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Logarithm Depreciation Word Problem

  1. May 10, 2015 #1
    1. The problem statement, all variables and given/known data

    A vehicle purchase for $32,000 depreciates at a rate of 75% every 6 years. Another vehicle purchased for $16,000 depreciates at a rate of 50% every 4 years. Create an exponential function for each situation, and use the functions to algebraically determine the amount of time it would take for the vehicles to be equal in value.


    2. Relevant equations

    [tex] A = P(1 \pm i )^{n} [/tex]

    A is the future amount
    P is the present/principal amount
    i = interest rate per compounding period in decimal form
    n = number of compounding periods


    3. The attempt at a solution

    Algebraically:

    [tex] A_{1} = A_{2} [/tex]

    [tex] P_{1}(1-i_{1})^{n} = P_{2}(1-i_{2})^{n} [/tex]

    [tex] \log P_{1} + n\log (1-i_{1}) = \log P_{2} + n\log (1-i_{2}) [/tex]

    [tex] n\log (1-i_{1}) - n\log (1-i_{2}) = \log P_{2} - \log P_{1} [/tex]

    [tex] n = \frac{ \log P_{2} - \log P_{1}}{\log (1-i_{1}) - \log (1-i_{2}) } [/tex]

    My issue is I don't know how to use the 6 years and the 4 years. I've done it with semi-annually, annually, quarterly, etc, but this one deals with years > 1 . For instance, if this was semi, it would be (75% / 2) /100 , to get i.

     
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  3. May 10, 2015 #2

    SteamKing

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    I think you have overlooked the key suggestion in the problem statement: Create an exponential function which describes the depreciation in value of each car.
    This means that the value of each car is constantly declining, not every six months, not every month, not even every day, but all the time.

    Using the standard formulas for compounding interest over discrete time periods does not meet the criterion of establishing an exponential function to describe this process.

    You want a function which relates the final value of each car V(t) to its initial value V(0) over a certain time; in other words,

    V(t) = V(0) * e-rt

    The rate r at which the value depreciates can be determined from the initial and the final values of each car and the time over which the depreciation occurs.
     
  4. May 10, 2015 #3
    While I have no doubt that is correct, the given equation is the extent to this curriculum (it's a special word problem in the logarithm chapter). Is there anyway to do it using the standard formula?
     
  5. May 10, 2015 #4

    SteamKing

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    Not that I am aware of.

    You should also remember that logarithms have a special relationship with exponential functions, namely

    log (ex) = x, where "log" stands for the "natural logarithm", and is sometimes written ln (ex) = x.
     
  6. May 10, 2015 #5
    You will be able to solve the problem even without logarithms, but since the question wants you to do it exponentially, it is recommended you do so.
    ##i## would be ##\frac{\frac{75}{2}}{100}## if the rate was 75% annually and compounded semi-anually. What is the time period for the rate in this case ?
     
  7. May 10, 2015 #6
    Also note that this is a clear indication that the period of compounding is 6 years, not 1. Therefore you must count ##n## with multiplicity.
    [EDIT:- and accordingly for the other case.]
     
  8. May 10, 2015 #7

    Ray Vickson

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    One approach is to use the 1-year depreciation rate for each car. For Car 1 (the $32,000 one), say the depreciation is ##100\, r_1## percent per year, while for Car 2 it is ##100\, r_2## per year. Then you need to solve
    [tex] \frac{32000}{(1+r_1)^n} = \frac{16000}{(1+r_2)^n}. [/tex]
    Do you see how to get ##r_1## and ##r_2## from the information given in the problem?
     
  9. May 11, 2015 #8

    SteamKing

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    IMO, I don't think that's the way to solve this problem.

    Based on the problem statement, this is an example of continuous compounding with a twist: instead of the principal increasing in value over time, an asset is depreciating.

    http://cs.selu.edu/~rbyrd/math/continuous/

    The formula for continuous compounding involves the exponential function (ex) as described in the problem statement as the method of solution. By using a negative exponent, the continuous compounding formula is converted into a continuous depreciation formula. :wink:
     
  10. May 11, 2015 #9
    Hi Ray,
    Since this is compound interest, and the period of compounding is 6 years and 4 years respectively, I don't think you will be able to find the depreciation rate for 1 year for compound interest.
    But the problem can still be solved by using simply ##A=P(1\pm i)^n##.
    Cheers.
     
    Last edited: May 11, 2015
  11. May 11, 2015 #10

    Ray Vickson

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    In post #3 the OP (sort-of) indicated that the exponential formula was unfamiliar and (more-or-less) outside the coverage of the course up to then. So, while I personally agree with you, I tried to put the method into a form that would be familiar to the OP, hence wrote ##1/(1+r)^n## instead of ##e^{-vn}##. But, of course, ##e^{-v} = 1/(1+r)##. The remaining issue is whether ##n## turns out to be integer, or not.
     
  12. May 11, 2015 #11

    Ray Vickson

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  13. May 12, 2015 #12

    Ray Vickson

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    Well, if we write ##1-i = 1/(1+r)## they are exactly the same. The issue of finding ##r## is the same as the issue of finding ##i##.
     
  14. May 12, 2015 #13

    SteamKing

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    This problem, as stated, does not necessarily require the use of the compound interest formula or even writing an exponential function to solve. Given the present values of the two cars, and knowing by how much each depreciates over a fixed term, one can simply set up a calculation using the stated amount of depreciation and the time involved in order to determine when the two cars reach equal value. :smile:

    I suspect the purpose of this exercise was not so much for the student to calculate how long it takes for equal value to be obtained, but in how well the student follows directions, especially since the problem statement was quite specific in the manner in which it desired the solution to be obtained. :wink:
     
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