# Homework Help: Exponents in a radicand

1. Jan 27, 2010

### Cornraker

1. The problem statement, all variables and given/known data

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

2. Relevant equations

∛(x^6 y^4 )

3. The attempt at a solution

x^2 y^4 ?

2. Jan 27, 2010

### sylas

$$\sqrt[3]{x^6 y^4}$$​

Given your bracketing the y4 is inside the cube root as well. Note that you can write
$$\sqrt[3]{x^6 y^4} = \sqrt[3]{x^6} \sqrt[3]{y^4}$$​

3. Jan 27, 2010

### 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.

4. Jan 27, 2010

### sylas

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
$$\sqrt[n]{a} = a^{1/n}$$​

5. Jan 28, 2010

### HallsofIvy

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 $$(x^6y^4)^{1/3}= x^{6/3}y^{4/3}= x^2y^{1+ 1/3}= x^2yy^{1/3}$$