How do you solve $log_2x=log_4 25$ ?

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Answer 1 · 2016-03-15T20:25:32

It is

$log_2 x=log_(2^2) 25=>(logx/log2) = (log25/log2^2)=> logx/log2=(2log5)/(2log2)=>logx=log5=>x=5$

How do you solve log_2x=log_4 25log_2x=log_4 25?