SOLUTION: Solve (x² - 7x + 13)^(x² - 3x + 2) = 1

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Question 1146365: Solve
(x² - 7x + 13)^(x² - 3x + 2) = 1
Found 2 solutions by mini_, ikleyn:
Answer by mini_(3)   (Show Source): You can put this solution on YOUR website!
According to the Zero Power Rule: any non-zero number that raised to the zero power equals 1 (one).
So if the result of an exponentiation (power raising) is 1, then the exponent (the power value) might be equal only to 0 (zero).
So (x^2 -3*x + 2) = 0.
Let us solve the given quadratic equation:
D = (-3)^2 - 4*1*2 = 9 - 8 = 1.
We have 2 roots:
x1 = (-(-3) + 1) / 2 = (3+1)/2 = 2;
x2 = (-(-3) - 1) / 2 = (3-1)/2 = 1.
Answer by ikleyn(52834)   (Show Source): You can put this solution on YOUR website!
.
Either


    x^2 - 7x + 13 = 1,      (1)


or


    x^2 - 3x + 2 = 0.       (2)


You should consider BOTH opportunities and find the set of solutions for each of the two equations (1) and (2).


Their union will be the solution set for the entire problem.


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