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: It is $$49^{-1/2}=(7^2)^{-1/2}=7^{-1}$$
A: $$
a^{\frac{1}{2}}\cdot a^{\frac{1}{2}}=
a^{\frac{1}{2}+\frac{1}{2}}=
a^1=a.
$$
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.
A: I see that this answer has been down-voted. I would like to know what is wrong with it. I have already made a revision and I assume that this was the only error. Thanks for your help.
$$49^{-\frac{1}{2}}=$$
$$\frac{1}{\sqrt{49}}$$
You know that: $\sqrt{49}= 7$ (by definition of square root, we only consider the positive value not the $-7$) so,
$$\frac{1}{\sqrt{49}}=+\frac{1}{7}$$ so,
$$49^{-\frac{1}{2}}=\frac{1}{\sqrt{49}}=+\frac{1}{7}$$ 
Note/Edit: I have removed the value: $-\frac{1}{7}$ from the above answer as I have reviewed the comments below and revisited that convention for $\sqrt{}$, which denotes only positive value. Also, raising a number to a positive value requires that the number be positive otherwise, the result would be a complex number, which is also not the case here.
