sold again in 1990 for $483. 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 = 272. k = 0.041 (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) = 272 * e^(0.041t) c) Estimate the value of the painting in 2011. $ [ ] (Round to the nearest dollar.)

Solution

A painting sold for $272 in 1976 and was sold again in 1990 for $483. The growth in the value \( V \) of the collector’s item is assumed to be exponential. a) Find the value \( k \) of the exponential growth rate. Assume \( V_0 = 272 \). The formula for exponential growth is: \[ V = V_0 \cdot e^{kt} \] Given: - \( V = 483 \) - \( V_0 = 272 \) - \( t = 1990 - 1976 = 14 \) Substitute the known values into the formula: \[ 483 = 272 \cdot e^{14k} \] Divide both sides by 272: \[ \frac{483}{272} = e^{14k} \] Take the natural logarithm of both sides: \[ \ln\left(\frac{483}{272}\right) = 14k \] Solve for \( k \): \[ k = \frac{\ln\left(\frac{483}{272}\right)}{14} \approx 0.041 \] b) Find the exponential growth function in terms of \( t \). The exponential growth function is: \[ V(t) = 272 \cdot e^{0.041t} \] c) Estimate the value of the painting in 2011. Calculate \( t \): \[ t = 2011 - 1976 = 35 \] Substitute \( t \) into the growth function: \[ V(35) = 272 \cdot e^{0.041 \times 35} \] Calculate the value: \[ V(35) \approx 272 \cdot e^{1.435} \approx 272 \cdot 4.198 \] \[ V(35) \approx 1141 \] The estimated value of the painting in 2011 is $1141 (rounded to the nearest dollar).