How do you evaluate #log_(1/3) (1/81)#?

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Answer 1 · 2016-12-16T02:06:40

#4#

#1/81# can be written as #(81)^-1#:

#= log_(1/3) (81^-1)#

Use the change of base rule #log_a n = logn/loga#:

#= log81^-1/log(1/3)#

#=log(3^4)^-1/log(3^-1)#

#= log(3^-4)/log3^-1#

Use the exponent rule that #loga^n = nloga#:

#=(-4log3)/(-1log3)#

#=4#

Hopefully this helps!