How do you simplify $\frac{\sqrt[5]{x^{3}}}{\sqrt[7]{x^{4}}}$ $root5(x^3)/root7(x^4)$ ?

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How do you simplify $\frac{\sqrt[5]{x^{3}}}{\sqrt[7]{x^{4}}}$ $root5(x^3)/root7(x^4)$ ?

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Answer 1 · 2016-07-30T14:51:22

$\frac{x^{\frac{3}{5}}}{x^{\frac{4}{7}}} = x^{\frac{1}{35}} = \sqrt[35]{x} \leftarrow 35th root$ $" "x^(3/5)/x^(4/7) = x^(1/35) = root(35)(x) larr" 35th root"$

Explanation:

Write as: $\frac{x^{\frac{3}{5}}}{x^{\frac{4}{7}}}$ $(x^(3/5))/(x^(4/7)$

This is the same as: $x^{\frac{3}{5} - \frac{4}{7}}$ $x^(3/5-4/7)$

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Consider the $\frac{3}{5} - \frac{4}{7}$ $3/5-4/7$

Write as $(\frac{3}{5} \times 1) - (\frac{4}{7} \times 1)$ $(3/5xx1)-(4/7xx1)$

This is the same as: $(\frac{3}{5} \times \frac{7}{7}) - (\frac{4}{7} \times \frac{5}{5})$ $(3/5xx7/7)-(4/7xx5/5)$

$= \frac{21}{35} - \frac{20}{35} = \frac{21 - 20}{35} = \frac{1}{35}$ $=21/35-20/35 = (21-20)/35= 1/35$
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So $\frac{x^{\frac{3}{5}}}{x^{\frac{4}{7}}} = x^{\frac{1}{35}} = \sqrt[35]{x} \leftarrow 35th root$ $" "x^(3/5)/x^(4/7) = x^(1/35) = root(35)(x) larr" 35th root"$

Answer 2 · 2016-07-30T14:57:35

$x^{\frac{1}{20}} = \sqrt[20]{x}$ $x^(1/20) = root20(x)$

Explanation:

If both roots were the same we could have combined them into the root of a single fraction. But they are different.

Change to index form using : $\sqrt[q]{x^{p}} = x^{\frac{p}{q}}$ $" "rootq(x^p) = x^(p/q)$

$\frac{\sqrt[5]{x^{3}}}{\sqrt[7]{x^{4}}} = \frac{x^{\frac{3}{5}}}{x^{\frac{4}{7}}} simplify using \frac{x^{m}}{x^{n}} = x^{m - n}$ $root5(x^3)/root7(x^4) = x^(3/5)/x^(4/7) " simplify using" x^m/x^n = x^(m-n)$

= $x^{\frac{3}{5} - \frac{4}{7}}$ $x^(3/5-4/7)$

= $x^{\frac{21 - 20}{35}}$ $x^((21-20)/35)$

= $x^{\frac{1}{35}} = \sqrt[35]{x}$ $x^(1/35) = root35(x)$

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How do you simplify $\frac{\sqrt[5]{x^{3}}}{\sqrt[7]{x^{4}}}$ $root5(x^3)/root7(x^4)$ ?

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