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Math Help - Condensing to One Radical

  1. #1
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    Condensing to One Radical

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

    <br />
\sqrt[3]{6}\sqrt{2}<br />

    Here's my steps for solving it:

    <br />
6^{1/3} * 2^{1/2}<br />

    <br />
12^{5/6}<br />

    <br />
\sqrt[6]{12^5}<br />

    However, the book gives the answer as \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
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  2. #2
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    Quote Originally Posted by Noegddgeon View Post
    Hello, everybody. A problem I was assigned to solve was the following, where I needed to condense the expression into one single radical:

    <br />
\sqrt[3]{6}\sqrt{2}<br />

    Here's my steps for solving it:

    <br />
6^{1/3} * 2^{1/2}<br />

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

    <br />
\sqrt[6]{12^5}<br />

    However, the book gives the answer as \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,

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

    =\left(36(8)\right)^{\frac{1}{6}}
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  3. #3
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    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
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  4. #4
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    Sure,

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

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

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

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

    Again we have an index where we obtain \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 12^2=144

    as 12^2=[4(3)]^2=4^23^2=16(9)=144
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  5. #5
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    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:

    <br />
x^{5/6} * x^{2/3}<br />

    <br />
x^{9/6}<br />

    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
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  6. #6
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    Quote Originally Posted by Noegddgeon View Post
    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:

    <br />
x^{5/6} * x^{2/3}<br />

    <br />
x^{9/6}<br />

    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....

    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......

    2^22^3=(2*2)(2*2*2)

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

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

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

    x^a*x^b=x^{a+b}

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

    Also, as in your earlier example

    x^a*y^a=(xy)^a

    since 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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  7. #7
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    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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