# pre-cal

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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.

• pre-cal -

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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