We are asked to solve `2^(2x)-12*2^x+32=0 ` :

Rewrite as `(2^x)^2-12*2^x+32=0 ` and let `y=2^x ` ; then

`y^2-12y+32=0 ` and

(y-8)(y-4)=0 so y=8 or y=4.

y=8 ==> ` 2^x=8 ==> x=3 `

y=4 ==> `2^x=4 ==> x=2 `

Checking we find that these are indeed solutions.

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