How do you express $\frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}}$ $x^(1/2) / x^(1/3)$ in radical form?

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How do you express $\frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}}$ $x^(1/2) / x^(1/3)$ in radical form?

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Answer 1 · 2016-09-04T09:09:59

$\frac{\sqrt[2]{x}}{\sqrt[3]{x}} or \sqrt[6]{x}$ $root(2)x/root(3)x or root(6)x$

Explanation:

$\frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}} = x^{(\frac{1}{2} - \frac{1}{3})} = x^{\frac{1}{6}} = \sqrt[6]{x}$ $x^(1/2)/x^(1/3) = x^((1/2-1/3))= x^(1/6)= root(6)x$

Answer 2 · 2016-09-04T09:17:11

$\sqrt[6]{x}$ $root6 x$

Explanation:

Use the law of indices: $\frac{x^{m}}{x^{n}} = x^{m - n}$ $" "x^m/x^n = x^(m-n)$

$\frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}} = x^{\frac{1}{2} - \frac{1}{3}}$ $x^(1/2)/x^(1/3) = x^(1/2-1/3)$

= $x^{\frac{3 - 2}{6}}$ $x^((3-2)/6)$

= $x^{\frac{1}{6}}$ $x^(1/6)$

Use the law of indices $x^{\frac{p}{q}} = {\sqrt[q]{x}}^{p}$ $x^(p/q)= rootq x^p$

= $\sqrt[6]{x}$ $root6 x$

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