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Math Help - Calculating depreciation using given model?

  1. #1
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    Calculating depreciation using given model?

    Depreciation model for the price (P) in of the car a person wishes to buy is given by P=45520e^-0.1t, where t is the number of years since the car was brand new. The savings model for the amount (A) in dollars, this person has accumulated in the same t years is given by A=10,000(1.05)^t.

    How many years (to 2 dps) will it take for this person to be able to buy the car (ie. where A=P) using his savings only?

    The answer in the book is 10.19 years, but i dont know how to get there.
    Thank you in advance
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  2. #2
    MHF Contributor Unknown008's Avatar
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    P = 45520e^{-0.1t}

    A = 10000(1.05)^t

    When P = A;

    45520e^{-0.1t} = 10000(1.05)^t

    Can you solve this?
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  3. #3
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    45520e^{-0.1t} = 10000(1.05)^t  \setminus \cdot \frac{e^{0.1t}}{10000}
    (1.05)^t\cdot e^{0.1t}=\frac{45520}{10000}
    (1.05\cdot e^{0.1})^t=4.552
    \log \left((1.05\cdot e^{0.1})^t\right)=\log 4.552
    t\cdot \log \left(1.05\cdot e^{0.1}\right)=\log 4.552
    t=\frac{\log 4.552}{\log \left(1.05\cdot e^{0.1}\right)}\approx 10.1859
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  4. #4
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    Thank you MathoMan, however can you please explain the first line (in particular \.e^.1t/10000)
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  5. #5
    MHF Contributor Unknown008's Avatar
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    This is how I do it:

    45520e^{-0.1t} = 10000(1.05)^t

    ln(45520e^{-0.1t}) = ln(10000(1.05)^t)

    ln(45520) - 0.01tln(e) = ln(10000) + tln(1.05)

    ln(45520) - ln (10000) = tln(1.05) + 0.01t

    ln(45520) - ln (10000) = t(ln(1.05) + 0.01)

    t = \frac{ln(45520) - ln (10000)}{ln(1.05) + 0.01}

    t = \frac{ln(4.5520)}{ln(1.05) + 0.01}

    t = 10.185
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  6. #6
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    Thank you heaps
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  7. #7
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    Quote Originally Posted by daniel52947 View Post
    Thank you MathoMan, however can you please explain the first line (in particular \.e^.1t/10000)
    It means that the whole equation is multiplied by the expression \frac{e^{0.1t}}{10000}
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