# 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.

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from High School Algebra I: Help and Review

Chapter 6 / Lesson 10