Story Problem Help

Question

A painting sold for $268 in 1980 and was sold again in 1989 for $440. 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=268

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 painting in 2008.

d)Find the amount of time after which the value of the painting will be $1985

Answer 1

V = k*B^t with (0,268) and (9,440)

268 = k*B^0 ---> k=268

so V = 268 * B^t

440 = 268* B^9

440/268 = B^9

110/67 = B^9

log(110/67) = 9 log B

(1/9) log (110/67) = log B

so the base is B= 10^((1/9)(log(110/67))

= 10^((log(110/67)(1/9))

= 10^(log(110/67))^(1/9)

= (110/67)^(1/9)

which is APPROXIMATELY B=1.0566330949524018266995948794498...

V = 268*[(110/67)^(1/9)]^t

For 2008, x = 28 , so the value is approximately 1253.177678

Finally, for the value to be $1985:

1985 = 268 * [(110/67)^(1/9)]^t

1985/268 = [(110/67)^(1/9)]^t

log(1985/268) = t * log((110/67)^(1/9))

log(1985/268) divided by log((110/67)^(1/9)) = t

t = 36.34919387

that's May 7, 2016

[ 2016 is a leap year and 0.35*366 = 128.1 days, 121 of which is

Jan-April including the leap day Feb 29]