Question: A painting sold for $283 in 1980 and was sold again in 1989 for $461. Assume that the growth in the…

A painting sold for $283 in 1980 and was sold again in 1989 for $461. 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 = 283. k = 0.054 (Round to the nearest thousandth.)

b) Find the exponential growth function in terms of t, where t is the number of years since 1980. V(t) = 283 * e^(0.054t)

C) Estimate the value of the painting in 2008. $ 1284 (Round to the nearest dollar.)

d) What is the doubling time for the value of the painting to the nearest tenth of a year? 12.8 years (Round to the nearest tenth.)

e) Find the amount of time after which the value of the painting will be $2172. years (Round to the nearest tenth.)

Solution

A painting sold for $283 in 1980 and was sold again in 1989 for $461. 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 \( V_0 = 283 \). Start with the formula for exponential growth: \[ V(t) = V_0 \cdot e^{kt} \] In 1989, \( t = 9 \) and \( V(9) = 461 \): \[ 461 = 283 \cdot e^{9k} \] Solve for \( k \): \[ \frac{461}{283} = e^{9k} \] \[ \ln\left(\frac{461}{283}\right) = 9k \] \[ k = \frac{\ln\left(\frac{461}{283}\right)}{9} \] Calculate: \[ k \approx 0.054 \] b) Find the exponential growth function in terms of \( t \), where \( t \) is the number of years since 1980. Using the calculated \( k \): \[ V(t) = 283 \cdot e^{0.054t} \] c) Estimate the value of the painting in 2008. For 2008, \( t = 28 \): \[ V(28) = 283 \cdot e^{0.054 \times 28} \] Calculate: \[ V(28) \approx 1284 \] d) What is the doubling time for the value of the painting to the nearest tenth of a year? Use the formula for doubling time \( T \): \[ T = \frac{\ln(2)}{k} \] Calculate: \[ T \approx \frac{\ln(2)}{0.054} \approx 12.8 \text{ years} \] e) Find the amount of time after which the value of the painting will be $2172. Use the formula to solve for \( t \): \[ 2172 = 283 \cdot e^{0.054t} \] \[ \frac{2172}{283} = e^{0.054t} \] \[ \ln\left(\frac{2172}{283}\right) = 0.054t \] \[ t = \frac{\ln\left(\frac{2172}{283}\right)}{0.054} \] Calculate: \[ t \approx 42.4 \text{ years} \]