i am willing to do all the work, i just dont know where to start. i did find the value of k (1/39 ln 2700/800)= .0902 and i think the exponentiial growth is N(0)=800e.0902t

In 1967 a 10 year-old Golden Hawk sold for only $800, and in 2006, this car, was worth about $27,000.
Question: Estimate the value of the car in 2008.

Question

i am willing to do all the work, i just dont know where to start. i did find the value of k (1/39 ln 2700/800)= .0902 and i think the exponentiial growth is N(0)=800e.0902t

In 1967 a 10 year-old Golden Hawk sold for only $800, and in 2006, this car, was worth about $27,000. 
Question: Estimate the value of the car in 2008.

anonymous · Answer

800=N(0)* e^(10*k)    1st equation
27000=N(0)* e^(49*k) 2nd equation

find N(0) and k from above two equations and then use price2008=N(0)*e^(k*51)