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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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    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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HINT

  1. $\log_x(a)$ answers the question "$x$ to the what is $a$?"
  2. You are, in fact, taking it to the power of $x$
  3. While you solve, you should also remember the domain of the logarithm.
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    I suppose there are many ways to take any problem out of its original conte$x$t to present different "solutions," but I know little of different number systems.2012-08-11
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Hint: $a^{log_a(b)}=b$ (to see why this is true look at mixdmath answer)

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    If we limit ourselves to simple answers, then we're left with "no solution in the real or complex number system".2012-08-11
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Since you are using $\log_x$, you must exclude $x = 0$ in the real or complex domain. Now let $w, z\in C$. Then we have $w^z = \exp(z L(w)), $ where $L$ is some branch of the logarithm. Note that $0$ is not in the range of the exponential function. Hence you are sunk here. There are no solutions, no matter what branch of the log you choose.

If you are in the real domain only, the logic is even simpler. You cannot raise a nonzero number to a power and get 0.

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    If you are in the real domain only, the logic is even simpler.2012-08-11