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

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We have to solve for x: 2^( x^2 - 3x + 2) = 4^3

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

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

as the base 2 is equal we equate the exponent.

x^2 - 3x + 2 = 6

=> x^2 - 3x -4 = 0

=> x^2 - 4x + x - 4 = 0

=> x(x - 4) +1(x - 4) = 0

=> (x + 1)(x - 4) = 0

**=> x = -1 and x = 4**

2^(x^2-3x+2)=64 we know 2^6=64.

so, x^2-3x+2=6

**transfer 6 to left hand side.**

so, x^2-3x-4=0

so, x^2-4x + x-4=0

**take common.**

x(x-4)+1(x-4)=0

so, (x+1)(x-4)=0

**then it must have 2 solutions.**

x={-1,4}. Check which one helps you in solving the sum.

THANKS.

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}.**