How do you find the value of x: #log_x (121/289) = 2#?

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How do you find the value of x: #log_x (121/289) = 2#?

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Answer 1 · 2018-03-14T18:09:21

#x=11/17#

Explanation:

#color(red)((1)log_aX=n<=>X=a^n)#
#color(red)((2)log_aX^n=nlogX)#
We have,
#log_x(121/289)=2=>log_x(11/17)^2=2#
#=>2*log_x(11/17)=2#
#=>log_x(11/17)=1#
#=>11/17=x^1=>x=11/17#

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How do you find the value of x: #log_x (121/289) = 2#?

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