What are the solutions of the equation 4*2^(x^2)/2^(3x)=64?

We'll divide both sides by
4:

2^(x^2)/2^(3x) = 16

We'll
re-write the equation, using quotient property of exponentials, in this
way:

2^(x^2-3x)= 2^4

Since the
bases are matching now, we'll apply one to one rule and we'll
get:

(x^2-3x) = 4

We'll
subtract 4 both sides:

x^2 - 3x - 4 =
0

We'll apply quadratic
formula:

x1=[(-3)+sqrt(9+16)]/2

x1=(3+5)/2

x1=4

x2=[(-3)-sqrt(9+16)]/2

x2=(3-5)/2

x2=-1

The
real solutions of the given exponential equation are: {-1 ;
4}.