# Writing an Exponential Model

1. Jan 7, 2012

### darshanpatel

1. The problem statement, all variables and given/known data

A new car that costs $30,000 has a book value of$18,000 after 2 years.

a) Find a linear model that represents the value 'V' of the care after 'x' years

b) Find an exponential model of V= ab^(x) that represents the value 'V' of the car after 'x' years.

2. Relevant equations

-None-

3. The attempt at a solution

I found the linear model just by basic graphing knowledge and using slope formula.

Linear Model: V= -6,000x + 30,000

I don't know how to find the exponential function, I know its a decay so it has to be a fraction, but how do I start it, and what is it supposed to be?

2. Jan 7, 2012

### LCKurtz

When you did your linear model, you had an unknown equation $y = mx + b$. You are probably familiar enough with that form that you knew the two unknowns are the slope and y intercept. If you hadn't known that, you would have plugged in the two points giving you two equations in two unknowns a and b.

Do the exponential the same way. Plug in the two points to get two equations in two unknowns.

3. Jan 7, 2012

### darshanpatel

What would I plug the points in to? I am getting confused with the variable 'a' and 'b' from y=ab^(x)

This is my shot at what you said:

I plugged the points (0,30000) in first and got 'a' to be 30000 then plugged that 'a' value into y=ab^x again, for the point (2,18000)

30000=a*b^0
30000=a*1
30000=a

Then...

18000= 30000*(b)^2
18/30 = b^2
(√15)/5

then for the final equation to model the situation, I got this:

V= 30000*((√15)/5)^x

Would that be correct?

4. Jan 7, 2012

### LCKurtz

All you have to do to see if it is correct is to check whether it passes through your two points (0,30000) and (2,18000).

5. Jan 7, 2012

### darshanpatel

oh yeah hahahah, I just remembered that while I was reading your post, lol sorry