$Exponential growth word problem$

Question

$Exponential growth word problem$

A coin sold for $\$ 269$ in 1980 and was sold again in 1989 for $\$ 477$. 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 $V_0$ equals=269.
b) Find the exponential growth function in terms of t, where t is the number of years since 1980.
c) Estimate the value of the coin in 2013.
d) What is the doubling time for the value of the coin to the nearest tenth of a year?
e) Find the amount of time after which the value of the coin will be $ \$ 2878$.

Word Problem Equations

Fykl

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer.

Answer 1

Answers can be viewed only if

The questioner was satisfied and accepted the answer, or
The answer was disputed, but the judge evaluated it as 100% correct.

View the answer

Since the growth is exponential,
$$V(t)=V_0 e^{a t}=269 e^{kt}.$$

a)
$$V(9)=269e^{9 k}=477 \Rightarrow e^{9k}=\frac{477}{269} \Rightarrow 9k=\ln \frac{477}{269} .$$
So
$$k=\frac{1}{9}\ln (\frac{477}{269})=0.064.$$

b) $V(t)=269 e^{0.064t}.$
c) We have 2013-1980=33.

$$v(33)=269 e^{0.064 \times 33}=269*8.17=\$ 2196.7.$$

d) To find the doubling time we must have $V(t)=2v(0)$. Hence
$$269 e^{kt}=2 \times 269 \Rightarrow e^{kt}=2 \Rightarrow kt= \ln 2 $$
\[\Rightarrow t=\frac{\ln 2}{k}=\frac{\ln 2}{0.064}=10.8.\]

e) \[V(t)=269 e^{kt}=2878 \Rightarrow e^{kt}=\frac{2878}{269} \Rightarrow kt =\ln (\frac{2878}{269})=2.37.\]
Hence
\[0.064 t= 2.37 \Rightarrow t= 37.03. \]
So the value of the coin will be $\$ 2878$ in the year 1980+37.03=2017.03.

Erdos

$Exponential growth word problem$

Answer

Related Questions

Search