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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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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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    Unless you're suggesting an approach differing from @RaymondManzoni in a comment to the question, this leaves you with having to take the log of 0. I.e., it's like being given $x/x = 0$ and trying to solve by multiplying both sides by $x$—one first has to assert that $x$ is not zero.2012-08-11
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    @Ben: I would argue that I'm taking the precalculus approach. You find out that you are solving for the roots of the quadratic, so that if there is a solution, it would be of that form. But $0$ is not in the domain of the logarithm, and thus it's not a solution afterall. Thus there is no solution. Note that $x = 0$ is not the 'proposed' solution - we 'want' $x = 1,-2$. Neither work.2012-08-11
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    Personally, I'm interested in seeing a rigorous solution flushed out in a consistent number system where there is a solution, if such a number system exists. I suppose one trick is to do use the extended reals that @copper.hat referenced and define ${\log}_1(0)$ to be ${\log}_1(0^+)$ (which would at least give you a solution for x=1), but then I'm not 100% convinced that such a number system is internally consistent.2012-08-11
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    Of course, we could always try a different number system where x=0 is a valid solution by trying to create a definition for ${\log}_0(-2)$.2012-08-11
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    I suppose there are many ways to take any problem out of its original context 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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    This is true as long as $b$ doesn't equal 0. Unfortunately, in this case, it *does* equal 0.2012-08-11
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    @BenHocking - $log(0)$ is undefined, you should only consider values of $x$ where the expression is defined - that is $x>0$ and $x^2+x-2>0$2012-08-11
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    That's what I was getting at. (However, ${\log}(x)$ for x < 0 is well defined in the complex number system, and ${\log}(0^+)$ is defined in the extended real number system. We're still left with the problem that the right hand side is $0$ and not $0^+$)2012-08-11
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    @BenHocking - this was tagged algebra-**precalculus**2012-08-11
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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