Equation function

Question

A car was valued at $39,000 in the year 1993. The value depreciated to $12,000 by the year 2006.A) What was the annual rate of change between 1993 and 2006? =  Round the rate of decrease to 4 decimal places.B) What is the correct answer to part A written in percentage form? =  %.C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2010 ?value = $  Round to the nearest 50 dollars

Answer 1

Equation for Future value, F, in terms of Present value, P, and annual interest rate (as a decimal), i:

F = P(1+i)^t where t is in years

1) You are solving for I given P, and F

2) You are solving for F given P and the i from (a) (You can pick 1993 or 2006 as your starting point - just the number of years changes to 2010.

I'll leave you part (2) which is just plugging into the first equation.

Part 1, I would solve for i and then plug in (Better form than using numbers and then solving for many reasons)

F/P = (1+i)^t

ln(F/P) = t ln(1+i)

1/t ln(F/P) = ln(1+i)

e^1/t(F/P) - 1 = i where t is 2006-1993 - 16 yrs and F = 12000 and P = 39000 (note that i will be negative because depreciation requires the value of 1+i to be less than 1.