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In 2004, an art collector paid $92,458,000 for a particular painting. The same painting sold for...

Question:

In 2004, an art collector paid $92,458,000 for a particular painting. The same painting sold for $33,000 in 1950.

Find the exponential growth rate k, to three decimal places, and determine the exponential growth function V, for which V(t) is the painting's value, in dollars, t years. (Do not round until the final answer. Then round to three decimal places as needed.)

Exponential Growth

An exponential growth function can be used to model quantities that are changing in time. This type of function is a function that describes a quantity that whose rate of change is proportional to the quantity itself. This function satisfies the differential equation {eq}\displaystyle \frac{dV}{dt} = kV {/eq} where k is the growth constant and V is the exponential growth function.

Answer and Explanation:


We start with the fact that an exponential growth function is given by the equation {eq}V(t) = V_0e^{kt} {/eq} where {eq}V_0 {/eq} and {eq}k {/eq} are constants.


To express an exponential growth function for the painting's value, we set 1950 as {eq}t = 0 {/eq} and {eq}V(0) =\$ 33000 {/eq} as the initial value.


Plugging this initial value into the exponential growth equation.


{eq}\displaystyle \begin{align*} V(t) &= V_0e^{kt}\\ \$\ 33000&=V_0 e^{k(0)}\\ V_0 &= \$\ 33000 \end{align*} {/eq}


Now, we need to solve for k, using the fact that in 2004, the value of the painting is {eq}\$\ 92458000 {/eq}.


We need to solve for t.


{eq}\begin{align*} t &= 2004-1950\\ t &=54\ years \end{align*} {/eq}


To solve for k or growth rate, we plug the following values into the exponential growth equation.


{eq}\displaystyle \begin{align*} V(t) &= V_0e^{kt},\ V(54) =\$\ 92458000,\ t = 54\ years\\ \$\ 92458000 &= \$\ 33000e^{k(54\ years)}\\ \frac{92458}{33} &= e^{k(54\ years)}\\ \end{align*} {/eq}

Solving for k by taking the natural logarithm.


{eq}\displaystyle \begin{align*} \ln \bigg( \frac{92458}{33} &= e^{k(54\ years)} \bigg) \\ \ln \frac{92458}{33} &= k(54\ years)\\ k &=\boxed{ \frac{1}{54} \ln \frac{92458}{33}\ year^{-1} \approx 0.147\ year^{-1}} \end{align*} {/eq}


Therefore the growth rate is approximately {eq}\boxed{ k \approx 0.147\ year^{-1}} {/eq} and the exponential growth function is {eq}\boxed{V(t) = \$33000e^{0.147t}} {/eq}, where t is time in year. The initial time is set to be the year 1950.


Learn more about this topic:

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Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10
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