How do you evaluate #log_36 (1/6)#?

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Answer 1 · 2016-10-26T16:11:34

#log_36(1/6) = -1/2#

Recall that #1/6# can be written as #6^-1#.

#=log_36(6^-1)#

Now, use the change of base rule #log_a(n) = logn/loga#.

#=log(6^-1)/log36#

#=log(6^-1)/log(6^2)#

Use the rule #loga^n = nloga#.

#=(-1log6)/(2log6)#

#=-1/2#

Hopefully this helps!