<s>[INST] What is the number of real solutions of the equation $ | x - 3 | ^ { 3x^2 -10x + 3 } = 1 $? 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?
 [/INST] First of all notice that $0^0\ne 1$
$$|x-3|^{3x^2-10x+3}=1\Longleftrightarrow$$
$$\ln\left(|x-3|^{3x^2-10x+3}\right)=\ln(1)\Longleftrightarrow$$
$$\ln\left(|x-3|\right)\left(3x^2-10x+3\right)=0\Longleftrightarrow$$
$$\ln\left(|x-3|\right)\left(x-3\right)\left(3x-1\right)=0$$

Split $\ln\left(|x-3|\right)\left(x-3\right)\left(3x-1\right)$ into seperate parts with additional assumptions:

So we got:


*

*$$\ln\left(|x-3|\right)=0\Longleftrightarrow$$
$$|x-3|=1\Longleftrightarrow$$
$$x-3=\pm 1\Longleftrightarrow$$
$$x=\pm 1+3$$

*$$x-3=0\to\text{not a valid solution}$$

*$$3x-1=0\Longleftrightarrow$$
$$3x=1\Longleftrightarrow$$
$$x=\frac{1}{3}$$


So at the end we got 3 solutions:
$$x_1=\frac{1}{3}$$
$$x_2=2$$
$$x_3=4$$
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