Exponential equation solving

$3$ cases.

$1.$ $x\neq 0$ , $x^2-5x+6=0$ gives $x=2,3$ .

$2.$ $x=1$ , $x^2-5x+6=2$ .

$3.$ $x=-1$ , $x^2-5x+6=12$ which is even.

Answer: $2+3+1+1=\fbox 7$ .

To have $a^b=1$ , we may have one of three possibilities; (1) $a=1$ , (2) $a=-1$ with $b$ even or (3) $b=0$ , provided $a\neq 0$ .

Case (1): we have $x=1$ as a solution trivially.

Case (2): we take $x=-1$ and verify that the exponent is even:

$(-1)^2-5(-1)+6=1+5+6=12$ , which is even, so $x=-1$ is a solution.

Case (3): we solve $x^2-5x+6=0$ . Factoring, $(x-2)(x-3)=0$ , giving $x=2$ and $x=3$ , which are both nonzero, thus also solutions.

We answer: $|-1|+|1|+|2|+|3|=1+1+2+3=\boxed{7}$

$\textbf{Solution Outline:}$

Taking log on both sides we get $(x^{2}-5x+6)ln(x) = 0$

So either we have $x^{2}-5x + 6 = 0$ or $ln(|x|)$ = 0.