Why is $49^{-\frac{1}{2}}=\frac{1}{7}$?

Question

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 $\frac{1}{7}$"

Answer 1

It is $$49^{-1/2}=(7^2)^{-1/2}=7^{-1}$$

Answer 2

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.

Answer 3

$$ a^{\frac{1}{2}}\cdot a^{\frac{1}{2}}= a^{\frac{1}{2}+\frac{1}{2}}= a^1=a. $$ 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:

$$ 49^{-\frac{1}{2}}= \left(49^{\frac{1}{2}}\right)^{-1}= \frac{1}{\sqrt{49}}=\frac{1}{7}. $$

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$$x^2=\frac{1}{49}\implies x=\pm\frac{1}{7}.$$ 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.