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#