pre-cal

Use logarithms to solve log 8 / log x < 1.

I understand that there are 2 cases: when x is negative or positive.

I don't understand how it affects the answer though.

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  1. log8<logx if logx is positive
    10^8<x

    Log8>logx if logx is negative.
    10^8>x if x is less one

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    bobpursley

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