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:

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

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

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

$loga^b=bloga$

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

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