Question: A painting sold for $272 in 1977 and was sold again in 1987 for $416. Assume that the growth in the…

A painting sold for $272 in 1977 and was sold again in 1987 for $416. 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 V₀ = 272. k = 0.042 (Round to the nearest thousandth.) b) Find the exponential growth function in terms of t, where t is the number of years since 1977. V(t) = 272 * e^(0.042t) c) Estimate the value of the painting in 2015. $ (Round to the nearest dollar.)

Solution

A painting sold for $272 in 1977 and was sold again in 1987 for $416. Assume the growth in value \( V \) of the collector’s item was exponential. a) Find the value \( k \) of the exponential growth rate. Assume \( V_0 = 272 \). We know that the formula for exponential growth is: \[ V = V_0 \cdot e^{kt} \] Given \( V_0 = 272 \), \( V = 416 \), and \( t = 10 \) years (from 1977 to 1987), we have: \[ 416 = 272 \cdot e^{10k} \] Divide both sides by 272: \[ \frac{416}{272} = e^{10k} \] Calculate the left side: \[ 1.5294 = e^{10k} \] Take the natural logarithm of both sides to solve for \( k \): \[ \ln(1.5294) = 10k \] \[ 0.4242 = 10k \] Divide by 10: \[ k = 0.0424 \] b) Find the exponential growth function in terms of \( t \), where \( t \) is the number of years since 1977. The growth function is: \[ V(t) = 272 \cdot e^{0.0424t} \] c) Estimate the value of the painting in 2015. From 1977 to 2015, \( t = 38 \) years. Use the growth function: \[ V(38) = 272 \cdot e^{0.0424 \times 38} \] Calculate the exponent: \[ 0.0424 \times 38 = 1.6112 \] Then the value of the painting is: \[ V(38) = 272 \cdot e^{1.6112} \] Calculate \( e^{1.6112} \) and multiply by 272: \[ e^{1.6112} \approx 5.009 \] \[ V(38) \approx 272 \cdot 5.009 \approx 1362 \] Therefore, the estimated value of the painting in 2015 is \$1362 (rounded to the nearest dollar).