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If I raise both sides of an equation to the same negative exponent will the equality remain? $$a = b$$

Jean Marie
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Omicron
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    Short answer: yes – Jacob Claassen May 06 '17 at 22:29
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    as long as it is not the case: a=b=0 yes. – Xaqron May 06 '17 at 22:31
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    If $a=b$ then $f(a)=f(b)$ for any and all functions $f$. This is one of the requirements for something to be defined as a function in the first place. All that remains is noting that $f(x)=x^{\alpha}$ is a welldefined function from the positive reals to the reals or from the complex numbers to the complex numbers for any and all real values of $\alpha$ including negative values. – JMoravitz May 06 '17 at 22:33
  • I think, @JMoravitz, that for $\alpha=1/2$, the function $x\mapsto x^\alpha$ is not well defined on the complex numbers. – Lubin May 06 '17 at 22:51
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    Your difficulty, @Omicron, is that your mathematics teachers did not stress enough the meaning of the symbol “$=$”. When we write “$a=b$”, we mean that “$a$” and “$b$” are two different names for the same thing; in your case, for the same number. So we are not talking about two numbers here, but one number that happens to have two names. Therefore, when you apply a well-defined function to this number, something happens to it, but whatever happens to it is not dependent on what you happened to name it. – Lubin May 06 '17 at 22:55
  • @lubin, unless specified I was under the impression that the notation without any additional context implies that we refer to the output within the principle branch, e.g. $\sqrt{i}=\frac{\sqrt{2}}{2}(1+i)$, which should make it well defined, the same way we do so for the mapping from the positive reals to the reals. – JMoravitz May 06 '17 at 22:57
  • @JMoravitz, that’s a mere convention. The artificiality of this convention becomes apparent when we try to cross the cut that turns $\Bbb C\setminus{0}$ into a simply connected domain. For special purposes, we may want to locate the slit elsewhere than where it’s most often placed. Then we get a well-defined power-function $x^\alpha$, but a different one to the one that you’re thinking of. I suppose, though, that it may boil down to a philosophical difference between us. – Lubin May 06 '17 at 23:03

2 Answers2

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Yes. Let's say $a, b, \alpha$ are all nonzero real numbers, and $\alpha$ is negative. Then, if $a = b$, then $a^\alpha = b^\alpha$ as well.

Remember also that $$a^\alpha = \frac{1}{a^{|\alpha|}}.$$

For example, let's say we're trying to solve $$x^2 + 3 = 7.$$ I suppose the obvious thing to do here would be to subtract 3 from both sides. But instead let's raise both sides to the power of $-2$: $$(x^2 + 3)^{-2} = 7^{-2}$$ $$\frac{1}{(x^2 + 3)^2} = \frac{1}{49}$$ Then $$(x^2 + 3)^2 = 49$$ $$\sqrt{(x^2 + 3)^2} = 7$$ $$x^2 + 3 = 7.$$

Well, we're back where we started. Does this make any sense or did I just confuse you further?

Robert Soupe
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    Not that I question your knowedge of this, but I think it should be mentioned that your last step works only because $x^2+3 \ge 0 , \forall x \in \mathbb R$. If you had e.g. $x^2-3=2$ instead, you'd get $|x^2-3| = 2$, which has four solutions - $\pm 1, \pm \sqrt 5$ and not two, as was in the beginning. – Al.G. Mar 28 '20 at 21:43
  • Thank you for pointing that out. I would've thanked you sooner, but I'd been off the grid for a long time and I'm still catching up with online stuff. – Robert Soupe Apr 09 '21 at 23:05
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Given $$a=b$$ you want to know if $$a^n = b^n$$ for some negative value of $n.$

Since $a=b$, let's replace them both with $c$ in the first equation: $$c=c$$ Now let's put both sides of the equation to the power of $n$: $$c^n=c^n$$ Finally, let's replace the first $c$ with $a$ (because $a=c$) and the second $c$ with $b$ (same reasoning​): $$a^n=b^n$$ Looks good.

MathAdam
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