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Let's say we have 2 = 2^(2/2), then using the multiplicative property of exponents to turn it into a square root: ((2)^2)^(1/2) = 2, it just works.

But in the case of -2 = (-2)^(2/2), if we use the property: ((-2)^2)^(1/2) = 2, because the ^1/2 notation means the principal root, we must pick the positive result, which is 2.

Basically (-2)^(2/2) != ((-2)^2)^(1/2), which proves that the multiplicative property of exponents doesn't apply to all bases, what is bothering me is where this property doesn't work? For example, -2 = (-2)^(3/3) = ((-2)^3)^(1/3) = -2, it works. I have a feeling that it is when the denominator is even and the base is negative, but i can't prove it or make a logical reasoning to why and where it doesn't holds.

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In general, the rule $a^{bc}=(a^b)^c$ only applies for non-negative values of $a$ (and always remember that $0^0$ is undetermined). This is because general exponentiation is usually defined via

$$a^b:= \exp(b\cdot\log(a))$$

where $\exp$ and $\log$ can be defined by e.g. power series or differential equations. The natural logarithm $\log(a)$ has a canonical value only for positive $a$, and we can extend the defnition to $a=0,b\not=0$ by setting $0^b=0$.

Writing e.g. $(-2)^{1/2}$ requires more explanation and some arbitrary choices in order to uniquely determine the value. It is dangerous to write it this way because of this ambiguity. One preferably talks about "the two solutions of $x^2=-2$" instead.


One other situation in which the rule $a^{bc}=(a^b)^c$ can be applied to all $a$, is when $b$ and $c$ are integers (still $0^0$ is excluded). That this works out well with the definition of $a^b$ given above requires some knowledge about complex numbers (I can add the reasoning if requested).

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  • $\begingroup$ Thanks for the answer, i should note that i'm not familiar with complex analysis, but a friend told me my question is related to it, that's why the tag. It seems strange to me to define exponentiation with a logarithm, but it kinda makes sense, guess i should finish calculus first. $\endgroup$ – Karine Silva Jun 17 '18 at 20:29
  • $\begingroup$ What is bothering me is that in some cases like: -2 = (-2)^(3/3) = ((-2)^3)^(1/3) = -2 the property works as intended. $\endgroup$ – Karine Silva Jun 17 '18 at 20:45
  • $\begingroup$ @KarineSilva Also in the case $(-2)^{1/3}$ there are several possible "solutions". One is the real valued one $-1.2599...$ that you probably thought about, the other two are complex valued, so this might be the reason "you cannot see them". This also happens for positive numbers, e.g. $2^{1/3}$ also has three "solutions", but there is consensus to only take the positive one (the one the calculator spits out), which always exists (for positive numbers). $\endgroup$ – M. Winter Jun 17 '18 at 21:06
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It does not work, because $$\sqrt{x^2} \neq \sqrt{x}^2$$ but $$\sqrt{|x|^2}=\sqrt{|x|}^2$$ Because the $\sqrt{x}$ function is the inverse of $f:\mathbb{R}_+\cup\{0\} \to \mathbb{R}_+\cup\{0\}$, $f(x)=x^2$.

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Actually the multiplicative property of exponents is valid regardless of the base but I presume that you would require prior knowledge about complex number analysis. What you say is correct if we are dealing with numbers that are strictly real and therefore can't have imaginary or complex solutions. Hope this helps...

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