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Math Help - Depreciation.

  1. #1
    Senior Member Mukilab's Avatar
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    Depreciation.

    Each year a car depreciated by 9%. Calculate the number of years after which the value of the car is 47% of it's original value.

    I'm aware of the compound interest formula but I don't know if it would be any use here or how to reverse it because I'm not looking for a precise number. Should I create a number such as x and try and somehow change the compound interest formula so it's inverse?
    Last edited by mr fantastic; June 1st 2010 at 05:50 AM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by Mukilab View Post
    Each year a car depreciated by 9%. Calculate the number of years after which the value of the car is 47% of it's original value.
    i is negative, so: (1 + i)^n = (1 - .09)^n = .91^n

    .91^n = .47
    n = log(.47) / log(.91) = 8.005695...purty close to 8 years
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  3. #3
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    Quote Originally Posted by biancadee View Post
    That was very well presented! I am learning a lot here.
    Hmmm...all your posts are similar, Biancadee.
    Sneaky way to get visitors to your homepage
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  4. #4
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by Wilmer View Post
    i is negative, so: (1 + i)^n = (1 - .09)^n = .91^n

    .91^n = .47
    n = log(.47) / log(.91) = 8.005695...purty close to 8 years
    anyway to do this without logs?
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  5. #5
    Senior Member Mukilab's Avatar
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    Bump. Still no closer to a clue and its been ages. Hope the admin doesn't mind me bumping this.
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  6. #6
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    Quote Originally Posted by Mukilab View Post
    anyway to do this without logs?
    No, you must use logarithms but I cannot imagine why you could not just use a calculator
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  7. #7
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    Quote Originally Posted by Mukilab View Post
    Bump. Still no closer to a clue and its been ages. Hope the admin doesn't mind me bumping this.
    Well, to be blunt, your original post doesn't make much sense;
    what's "reverse" and "inverse"?

    Once more, to find the number of periods, logs ARE REQUIRED.

    Finding the "value left" after a GIVEN period is a different story: (1 + i)^n
    With i = -.09 and n = 8: (1 - .09)^8 = .91^8 = .47025...
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