For $x, y, z \in \mathbb{R}$, $(x^y)^z = x^{yz}$?

Question

Is it always true that for $x, y, z \in \mathbb{R}$, $(x^y)^z = x^{yz}$, whenever both expressions are defined? Assume all are nonzero.

I think this isn't true in general, because for instance

$$-1 = (-1)^{2/2}$$

but

$$((-1)^2)^{1/2} = 1^{1/2} = 1$$

Answer 1

The answer is no. The problem is that fractional powers do not have a unique definition, i.e. they are not single-valued functions. It does not make sense to say the definition of $x^{1/5}$.

If $x$ is real and positive, there is a unique positive real whose fifth power is $x$. There are also nonreal complex numbers whose fifth power is $x$, and they are arguably better choices than the real version.

Because roots are not single-valued, we must make a choice whenever we take a root. For the square root of a positive real number, by convention we normally take the positive real square root; however there is no convention for more general roots.

In the problem posed, there are three opportunities to take a root: via $y$, via $z$ and via $yz$. Depending on the choices made, the two sides may agree or not.

Answer 2

There are two definitions for powers:

• Algebraically, if the exponent is an integer, we use repeated multiplication (or reciprocals thereof). And algebraically we also have $0^0=1$, by the way.
• Analytically, $a^b=\exp(b\ln a)$ if $a>0$.

In both cases, we obtain the law $(x^y)^z=x^{yz}$, fo example here's the proof for the ananlytical definition: $$(x^y)^z=\exp(z\cdot \underbrace{\ln(\exp}_{\text{cancel}}(y\ln x)))= \exp(zy\ln x))=x^{yz}.$$ And here the algebraic (here only for positive $y,z$): $$ (x^y)^z=\underbrace{(\underbrace{x\cdot\ldots \cdot x}_y)\cdot\ldots\cdot (\underbrace{x\cdot\ldots \cdot x}_y)}_z=\underbrace{x\cdot\ldots\cdot x}_{yz}=x^{yz}.$$

The algebraic and the analytic definition agree wherever both apply, but mixing both nevertheless does not work and you gave a suitable example: $(-1)^2=1$ requires the algebraic, $1^{\frac12}$ the analytic definition. Neither the algebraic nor the analytic proof apply so that we cannot conclude that $((-1)^2)^{\frac12}$ equals $(-1)^{2\cdot\frac12}$.