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17 December, 19:04

What value will x have to for the first number to reach zero 153•0.92^x

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  1. 17 December, 21:36
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    Step-by-step explanation:

    153•0.92^x is a decaying exponential function.

    In theory these functions never reach zero.

    Suppose that you were willing to change the problem to read:

    "What value will x have to for the first number to come within 0.0001 of zero?" Solve 153•0.92^x = 0.0001.

    To do this, take the common log of both sides, obtaining

    log 153 + x*log 0.92 = log 0.0001

    Note that log 0.0001 = - 4; log 153 = 2.18469; and log 0.92 = - 0.03621.

    Then we have:

    2.18469 + x (-0.03621) = - 4.

    Isolate the 2nd term. To accomplish this, subtract 2.18469 from both sides, obtaining:

    -0.03621x = - 6.18469

    Isolate x by dividing both sides by - 0.03621:

    x = 170.8

    This tells us that as x approaches + 179, the quantity 153·0.92x will be within 0.0001 of zero.
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