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6 April, 00:55

The depreciation of the value for a car is modeled by the equation y = 100,000 (.85) ^x for x years since 2000.

a. In what year was the value of the $61,412.50?

b. In what year will the value of the car reach 1/4 of its original value?

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  1. 6 April, 01:54
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    Step-by-step explanation:

    The depreciation of the value for a car is modeled by the equation y = 100,000 (.85) ^x for x years since 2000

    a) for the value to be $61,412.50, then

    61412.50 = 100000 (.85) ^x

    61412.50/100000 = 0.85^x

    0.614125 = 0.85^x

    We would take log to base 10 of both sides. It becomes

    Log 0.614125 = log 0.85^x =

    x log0.85

    - 0.2117 = x * - 0.07058

    - 0.2117 = - 0.07058x

    x = - 0.2117 / - 0.07058

    x = 3 years

    b) 1/4 * 100000 = 25000

    Therefore,

    25000 = 100000 (.85) ^x

    25000/100000 = 0.85^x

    0.25 = 0.85^x

    We would take log to base 10 of both sides. It becomes

    Log 0.25 = log 0.85^x =

    x log0.85

    - 0.602 = x * - 0.07058

    - 0.602 = - 0.07058x

    x = - 0.602 / - 0.07058

    x = 9 years
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