Is the power of 1/2 same thing as principal square root? $\sqrt{9} = 3$
9 has 2 square roots: 3 and -3.
What is $9^\frac12$? Is $9^\frac12 = \sqrt{9} = 3$ or is $9^\frac12 = \pm3$?
 A: To tell you the truth, it all depends on the given context, because the $n^{th}$ root of a certain number can be any of the n complex solutions to the equation $x^n=a$, so each radical is a choice. If your context is $\mathbb{R}$, then you look for the single real solution if n is odd, or for the positive one if n is even. But if you dabble into something a bit more serious, like $\mathbb{C}$, cubics, quartics, or Galois theory, then you have to be very careful each step of the way that the values you ascribe to each of these roots fulfill certain criteria. For instance, all the various formulae for the zeroes of the cubic or quartic equations rely heavily on which choices one makes for each radical, and if these choices fail to meet certain demands, then the formulae yield false results. E.g., at a certain point, when $\Delta_0=0$, one has to choose the value of $\sqrt{\Delta_1^2\!-\!4\Delta_0^3}=\!\sqrt{\Delta_1^2}=\!\Delta_1$ , even if the quantity in question is negative, otherwise the formula simply doesn't work !
A: $a^b$ is always positive for $a\in\Bbb R^+$ and $b\in\Bbb R$.  This is because $f(x)=a^x$ satisfies the functional equation $f(x+y)=f(x)f(y)$ (i.e. $a^{x+y}=a^xa^y$).  It follows that $f(2x)=f(x)^2,$ or $$a^x=f(x)=f\left(\frac x2\right)^2\ge 0$$
Consequently, $9^\frac12\ge 0\implies 9^\frac12=3$
A: $9^\frac{1}{2}=3.$ not $-3$. This is because $9^{\frac{1}{2}}$ is positive.
A: This previous answer of mine talks about the fact there are actually multiple exponentiation operators. The meaning of $9^{\frac{1}{2}}$ depends on which one you are actually using.
