<s>[INST] Exponential equation with same, unknown bases I have the following equation:
$(x-3)^{(x^2-x)} = (x-3)^2$
The book says solutions are: $x_1 = -1, x_2 = 2, x_3 = 3, x_4 = 4$
I was only able to get -1, 2, 4 by doing this:
$(x-3)^{(x^2-x)} = (x-3)^2 / \log$
$(x^2-x) \cdot \log(x-3) = 2\log(x-3)$ (exponents out)
$(x^2-x) \cdot \log(x-3) - 2\log(x-3) = 0$
$\log(x-3)(x^2-x - 2) = 0$
$\log(x-3) = 0 \; \lor \; x^2-x-2=0 $
$x-3=1 \; \lor \; x^2-x-2=0 $
$x_1 = 4, \; x_2 = -1, \; x_3 = 2$
The explanation in the book says only this:
$x-3 = 1 \; \lor \; (x-3 = 0 \; \land \; x^2-x \ne 0) \; \lor \; (x-3 \ne 0 \; \land \; x^2-x=2)$
What exactly am I missing here? I don't quite undersand the book's explanation though.
 [/INST] If $a^m=a^n$
either $m=n$
or $a=0, mn>0$
or $a=1$
or $a=-1,m-n=$  some even integer
By taking logarithm, we implicitly assume $a\ne0$
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