Condensing to One Radical

Nov 2009
21
0
California, United States
Hello, everybody. A problem I was assigned to solve was the following, where I needed to condense the expression into one single radical:

\(\displaystyle
\sqrt[3]{6}\sqrt{2}
\)

Here's my steps for solving it:

\(\displaystyle
6^{1/3} * 2^{1/2}
\)

\(\displaystyle
12^{5/6}
\)

\(\displaystyle
\sqrt[6]{12^5}
\)

However, the book gives the answer as \(\displaystyle \sqrt[6]{288}\). I don't understand why this is, and I would greatly appreciate any help in figuring so out. :] Thank you much.

Colton
 
Dec 2009
3,120
1,342
Hello, everybody. A problem I was assigned to solve was the following, where I needed to condense the expression into one single radical:

\(\displaystyle
\sqrt[3]{6}\sqrt{2}
\)

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!

\(\displaystyle
\sqrt[6]{12^5}
\)

However, the book gives the answer as \(\displaystyle \sqrt[6]{288}\). I don't understand why this is, and I would greatly appreciate any help in figuring so out. :] Thank you much.

Colton
Hi Colton,

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

\(\displaystyle =\left(36(8)\right)^{\frac{1}{6}}\)
 
Nov 2009
21
0
California, United States
Archie Meade,

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
 
Dec 2009
3,120
1,342
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 \(\displaystyle 12^2=[4(3)]^2=4^23^2=16(9)=144\)
 
Nov 2009
21
0
California, United States
Archie Mead,

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
 
Dec 2009
3,120
1,342
Archie Mead,

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
 
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Nov 2009
21
0
California, United States
Archie Meade,

Thank you very much. It makes sense to me. These little rules are things I tend to forget every once in a while if I don't use them or think about them as I should, but I'm hoping they stick if I keep practicing. :] I appreciate all your help and I'm sure you'll be seeing more from me in the future... haha :]

Colton
 
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