Solve for x 2^(x^2-3x+2) = 4^3

We'll create matching bases both sides. For this reason,
we'll re-write 4^3 = (2^2)^3 = 2^2*3 = 2^6

We'll re-write
the equation in this way:

2^(x^2-3x+2)=
2^6

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

(x^2-3x+2)=6

We'll
subtract 6 both
sides:

x^2-3x+2-6=0

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