Question

A doll sold for ?$215 in 1975 and was sold again in 1989 for $488. 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 Vo=215.

?(Round to the nearest? thousandth.)

?b) Find the exponential growth function in terms of? t, where t is the number of years since 1975

?V(t)=

?c) Estimate the value of the doll in 2015.

?(Round to the nearest? dollar.)

?d) What is the doubling time for the value of the doll to the nearest tenth of a? year?

?(Round to the nearest? tenth.)

?e) Find the amount of time after which the value of the doll will be ?$3037

?(Round to the nearest? tenth.)

Answer 1

a). Let the exponential growth function in terms of t, where t is the number of years since 1975, be V(t) = V₀e^kt, where V₀ is the initial value of the doll in 1975, V(t) is the value of the doll t years after 1975 and k is the exponential growth rate. Here, V₀ = $ 215, t = 14 and V(14) = $488. Hence 488 = 215e^14k or, e^14k =488/215. Now, on taking natural log of both the sides, we get ln(e^14k) = ln(488/215) or, 14k ln e = ln 488-ln 215 or,14k=6.910315406-5.370638028 =0.819677377 so that k=0.819677377/14=0.058548384 =0.059(on rounding off to the nearest thousandth). ( ln mⁿ = n ln m , ln e = 1, and ln (m/n)=ln(m)–ln (n)).

b). The exponential growth function is V(t) = 215e^0.059t, where V(t),V₀ and t are as in part a) above.

c). In 2015, t = 2015-1975 = 40 so that the value of the doll in 2015 is 215 e^{40*0.058548384} = 215 e^2.2193536 = 215* 9.20138117 = $ 1978 (on rounding off to the nearest dollar).

d).Let the doubling time for the value of the doll be t years. Then 2*215=215e^0.058548384t or, e^0.058548384t =2. Now, on taking natural log of both the sides, we get ln(e^0.058548384t ) = ln 2 or, 0.058548384t = 0.69314718 so that t =0.69314718/0.058548384 = 11.83887809 = 11.8 years (on rounding off to the nearest tenth of an year).

e). Let the value of the doll be $ 3037, t years after 1975. Then, 3037 = 215 e^0.058548384t or, e^0.058548384t   = 3037/215. Now, on taking natural log of both the sides, we get ln(e^0.058548384t ) = ln (3037/215) or, 0.058548384t = 8.018625465-5.370638028= 2.647987437 so that t = 2.647987437/0.058548384 = 45.22733603 =   45.2 years(on rounding off to the nearest tenth of an year).