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Question Number 171828 by Mikenice last updated on 21/Jun/22
solve for x: 2log_x 4=log_2 x
solveforx:2logx4=log2x
Commented by kaivan.ahmadi last updated on 21/Jun/22
log_x 16=log_2 x⇒((log_2 16)/(log_2 x))=log_2 x⇒ 4=(log_2 x)^2 ⇒{_(log_2 x=−2⇒x=(1/4)) ^(log_2 x=2⇒x=4)
logx16=log2x⇒log216log2x=log2x⇒4=(log2x)2⇒{log2x=−2⇒x=14log2x=2⇒x=4
Commented by Mikenice last updated on 21/Jun/22
thanks sir
thankssir

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