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

Feb 12, 2017

$x = \left\{- 5 , 7\right\}$

#### Explanation:

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

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

Finally the equation is meaningful for

$x = \left\{- 5 , 7\right\}$