#### Noegddgeon

Hello, everybody. A problem I was assigned to solve was the following, where I needed to condense the expression into one single radical:

$$\displaystyle \sqrt{6}\sqrt{2}$$

Here's my steps for solving it:

$$\displaystyle 6^{1/3} * 2^{1/2}$$

$$\displaystyle 12^{5/6}$$

$$\displaystyle \sqrt{12^5}$$

However, the book gives the answer as $$\displaystyle \sqrt{288}$$. I don't understand why this is, and I would greatly appreciate any help in figuring so out. :] Thank you much.

Colton

Hello, everybody. A problem I was assigned to solve was the following, where I needed to condense the expression into one single radical:

$$\displaystyle \sqrt{6}\sqrt{2}$$

Here's my steps for solving it:

$$\displaystyle 6^{1/3} * 2^{1/2}$$

$$\displaystyle 12^{5/6}$$ You cannot multiply the 6 by the 2 and add the indices like this!

$$\displaystyle \sqrt{12^5}$$

However, the book gives the answer as $$\displaystyle \sqrt{288}$$. I don't understand why this is, and I would greatly appreciate any help in figuring so out. :] Thank you much.

Colton
Hi Colton,

$$\displaystyle 6^{\frac{1}{3}}2^{\frac{1}{2}}=\left(36^{\frac{1}{2}}\right)^{\frac{1}{3}}\left(8^{\frac{1}{3}}\right)^{\frac{1}{2}}$$

$$\displaystyle =\left(36(8)\right)^{\frac{1}{6}}$$

#### Noegddgeon

Thank you very much for your reply. However, I'm still a little unsure as to how you got the 36 and the 8 in the steps which you showed me. I understand now that I cannot add the indices as I had done. Further help would be greatly appreciated. :] Thanks again.

Colton

Sure,

$$\displaystyle 6^2=36\ \Rightarrow\ 6=36^{\frac{1}{2}}$$

The $$\displaystyle 6^{\frac{1}{3}}$$ can now be written with a power of $$\displaystyle \frac{1}{3}\ \frac{1}{2}$$

which we multiply to get $$\displaystyle \frac{1}{6}$$

Same with the 2, except we use $$\displaystyle 2^3=8\ \Rightarrow\ 2=8^{\frac{1}{3}}$$

Again we have an index where we obtain $$\displaystyle \frac{1}{6}$$

The final answer can then be obtained since both fractions are to the power of a sixth,
just as we may write $$\displaystyle 12^2=144$$

as $$\displaystyle 12^2=[4(3)]^2=4^23^2=16(9)=144$$

#### Noegddgeon

Thank you much for clarifying. Are these rules particular to these kinds of problems, with fractionized exponents? I was following the rule I saw in my textbook when I added the indices with that multiplication problem. The example is shown in my workbook as so:

$$\displaystyle x^{5/6} * x^{2/3}$$

$$\displaystyle x^{9/6}$$

However, I notice that they both use x and not different numbers like the example which I gave you. I would like for this to be as clear as possible for myself so that I won't have any trouble further down the road. :] Thank you very much for your help and your time.

Colton

Thank you much for clarifying. Are these rules particular to these kinds of problems, with fractionized exponents? I was following the rule I saw in my textbook when I added the indices with that multiplication problem. The example is shown in my workbook as so:

$$\displaystyle x^{5/6} * x^{2/3}$$

$$\displaystyle x^{9/6}$$

However, I notice that they both use x and not different numbers like the example which I gave you. I would like for this to be as clear as possible for myself so that I won't have any trouble further down the road. :] Thank you very much for your help and your time.

Colton
Hi colton,

that one is slightly different....

$$\displaystyle x^{\frac{5}{6}}x^{\frac{2}{3}}=x^{\frac{5}{6}}x^{\frac{4}{6}}=x^{\frac{5}{6}+\frac{4}{6}}$$

In that case you add the exponents because x is a particular value.

You need to know the difference between when you add or multiply the exponents.

For example......

$$\displaystyle 2^22^3=(2*2)(2*2*2)$$

That's 5 twos multiplied together, so it's $$\displaystyle 2^5$$

However $$\displaystyle \left(2^2\right)^3=(2*2)(2*2)(2*2)$$

which is 6 twos multiplied together, so it's $$\displaystyle 2^6$$

$$\displaystyle x^a*x^b=x^{a+b}$$

$$\displaystyle \left(x^a\right)^b=x^{ab}$$

Also, as in your earlier example

$$\displaystyle x^a*y^a=(xy)^a$$

since $$\displaystyle 2^2*3^2=2*2*3*3=2*3*2*3=(2*3)(2*3)=(2*3)^2$$ and so on

• Noegddgeon