in 2004, an art collector paid $69,099,000 for a particular painting. the same painting sold for $29,000 in…

in 2004, an art collector paid $69,099,000 for a particular painting. the same painting sold for $29,000 in 1950. complete parts (a) through (d). a) find the exponential growth rate k, to three decimal places, and determine the exponential growth function v, for which v(t) is the paintings value, in dollars, t years after 1950. v(t)=□ (type an expression. type integers or decimals for any numbers in the expression. round to three decimal places as needed.)

in 2004, an art collector paid $69,099,000 for a particular painting. the same painting sold for $29,000 in 1950. complete parts (a) through (d). a) find the exponential growth rate k, to three decimal places, and determine the exponential growth function v, for which v(t) is the paintings value, in dollars, t years after 1950. v(t)=□ (type an expression. type integers or decimals for any numbers in the expression. round to three decimal places as needed.)

Answer

Explanation:

Step1: Identify the exponential - growth formula

The exponential - growth formula is $V(t)=V_0e^{kt}$, where $V_0$ is the initial value, $k$ is the growth rate, and $t$ is the time. In 1950 ($t = 0$), $V_0=29000$. In 2004, $t = 2004 - 1950=54$ years and $V(54)=69099000$.

Step2: Substitute the values into the formula

Substitute $V_0 = 29000$, $t = 54$, and $V(54)=69099000$ into $V(t)=V_0e^{kt}$: $69099000=29000e^{54k}$.

Step3: Solve for $k$

First, divide both sides of the equation by 29000: $\frac{69099000}{29000}=e^{54k}$, so $2382.724 = e^{54k}$. Then, take the natural - logarithm of both sides: $\ln(2382.724)=\ln(e^{54k})$. Since $\ln(e^{x})=x$, we have $\ln(2382.724)=54k$. $k=\frac{\ln(2382.724)}{54}$. $\ln(2382.724)\approx7.771$, so $k=\frac{7.771}{54}\approx0.144$.

Step4: Determine the exponential - growth function

Substitute $V_0 = 29000$ and $k\approx0.144$ into $V(t)=V_0e^{kt}$: $V(t)=29000e^{0.144t}$.

Answer:

$V(t)=29000e^{0.144t}$