Question 948680
Given   {{{2^(2x)-3*2^(x)+2=0}}}


{{{(2^x)(2^x)-3*2^x+2=0}}}


{{{(2^x)^2-3*2^x+2=0}}}


LET {{{u=2^x}}}  and first solve for u.  You should find that the quadratic equation in u is factorable.


Continue with u replaced by {{{2^x}}} in the found roots of u.