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 V₀ = 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) = [ ]

Solution

To solve the problem, we will address each part step by step. a) Find the value of \( k \) of the exponential growth rate. Assume \( V_0 = 272 \). The exponential growth formula is given by: \[ V(t) = V_0 \cdot e^{kt} \] Given that in 1990, \( V(t) = 483 \), and \( t = 1990 - 1976 = 14 \) years after 1976. Substitute the values into the equation: \[ 483 = 272 \cdot e^{14k} \] Divide both sides by 272 to solve for \( e^{14k} \): \[ \frac{483}{272} = e^{14k} \] Take the natural logarithm on both sides: \[ \ln\left(\frac{483}{272}\right) = 14k \] Solve for \( k \): \[ k = \frac{\ln\left(\frac{483}{272}\right)}{14} \] Calculate \( k \): \[ k \approx \frac{\ln(1.7765)}{14} \approx \frac{0.57403}{14} \approx 0.041 \] b) Find the exponential growth function in terms of \( t \), where \( t \) is the number of years since 1976. Using the value of \( k \) from part a, the exponential growth function is: \[ V(t) = 272 \cdot e^{0.041t} \] Thus, the exponential growth function is: \[ V(t) = 272 \cdot e^{0.041t} \]