A toy tractor sold for $244 in 1975 and was sold again in 1989 for $473. Assume the growth in the value V of the collector's item was exponential.

Question

What is the value of k?  (Round to the nearest thousandth)

b) Estimate the value of the toy in 2012. $ (Round to the nearest dollar)

c) What is the doubling time? years (Round to the nearest tenth)

Answer 1

a) k is the growth rate of an exponential function, F=Ve^(kt) which is current value = starting value * e^(growth rate * time (in years))

so to solve this question we setup the equation: 473=244e^(k*(1989-1975)), 473/244=e^(14k), take the ln of both sides: ln(473/244)=ln(e^14k), use the erd property of logarithms to pull out the 14k: ln(473/244)=14k*ln(e), ln(e)=1 so: ln(473/244)=14k, k=ln(473/244)/14=.047

b) now that we have the growth rate we can simply plug in all the numbers and solve for F:

F=244e^(.047(2012-1975))=$1403

c) time it takes to double is simply solve for t when F=2V so going back to part a):

2=1e^(.047t), ln2=.047t*lne, t=ln(2)/.047=14.7 years

Answer 2

473-244=229

229/473=48.814.......% rise/24yrs

48.8/24=1.2/yr

2012-1989=23yrs

23*1.2=27.6

473+27.3=$500.30