How do you find all real solutions to the following equation: ${(x^{2} - 7 x + 11)}^{x^{2} - 2 x - 35} = 1$ $(x^2-7x+11)^(x^2-2x-35)=1$ ?

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How do you find all real solutions to the following equation: ${(x^{2} - 7 x + 11)}^{x^{2} - 2 x - 35} = 1$ $(x^2-7x+11)^(x^2-2x-35)=1$ ?

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Answer 1 · 2017-02-12T23:14:19

$x = {- 5, 7}$ $x={-5,7}$

Explanation:

We know that $a^{0} = 1$ $a^0=1$ so making $x^{2} - 2 x - 35 = 0$ $x^2-2x-35=0$ the solutions are $x = {- 5, 7}$ $x={-5,7}$ but we know also that $a$ $a$ must be non null. Solving now

$x^{2} - 7 x + 11 = 0$ $x^2 - 7 x + 11 =0$ we have $x = \frac{1}{2} (7 \pm \sqrt{5}) \neq {- 5, 7}$ $x = 1/2 (7 pm sqrt[5]) ne {-5,7}$

Finally the equation is meaningful for

$x = {- 5, 7}$ $x={-5,7}$

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How do you find all real solutions to the following equation: ${(x^{2} - 7 x + 11)}^{x^{2} - 2 x - 35} = 1$ $(x^2-7x+11)^(x^2-2x-35)=1$ ?

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Explanation:

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How do you find all real solutions to the following equation: (x^2-7x+11)^(x^2-2x-35)=1(x2−7x+11)x2−2x−35=1(x^2-7x+11)^(x^2-2x-35)=1?

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Explanation:

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How do you find all real solutions to the following equation: ${(x^{2} - 7 x + 11)}^{x^{2} - 2 x - 35} = 1$ $(x^2-7x+11)^(x^2-2x-35)=1$ ?