I have recently learned (discovered) that the exponent law $b^{mn} = {(b^m)}^n$ is not universally applicable. To demonstrate, if it were we could conclude that $(-1)^{\frac{3}{2}}$ (or by extension -1 to any power) is equal to 1.

$(-1)^{\frac{3}{2}} = (-1)^{2*\frac{3}{4}}$
= $((-1)^2)^{\frac{3}{4}}$
= $1^{\frac{3}{4}}$
= $1$

Under exactly what circumstances (natural/integer/rational/real bases or exponents) are the various exponent laws applicable? For example, can I use the above law ( $b^{mn} = {(b^m)}^n$ ) with a negative base when the exponents are required to be integers?

  • 1
    $\begingroup$ When $x<0$ and $n$ is even, $x^{1/n}$ creates a complex-number ambiguity that can lead to violation of the exponent rules. The same is true for $x^{m/n}$ when $n$ is even. By an approximation argument, the same is true for $x^r$ for any irrational $r$. $\endgroup$
    – Ian
    Nov 1, 2014 at 22:05

1 Answer 1


Basically, the one thing you need to remember is : $b^m$ does not have a signification when the two following condition are simultaneously met :

-b is negative

-m is not an integer

So in your example, $(-1)^{\frac{3}{2}}$ does not have any signification. Hence obviously, you should not write it.

Just remember that and you can only write correct equalities.

  • 1
    $\begingroup$ What do you mean "signification"? $(-1)^{3/2}$ has a meaning as a complex number, it's just that the exponent laws can't be used as freely. Also, avoid using "-" in a list like that, as it looks like you're saying "-b is negative", which would mean "b is positive" $\endgroup$
    – GFauxPas
    Nov 1, 2014 at 23:24

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