I did solve, I got four solutions, but the book says there are only 3.
I considered the cases $| x - 3 | = 1$ or $3x^2 -10x + 3 = 0$.
I got for $x\leq 0$: $~2 , 3 , \frac13$
I got for $x > 0$: $~4$
Am I wrong? Is $0^0 = 1$ or NOT?
Considering the fact that : $ 2^2 = 2 \cdot2\cdot 1 $
$2^1 = 2\cdot 1$
$2^0 = 1$
$0^0$ should be $1$ right?
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