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Writing an Exponential Model

  1. Jan 7, 2012 #1
    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. jcsd
  3. Jan 7, 2012 #2

    LCKurtz

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    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.
     
  4. Jan 7, 2012 #3
    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?
     
  5. Jan 7, 2012 #4

    LCKurtz

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    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).
     
  6. Jan 7, 2012 #5
    oh yeah hahahah, I just remembered that while I was reading your post, lol sorry
     
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