SOLUTION: I need help solving (2^x^2 / 2^x) = 64. I need to solve for x. THanks

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Question 21507: I need help solving (2^x^2 / 2^x) = 64. I need to solve for x. THanks
Answer by Earlsdon(6287) About Me  (Show Source):
You can put this solution on YOUR website!
I hope I'm reading the problem correctly.
Solve for x:
%282%5E%28x%5E2%29%29%2F2%5Ex+=+64 Simplify the left side.
2%5E%28x%5E2-x%29+=+64 Substitute: 64+=+2%5E6
2%5E%28x%5E2-x%29+=+2%5E6 The bases (2) are equal, therefore, the exponents are equal, thus:
x%5E2-x+=+6 Subtract 6 from both sides.
x%5E2-x-6+=+0 Solve by factoring.
%28x%2B2%29%28x-3%29+=+0 Apply the zero product principle.
x%2B2+=+0 and/or x-3+=+0
If x%2B2+=+0 then, x+=+-2
If x-3+=+0 then, x+=+3
Check:
x = -2
%282%5E%28-2%29%5E2%29%2F2%5E%28-2%29+=+%282%5E4%29%2F2%5E%28-2%29 Simplify:
%282%5E4%29%282%5E2%29+=+2%5E6 = 64 This checks.
x = 3
%282%5E3%5E2%29%2F2%5E3+=+%282%5E9%29%2F2%5E3 Simplify.
%282%5E9%29%282%5E%28-3%29%29+=+2%5E6 = 64 This checks.