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January 30, 2015

January 30, 2015

Posted by **Tammy** on Tuesday, July 24, 2007 at 9:39pm.

I am supposed to use the rule

a^(m/n)= n

(the n-root) (don't know how to type it) and under the root symbol a^m

and the other rule is (a^x)(a^b)= a^(x+b)

The book calls this Radical Ecponents and Radical Expretions.

My problem is that it only says X, another problem I had, I can solve

(I will use ,- to symbolise the root)

to give you an idea on another problem how this is supposed to be solved

[y^ (-1/4)][y(3/4)] (multiply the two terms)

Following the rule

4,- [y^(-1)][y^3] (the for stands for the 4th root and the rest is under the root symbol)

using the rule of multiplication of the powers you then get

4,- y^2

which is the same as the following and keeping in mind that there has to be some number to the power of 4.

4,- (y^(1/2)^4)

The fours cancel and then you have y^ (1/2)

Note that (1/2)4 is the same thing as 2 so it's like factoring too.

I just can't do the problem I showed you. I hope I gave you enough info to help me out!!

Thanks you!

(Original question below)

Please read and then help me!

(x^(-3/5))

----------- divide

x ^ (1/5)

I know the problem then needs to look like this:

(x^(-3/5))(x^(-1/5))

(multiply the two above)

Then it needs to look like this:

(I can't write it though, so I'll tell you)

The 5th root out of (x^(-3)(x^(-1)

Then that adds up to the 5th root out of x^(-4)

and this is were I need help. To get rid of the fifth root there neeeds to be something to the power of 5 under the root symbol, but how do I get there? If my math to this point isn't correct, please correct it. I need Help!

I thought I had answered your question before.

Your original expression of

(x^(-3/5))

----------- divide

x ^ (1/5)

reduces or can be simplified to x^(-4/5)

Other than writing this in several different forms such as (x^-4)^(1/5) or (1/x^4)^(1/5)

there is no need or no method to "get rid of" the fifth power unless you have a value for x

If you had an equation containing that expression there would be ways to solve for x by manipulating the fifth power using power rules.

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