Question 21507
I hope I'm reading the problem correctly.

Solve for x:

{{{(2^(x^2))/2^x = 64}}} Simplify the left side.
{{{2^(x^2-x) = 64}}} Substitute: {{{64 = 2^6}}}
{{{2^(x^2-x) = 2^6}}} The bases (2) are equal, therefore, the exponents are equal, thus:
{{{x^2-x = 6}}} Subtract 6 from both sides.
{{{x^2-x-6 = 0}}} Solve by factoring.
{{{(x+2)(x-3) = 0}}} Apply the zero product principle.
{{{x+2 = 0}}} and/or {{{x-3 = 0}}}
If {{{x+2 = 0}}} then, {{{x = -2}}}
If {{{x-3 = 0}}} then, {{{x = 3}}}

Check:
x = -2
{{{(2^(-2)^2)/2^(-2) = (2^4)/2^(-2)}}} Simplify:
{{{(2^4)(2^2) = 2^6}}} = 64 This checks.

x = 3
{{{(2^3^2)/2^3 = (2^9)/2^3}}} Simplify.
{{{(2^9)(2^(-3)) = 2^6}}} = 64 This checks.