How do you evaluate #log_(1/11) (1/121)#?

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Answer 1 · 2018-04-05T07:43:49

#color(blue)(2)#

Using change of base formula:

#log_(1/11)(1/121)=(ln(1/121))/ln(1/11)#

#log(a/b)=loga-logb#

#(ln(1/121))/ln(1/11)=(ln(1)-ln(121))/(ln(1)-ln(11))=(0-ln(121))/(0-ln(11))=ln(121)/ln(11)#

#loga^b=bloga#

#=(ln(11^2))/ln(11)=(2ln(11))/ln(11)=2#