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Answer 1 · 2015-11-25T12:57:10

#x ={ 2,4}#

Now i am making an assumption your looking for integral roots

Take log on both sides

#log2^x =logx^2#

#=> xlog2 =2logx#

switch it up;

#x/2 = log_ 2 x#

Now notice to find integral solution the value of x must be a multiple of 2

So lets start with 2{We cant use -ve numbers as log of a negative number is just crazy}

#2/2 = log_2 2# {correct}

#4/2 = log_2 4#{correct}

#cancel( 6/2 = log_6 4)#{wrong}

In fact the solution set does not go beyond 4;

I know this is not a traditional way of getting to the roots;But one can only approximate it ;

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