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What makes $\sqrt[7]{9}$ = $9^\frac{1}{7}$ Can this be explained using laws of indices?

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+1. I always took $\sqrt[n]{x} = x^{\frac{1}{n}}$ for granted! – user2468 Mar 13 '12 at 16:46
-1. What is your definition of $x^{1\over n}$? How is it related to $\sqrt[n]{x}$? Without knowing what is meant, there is no point in using various formulas! – Blah Mar 13 '12 at 18:29
up vote 4 down vote accepted

Well, do you agree that $\sqrt[7]9$ is a number such that $(\sqrt[7]9)^7=9$?

Now, suppose $$\begin{align*}\sqrt[7]9&=9^x\\ (\sqrt[7]9)^7&=(9^x)^7 \tag 1\\ 9^{7x}&=9 \tag 2\\7x&=1 \tag 3\\ x&=\dfrac 1 7\end{align*}$$

We have used in going from $(1)$ to $(2)$ that,


In going from $(2)$ to $(3)$ we use the fact that, $$x^a=x^b \implies a=b$$

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$(9^{1/7})^7 = 9^{(1/7) \cdot 7} = 9^1 = 9$

This shows that $9^{1/7}$ really is a 7th root of 9.

In general, $(x^{a})^b = x^{ab}$. Assuming $x \geq 0$ and $a > 0$ an integer, we can use this to see that, more generally,

$$\sqrt[a]{x} = x^{1/a}$$

Does that answer your question?

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how come $(9^\frac{1}{7})^7 = 9^[(\frac{1}{7})\dot7]$? – alok Mar 13 '12 at 16:28
@alok $(x^a)^b = x^{ab}$ for $x \geq 0$ and $a, b$ real numbers, is a general fact, the proof of which requires calculus. It is in most calculus books. At lower levels, it is just given as a fact. You can prove it for $a, b$ positive integers using induction. – Graphth Mar 13 '12 at 16:31

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