# Can I write $(x^p)^{q} = x^{pq}$, where either of $p$ or $q$ are rational.

I am confused with a very basic algebra question about the following law of exponents.

We know that $(x^n)^{m} = x^{nm}$, holds true for real $x$ and integer exponents $n, m$. I want to know whether this result holds for rational exponents aswell?

For example can I write $(x^p)^{q} = x^{pq}$, where either of $p$ or $q$ are rational.

Kindly pardon me for asking a very silly question.

Thanks for the help.

• Yes you can write like that. Jul 14, 2013 at 7:02
• You need $x$ to be positive, to avoid getting into complex numbers and branches. Is your $x$ positive? Jul 14, 2013 at 7:07
• @ShreevatsaR Ya $x$ is positive. Jul 14, 2013 at 7:08
• I believe you can do it, but keep in mind that in general it fails, say $(x^2)^\frac{1}{2} = \sqrt{x^2} = |x| \neq x = x^{2 \times \frac{1}{2}}$. But, when you restrict $x$ to be positive, then everything's fine, since $|x| = x$ for $x \ge 0$. Jul 14, 2013 at 7:50
• @user49685 thank you very much Jul 14, 2013 at 8:56

Let us first recall the conventional definition of rational exponentiation:$$x^{\frac{m}{n}} = \left ( x^{\frac{1}{n}} \right )^m = \left ( \sqrt[n]{x} \right ) ^m,$$where $$m, n \in \mathbb{Z}$$ and $$x \in \mathbb{R}$$ ($$x \neq 0$$ if need be). So, $$x^{\frac{m}{n}}$$ is defined whenever $$\sqrt[n]{x}$$ is defined ($$x\neq 0$$ if need be).

It should be noted that, according to the definition, we do not have the same result for different representations of a rational exponent; that is, $$\frac{m}{n} = \frac{m'}{n'}$$ does not necessarily imply that$$x^{\frac{m}{n}} = x^{\frac{m'}{n'}}$$($$x\neq 0$$ if need be) because one of the sides may not be defined, for example, while $$\frac{1}{3} = \frac{2}{6}$$, we do not have the identity$$(-1)^{\frac{1}{3}} = (-1)^{\frac{2}{6}},$$noting that the right hand side is not defined since $$\sqrt{-1}$$ is not defined.

Now, suppose that $$x^p = x^{\frac{m}{n}}$$ and $$x^q = x^{\frac{r}{s}}$$, where $$m, n, r, s \in \mathbb{Z}$$ and $$x\in \mathbb{R}$$ ($$x\neq 0$$ if need be), are defined. So we have \begin{align*}\left (x^p \right )^q & = \left ( x^{\frac{m}{n}} \right )^{\frac{r}{s}}\\ & = \left ( \left ( \sqrt[n]{x} \right )^m \right )^{\frac{r}{s}}\\ & = \left (\sqrt[s]{\left ( \sqrt[n]{x} \right )^m} \right )^r\\ & = \left ( \left ( \sqrt[s]{\sqrt[n]{x}} \right )^m \right )^r\\ & = \left (\sqrt[ns]{x} \right )^{mr}\\ & = x^{\frac{m}{n} \frac{r}{s}}\\ & = x^{pq} \end{align*} In the derivation above we used the following facts:

• the fact that$$\left (x^m \right )^n = x^{mn},$$where $$m, n \in \mathbb{Z}$$ and both sides are defined, which can be easily shown by induction;
• the fact that$$\sqrt[n]{\sqrt[m]{x}}=\sqrt[mn]{x},$$where $$m, n \in \mathbb{Z}$$ and both sides are defined, which follows directly from the previous fact and the definition of the $$n$$th root of a real number;
• the fact that$$\sqrt[n]{x^m}=\left (\sqrt[n]{x} \right )^m,$$where $$m, n \in \mathbb{Z}$$ and both sides are defined, which follows directly from the definition of the $$n$$th root of a real number.

Please note that although it is correct to say that the identity holds for positive real numbers $$x$$, from the derivation above we see that the identity holds whenever all the terms involved in the derivation, namely, $$\sqrt[n]{x}$$ and $$\sqrt[s]{\sqrt[n]{x}}=\sqrt[ns]{x}$$, are defined; for example, the identity$$\left ( (-1)^{\frac{1}{3}} \right )^{\frac{3}{5}} = (-1)^{\frac{1}{5}}$$holds because the terms $$\sqrt{-1}$$ and $$\sqrt{\sqrt{-1}}$$ are defined.

P.S. No question at any level which is asked to gain knowledge is silly.

• the fact that$$\left (x^m \right )^n = x^{mn},$$where $m, n \in \mathbb{Z}$ and both sides are defined." it is always defined right? Sep 25 at 11:06
• You said $$\left ( (-1)^{\frac{1}{3}} \right )^{\frac{3}{5}} = (-1)^{\frac{1}{5}}$$holds Bcz all terms are defined .But if we take $$\left ( (-1)^{\frac{2}{3}} \right )^{\frac{3}{2}} = (-1)^{1}$$ then all the terms are defined but clearly both are not equal . what is the problem with above one? Sep 25 at 11:25
• @MeetPatel First off, thank you for your comments. Regarding your first comment, $x^n$, where $n$ is an integer, is not defined when $x=0$ and $n<0$, avoiding division by zero. For your second comment, you are right; my wording is not proper here. Please let me edit it. Sep 25 at 14:17
• @MeetPatel Now the problem was resolved? Sep 25 at 14:31
• Yes , Thank you . Sep 25 at 15:19

As long as x is positive, the rule holds true for rational exponents p and q. When x is negative, you run into issues with complex numbers.