How do you solve $log_(2)x^2+log_(.5)x=5$ ?

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Answer 1 · 2017-02-16T15:08:09

$x=32$

$log_2x^2+log_0.5x=5$

Now $log_2x^2=2log_2x=(2logx)/log2$

and $log_0.5x=logx/log0.5=logx/(log(1/2))=logx/(log1-log2)$

= $-logx/log2$

Hence $log_2x^2+log_0.5x=5$

or $2logx/log2-logx/log2=5$

or $logx/log2=5$

or $logx=5log2=log2^5=log32$

Hence, $x=32$

How do you solve log_(2)x^2+log_(.5)x=5log_(2)x^2+log_(.5)x=5?