SOLUTION: Solve the problem. A rare baseball card was sold in 1990 for $285,000. The card was then resold in 1998 for $459,000. Assume that the card's value increases exponentially, and f

Question 212789: Solve the problem.
A rare baseball card was sold in 1990 for $285,000. The card was then resold in 1998 for $459,000. Assume that the card's value increases exponentially, and find an exponential function V(t) that fits the data. (Round decimals to three places.) (Points: 5)
These are my options
V(t) = 285e0.766t, where t is the number of years after 1990.
V(t) = 285e0.06t, where t is the number of years after 1990.
V(t) = 285,000e0.766t, where t is the number of years after 1990.
V(t) = 285,000e0.06t, where t is the number of years after 1990.
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
A rare baseball card was sold in 1990 for $285,000. The card was then resold in 1998 for $459,000. Assume that the card's value increases exponentially, and find an exponential function V(t) that fits the data. (Round decimals to three places.) (Points: 5)
These are my options
V(t) = 285e0.766t, where t is the number of years after 1990.
V(t) = 285e0.06t, where t is the number of years after 1990.
V(t) = 285,000e0.766t, where t is the number of years after 1990.
V(t) = 285,000e0.06t, where t is the number of years after 1990.
.
"current value" = "initial value" * e^(kt)
V(t) = Ie^(kt)
where
I is the initial worth in dollars
k is constant of variation
t is time (in years)
.
We need to find 'k' based on the given data. Plug in what we know and solve for 'k':
V(t) = Ie^(kt)
459000 = 285000e^(8k)
459000/285000 = e^(8k)
ln(459000/285000) = 8k
ln(459000/285000)/8 = k
0.060 = k
.
Therefore your equation is:
V(t) = 285,000e0.06t, where t is the number of years after 1990.

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