# Exponent Law Question: $\left(x^2\right)^\frac{1}{2} = |x|$

Quick easy question! I was always taught at school that $(x^m)^n=x^{mn}$. However, I've only just noticed when computing something (two years into an undergrad course..) that for example $(x^2)^{\frac{1}{2}}=|x|$ rather than $x$. I've never really thought about it before. A lot of the time when simplifying I would have blindly done things like $\sqrt{x^4+x^2}=x\sqrt{x^2+1}$ which is clearly wrong now I've looked at it.

Is there a general rule I have not been taught before?

• The rule $(x^m)^n=x^{mn}$ is true if $m$ and $n$ are integers. Dec 28, 2013 at 21:48
• I think that pretty much answers my question. Dec 28, 2013 at 21:50
• Alternatively, $(x^m)^n = x^{mn}$ is also true when $m$ and $n$ are arbitrary reals but $x > 0$. Dec 28, 2013 at 21:55
• Taking non-integer powers of non-positive numbers is often a non-safe thing to do. Caveat exponentiator. Dec 28, 2013 at 22:47

$(x^2)^{1/2} = |x|$ is correct for real $x$. But incorrect (in general) for complex $x$.
For example, if $x=i$, the equation would be $(-1)^{1/2}=1$ which is wrong.
For complex $x$, write $\overline{x}$ for the complex conjugate of $x$, then we do have $\left(x \;\overline{x}\right)^{1/2} = |x|$ for all complex $x$.
• It was more a question about what I (thought!) I was taught rather than the example, but a good answer nonetheless. In the complex example, would $\sqrt{z^4+z^2}=z\sqrt{z^2+1}$ be correct then? Dec 28, 2013 at 22:16