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
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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}.
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