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- anonymous

.

- katieb

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- anonymous

Convert them to exponents and then divide.
\[\sqrt[n]{a} = a^{1/n}\]

- anonymous

So for example
\[\sqrt[4]{x^3}\text{ divided by }\sqrt[3]{x^2} \]
would be
\[\frac{x^{\frac{3}{4}}}{x^{\frac{2}{3}}}\]
or \[x^{\frac{3}{4}-\frac{2}{3}}\]

- anonymous

Which would then be
\[x^{\frac{1}{12}}=\sqrt[12]{x}\]

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## More answers

- anonymous

Sometimes the test wants you to give you answer in terms of a radical, then you might have to convert it back. If they will accept the exponent version, then that is the way to go.

- anonymous

yes, what is \[\frac{x^{5/12}}{x^{1/6}}\]
Just because they want it in radical form as the answer doesn't mean that you can't do all the work in exponent form and then reconvert.
\[\frac{x^a}{x^b}=x^{a-b}\] This seems easier to me than trying to figure out the radical forms.

- anonymous

Um. No. What is \[\frac{5}{12}-\frac{1}{6}\]?

- anonymous

You are forgetting to put them into common denoms.
5\12 - 1\6 = 5\12 -2\12 = 3\12 = 1\4

- anonymous

When you divide same base different exponents you subtract the exponents from one another.
\[\frac{aaaaa}{aa}=\frac{a^5}{a^2}\]
\[\frac{aaa}{1}\frac{aa}{aa}=aaa = a^3= a^{5-2}=a^3\]

- anonymous

No, x^2/3 is wrong.
Sorry. If you want to do it in radical form you can.
\[\frac{\sqrt[12]{x^5}}{\sqrt[6]{x}}\]
can also be done realizing that
\[\sqrt[5]{a} = \sqrt[10]{a^2}\] so you can do it that way. You need to convert the index 6 to a twelve

- anonymous

Sure.

- anonymous

index 12√x^5 x index 6√x
doesn't it mean,
\(\huge \sqrt[12]{x^5} \times\sqrt[6]x\)

- anonymous

It does say to divide the expressions though.
\[\frac{\sqrt[12]{x^5}}{\sqrt[6]{x}}\to\frac{\sqrt[12]{x^5}}{\sqrt[12]{x^2}}\to \]
\[\sqrt[12]{\frac{x^5}{x^2}}\to\sqrt[12]{x^{5-2}}\to\sqrt[12]{x^3}\to \sqrt[4]{\sqrt[3]{x^3}}\to\sqrt[4]{x}\]

- anonymous

The main methods you need to know is that
1) \[\sqrt[5]{x} = \sqrt[10]{x^2}\]
2) \[\frac{\sqrt[12]{a}}{\sqrt[12]{b}}=\sqrt[12]{\frac{a}{b}}\]
Yes the fourth root of x is the answer in radical form.

- anonymous

The reason why I suggested immediately to use your exponent conversion is that, if this is algII, you probably learned the exponent rules prior to this and can then use them rather than memorize another set of rules for radicals.

- anonymous

for questions like these, you need to be able to write down the 4 or so rules of exponents and the radical methods as well. Good luck and don't give up. exponents give a lot of people confusion.

- anonymous

i hope to question was avtually to divide... its bit unclear....

- anonymous

*actually

- anonymous

yes it was. I am working on the fact that it said to divide the two expressions.

- anonymous

the 'x' in middle is throwing me off...

- anonymous

can u confirm the question abbie ?

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