How do you evaluate $log_(1/2) (1/8)$ ?

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Answer 1 · 2016-09-21T00:23:33

Log means you have to find the power to which you have to raise the base in order to get the argument of the log.

In other words: $x=log_g aharr g^x=a$

Or: $(1/2)^x=1/8$

Where $x=3$

How do you evaluate log_(1/2) (1/8)log_(1/2) (1/8)?