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30 June, 12:23

Suppose a new car that cost 20 000 depreciates by 15% each year. in how many year till the car costs 7500. use the equation 7500 = (20,000) (0.85) ^x

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  1. 30 June, 13:28
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    First we can simplify a bit by dividing by 2500:

    3 = 8 * (0.85) ^x = > 0.85^x = 3/8

    next, we can take the log () on both sides, and use the fact that log (a^b) = b*log (a)

    log (3/8) = log (0.85^x) = > log (3/8) / log (0.85) = x

    If you calculate this, you find x ≈ 6.03
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