A toy tractor sold for $\$ 272$ in 1976 and was sold again in 1987 for $\$ 429$. Assume that the growth in the value $\mathrm{V}$ of the collector's item was exponential
a) Find the value $k$ of the exponential growth rate. Assume $V_{0}=272$.
\[
k=
\]
(Round to the nearest thousandth.)
b) Find the exponential growth function in terms of $t$, where $t$ is the number of years since 1976.
\[
V(t)=
\]
c) Estimate the value of the toy tractor in 2006.
(Round to the nearest dollar.)
d) What is the doubling time for the value of the toy tractor to the nearest tenth of a year?
years
(Round to the nearest tenth.)
e) Find the amount of time after which the value of the toy tractor will be $\$ 1377$. years
(Round to the nearest tenth.)
Answer
Rounding to the nearest thousandth gives the final answer: \(\boxed{0.041}\).
Step 1 ：Given that the initial value of the toy tractor, \(V_0\), is $272 and the value of the toy tractor 11 years later, \(V(11)\), is $429, we can use the formula for exponential growth, \(V(t) = V_0 * e^{kt}\), to find the growth rate, \(k\).
Step 2 ：Rearrange the formula to solve for \(k\): \(k = \frac{1}{t} * \ln(\frac{V(t)}{V_0})\).
Step 3 ：Substitute the given values into the equation: \(k = \frac{1}{11} * \ln(\frac{429}{272})\).
Step 4 ：Solving the equation gives \(k = 0.04142316842109269\).
Step 5 ：Rounding to the nearest thousandth gives the final answer: \(\boxed{0.041}\).
