How Do You Handle Exponents Inside a Radical?

[Cornraker]

Homework Statement

When there are exponents in a radicand as shown below, do i just divide them out?

Homework Equations

^{∛(x^6 y^4 )}

The Attempt at a Solution

x^2 y^4 ?

 

[sylas]

Homework Statement

When there are exponents in a radicand as shown below, do i just divide them out?

Homework Equations

^{∛(x^6 y^4 )}

The Attempt at a Solution

x^2 y^4 ?

You are asking about

\sqrt[3]{x^6 y^4}​

Given your bracketing the y⁴ is inside the cube root as well. Note that you can write

\sqrt[3]{x^6 y^4} = \sqrt[3]{x^6} \sqrt[3]{y^4}​

 

[Cornraker]

But to simplify it further to i divide the exponents of the variables by the index of the radical? I'm sorry if my terminology is bad and if my equations aren't right I'm new to this forum so I'm trying to get it right.

 

[sylas]

But to simplify it further to i divide the exponents of the variables by the index of the radical? I'm sorry if my terminology is bad and if my equations aren't right I'm new to this forum so I'm trying to get it right.

Yes. Your original answer only divided one of the exponents, but from the brackets it appears both x^6 and y^4 are inside the radical. The radical notion is just another way of writing an inverted exponent. That is

\sqrt[n]{a} = a^{1/n}​

 

[HallsofIvy]

But to simplify it further to i divide the exponents of the variables by the index of the radical? I'm sorry if my terminology is bad and if my equations aren't right I'm new to this forum so I'm trying to get it right.

Notice that
\sqrt[3]{x^6y^4}= \sqrt[3]{(x^3)(x^3)(y^3)(y)}= \sqrt[3]{x^3}\sqrt[3]{x^3}\sqrt[3]{y^3}\sqrt[3]{y}= (x)(x)(y)(\sqrt[3]{y})= x^2y\sqrt[3]{y}

Yes, that is the same as (x^6y^4)^{1/3}= x^{6/3}y^{4/3}= x^2y^{1+ 1/3}= x^2yy^{1/3}