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Math Help - Decay

  1. #1
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    Decay

    I am stuck on question 17 on this paper. Here goes...

    Gwen bought a new car.
    Each year, the value of her car depreciated by 9%

    Calculate the number of years after which the value of her car was 47% of its value when new.


    I just don't get this question at all.. Anyone help?

    Thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Danielisew View Post
    I am stuck on question 17 on this paper. Here goes...

    Gwen bought a new car.
    Each year, the value of her car depreciated by 9%

    Calculate the number of years after which the value of her car was 47% of its value when new.


    I just don't get this question at all.. Anyone help?

    Thanks
    Say the initial value of the car is x_0. Then, for t in terms of years, the value x(t) of the car is
    x(t) = x_0(1 - 0.09)^t = x_0(0.91)^t

    So when the car has a value of 47% of its inital value, x(t) = 0.47x_0

    Thus
    0.47x_0 = x_0(0.91)^t

    0.47 = (0.91)^t

    Take the natural log (or log to any base in fact) of both sides:
    ln(0.47) = ln[(0.91)^t]

    ln(0.47) = t \cdot ln(0.91)

    t = \frac{ln(0.47)}{ln(0.91)} = 8.0057

    or about 8 years.

    -Dan
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  3. #3
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    Hello, Danielisew!

    Gwen bought a new car.
    Each year, the value of her car depreciated by 9%

    Calculate the number of years after which the value of her car
    is 47% of its original value.
    Let's take baby-steps . . .


    The original value is V.

    During the first year, the car loses 9% of its value.
    . . Hence, its value is: . v_1 \:=\:0.91V

    During the second year, the car loses 9% of that value.
    . . Hence, its value is: . v_2\:=\:(0.91)(0.91V) \:=\:(0.91)^2V

    During the third year, the car loses 9% of that value.
    . . Hence, its value is: . v_3\:=\:(0.91)(0.91^2V) \:=\:(0.91)^3V

    In general, after n years, the car's value is: . v_n\:=\:(0.91)^nV


    When is this 47% of the original value?

    . . \frac{(0.91)^nV}{V} \:=\:47\%\quad\Rightarrow\quad (0.91)^n \:=\:0.47

    Take logs: . \log(0.91^n) \:=\:\log(0.47)\quad\Rightarrow\quad n\!\cdot\!\log(0.91) \:=\:\log(0.47)

    Then: . n \:=\:\frac{\log(0.47)}{\log(0.91)} \:=\:8.005695522\;\cdots \text{ about 8 years}


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