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) About Me  (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
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​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
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Cheers,
Stan H.
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