SOLUTION: A coincoin sold for ​$239 in 1980 and was sold again in 1987 for $443. Assume that the growth in the value V of the​ collector's item was exponential.
​a) Find
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Question 1120708: A coincoin sold for $239 in 1980 and was sold again in 1987 for $443. 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 Vsubscript o=239.
k=?
(Round to the nearest thousandth.)
b) Find the exponential growth function in terms of t, where t is the number of years since 1980.
V(t)=?
c) Estimate the value of the coincoin in 2015.
$=?
(Round to the nearest dollar.)
d) What is the doubling time for the value of the coincoin to the nearest tenth of a year?
yrs=?
(Round to the nearest tenth.)
e) Find the amount of time after which the value of the coincoin will be $5729.
yrs=?
(Round to the nearest tenth.)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A coincoin sold for $239 in 1980 and was sold again in 1987 for $443. Assume that the growth in the value V of the collector's item was exponential.
V(t) = ak^t
a) Find the value k of the exponential growth rate. Assume Vsubscript o=239.
k=?
V0 = ak^0 = 239
a = 239
V(7) = 239*k^7 = 443
k^7 = 1.853556....
k = 1.092
(Round to the nearest thousandth.)
------------------------------------------
b) Find the exponential growth function in terms of t, where t is the number of years since 1980.
V(t)= 239*1.092^t
------------------------------------
c) Estimate the value of the coincoin in 2015.
V(15) = 239*1.092^15 = $894.82
(Round to the nearest dollar.)
------------------------------
d) What is the doubling time for the value of the coincoin to the nearest tenth of a year?
Solve 2 = 1.092^t
t = log(2)/log(1.092) = 7.9 yrs
yrs=?
(Round to the nearest tenth.)
-------------------------------------
e) Find the amount of time after which the value of the coincoin will be $5729.
yrs=?
(Round to the nearest tenth.)
Solve:: 239*1.092^t = 5729
1.092^t = 23.97
t = log(23.97)/log(1.092) = 36.1 yrs
----------------
Cheers,
Stan H.
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