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How do you solve #4logx= log (1/16)#?

1 Answer

Shwetank Mauria

Mar 17, 2016

#x=1/2#

Explanation:

As #4logx=log(1/16)#

#logx^4=log(1/2^4)# or

#x^4=2^(-1xx4)=(2^(-1))^4#

Hence #x=2^(-1)# or #x=1/2#

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