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}.