Why is $49^{-\frac{1}{2}}=\frac{1}{7}$? Why is $49^{-\frac{1}{2}}=\frac{1}{7}$?
I know that $\sqrt[n]{m^p}=m^{\frac{p}{n}}$ so I figured I can state the above is $-\sqrt{49}=-7$, but that is incorrect. I can't put the negative inside the root because the solution doesn't contain the imaginary number, $i$. So I'm stuck here.  
I also know I can just set the equation equal to $x$ and easily get the solution as follows
$$x=49^{-\frac{1}{2}}$$
$$x^2=49^{-1}$$
$$x=\pm\frac{1}{7}$$
But can anybody give me an intuitive explanation why this is the answer? I'd like to be able to just look at the expression and think "ah, that's obviously $\frac{1}{7}$"

$$
a^{\frac{1}{2}}\cdot a^{\frac{1}{2}}=
a^{\frac{1}{2}+\frac{1}{2}}=
a^1=a.
$$
It follows logically from the above that $a^{\frac{1}{2}}$ is that number which, when multiplied by itself, gives you $a$. This in turn means that $a^{\frac{1}{2}}$ must be $\sqrt{a}$ because $\sqrt{a}\cdot\sqrt{a}=a$ (also note that $a\ge0$ because the square root function is only defined for nonnegative numbers).
$$
b^{-1}\cdot b=b^{-1+1}=b^0=1\\
b^{-1}\cdot b=1\implies\\
b^{-1}=\frac{1}{b},\ b\ne0.
$$
Therefore:
$$
49^{-\frac{1}{2}}=
\left(49^{\frac{1}{2}}\right)^{-1}=
\frac{1}{\sqrt{49}}=\frac{1}{7}.
$$

$$x^2=\frac{1}{49}\implies x=\pm\frac{1}{7}.$$
This is indeed true (just plug in $\frac{1}{7}$ and $-\frac{1}{7}$ back into $x$). But that is different from $49^{-\frac{1}{2}}$. $x^2=\frac{1}{49}$ is an equation with two solutions and $49^{-\frac{1}{2}}$ is a single number. The statement $x=49^{-\frac{1}{2}}$ received an extra solution precisely at that moment when you squared both sides.
