# How to solve the exponential equation $x^{\log_x (x^2 + x - 2)}=0$ for $x$?

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 $\ \displaystyle e^{\log(x^2+x-2)}=x^2+x-2$ –  Raymond Manzoni Aug 11 '12 at 15:20
@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… –  Ben Hocking Aug 11 '12 at 15:22
@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! :-)). –  Raymond Manzoni Aug 11 '12 at 15:30
There is no solution in the reals or complex. Maybe you need to consider solutions in the extended real line? –  copper.hat Aug 11 '12 at 15:31

## 3 Answers

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. –  Ben Hocking Aug 11 '12 at 15:32
@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. –  mixedmath Aug 11 '12 at 15:37
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. –  Ben Hocking Aug 11 '12 at 15:44
Of course, we could always try a different number system where x=0 is a valid solution by trying to create a definition for ${\log}_0(-2)$. –  Ben Hocking Aug 11 '12 at 15:50
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. –  mixedmath Aug 11 '12 at 16:16

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. –  Ben Hocking Aug 11 '12 at 15:46
@BenHocking - $log(0)$ is undefined, you should only consider values of $x$ where the expression is defined - that is $x>0$ and $x^2+x-2>0$ –  Belgi Aug 11 '12 at 15:53
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 $0^+$) –  Ben Hocking Aug 11 '12 at 16:02
@BenHocking - this was tagged algebra-precalculus –  Belgi Aug 11 '12 at 16:11
If we limit ourselves to simple answers, then we're left with "no solution in the real or complex number system". –  Ben Hocking Aug 11 '12 at 16:15

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. –  ncmathsadist Aug 11 '12 at 16:14