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Thread: 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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    $\displaystyle P = 45520e^{-0.1t}$

    $\displaystyle A = 10000(1.05)^t$

    When P = A;

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

    Can you solve this?
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    $\displaystyle 45520e^{-0.1t} = 10000(1.05)^t \setminus \cdot \frac{e^{0.1t}}{10000}$
    $\displaystyle (1.05)^t\cdot e^{0.1t}=\frac{45520}{10000}$
    $\displaystyle (1.05\cdot e^{0.1})^t=4.552$
    $\displaystyle \log \left((1.05\cdot e^{0.1})^t\right)=\log 4.552$
    $\displaystyle t\cdot \log \left(1.05\cdot e^{0.1}\right)=\log 4.552$
    $\displaystyle t=\frac{\log 4.552}{\log \left(1.05\cdot e^{0.1}\right)}\approx 10.1859$
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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:

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

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

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

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

    $\displaystyle ln(45520) - ln (10000) = t(ln(1.05) + 0.01)$

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

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

    $\displaystyle t = 10.185$
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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 $\displaystyle \frac{e^{0.1t}}{10000}$
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