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How to solve this equation? $x$ can never be equal to $0$ nor its exponent. Am I right?

$$\large x^{\log_x (x^2 + x - 2)}=0$$

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    Hint: rewrite the L.H.S as $\ \displaystyle e^{\log(x^2+x-2)}=x^2+x-2$2012-08-11
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    @RaymondManzoni: That was my first thought, too, but if you plug those solutions back into the original equation you end up with $x^{log_x(0)} = 0$ (where x is now either -2 or +1), but you still have an issue with taking the log of 0…2012-08-11
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    @BenHocking: if you want an exponential (or a power) to return $0$ you need $-\infty$ as exponent and $\log(0^+)$ will produce it. Giving perfect mathematical sense to this problem is another matter that didn't interest me i.e. you may go as near as $0$ as you wish but won't go to $0$ (that's why I only commented! :-)).2012-08-11
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    There is no solution in the reals or complex. Maybe you need to consider solutions in the extended real line?2012-08-11

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