Solving with indices

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  1. krisshP's Avatar
    1 06-07-2012  17:35
    • Peer Of The TSR Realm
    • Posts: 1,806
    Solving with indices
    Here is a question which I'm stuck on:

    Solve the following equation:
    3x^\frac{2}{3}=6x

    Below is my working out:

    x^\frac{2}{3}=2x

    (\sqrt[3]{x})^2=2x

    \sqrt[3]{x}= \sqrt[2]{2x}

    x=(\sqrt[2]{2x})^3

    x=2x^\frac{3}{2}

    Now I do not know what i am supposed to do further.:confused::confused:

    Thanks for any help provided.
  2. Hasufel's Avatar
    2 06-07-2012  17:39
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    Re: Solving with indices
    2nd line down - cube it - don`t square it, you should end up with:

    x^2(8x-1)=0
  3. krisshP's Avatar
    3 06-07-2012  18:03
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    Re: Solving with indices
    But I did cube root the x. I squared the LHS as I have to - the power clearly is 2 over 3.
    Last edited by krisshP; 06-07-2012 at 18:06.
  4. raheem94's Avatar
    4 06-07-2012  18:11
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    Re: Solving with indices
    (Original post by krisshP)
    But I did cube root the x. I squared the LHS as I have to - the power clearly is 2 over 3.
    x^{ \frac23} = 2 x

    Cube both sides

    x^{ \frac23 \times 3 } = (2x)^3 \implies x^2 = 8x^3 \implies 8x^3 - x^2 = 0 \implies x^2 ( 8x -1 ) = 0
    Last edited by raheem94; 07-07-2012 at 14:04. Reason: Mistake corrected!
  5. krisshP's Avatar
    5 06-07-2012  18:20
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    Re: Solving with indices
    How do I know x from what is below:

    x^2(8x-1)=0

    ??
  6. raheem94's Avatar
    6 06-07-2012  18:33
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    Re: Solving with indices
    (Original post by krisshP)
    How do I know x from what is below:

    x^2(8x-1)=0

    ??
    x^2 ( 8x -1 ) = 0 \implies x^2 = 0 \text{ or } 8x-1 = 0

    x^2 = 0 \implies x = 0

    8x - 1 = 0 \implies 8x = 1 \implies x = \dfrac18

    \boxed{ x = 0 \text{ or } \dfrac18 }

    Btw, which grade are you in?
  7. krisshP's Avatar
    7 07-07-2012  11:49
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    Re: Solving with indices
    (Original post by raheem94)
    x^2 ( 8x -1 ) = 0 \implies x^2 = 0 \text{ or } 8x-1 = 0

    x^2 = 0 \implies x = 0

    8x - 1 = 0 \implies 8x = 1 \implies x = \dfrac18

    \boxed{ x = 0 \text{ or } \dfrac18 }

    Btw, which grade are you in?
    I'm at the start of AS Maths. I'm predicted an A in GCSE maths (studying C1 + C2 in summer), but I might just about get an A*.
    Last edited by krisshP; 07-07-2012 at 12:07.
  8. krisshP's Avatar
    8 07-07-2012  12:30
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    Re: Solving with indices
    I worked out the answer and I would like to share it:

    x^{ \frac23} = 2 x

    Divide both sides by the x gives:

    x^{\frac{2}{3}\times\frac{1}{1}} =2

    x^\frac{2}{3}=2

    (\sqrt[3]{x})^2=2

    \sqrt[3]{x}=\sqrt{2}

    x=(\sqrt{2})^3

    x=(2^\frac{1}{2})^3

    0.5 X 3=1.5, hence

    x=2^\frac{3}{2}
  9. raheem94's Avatar
    9 07-07-2012  13:52
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    Re: Solving with indices
    (Original post by krisshP)
    I worked out the answer and I would like to share it:

    x^{ \frac23} = 2 x

    Divide both sides by the x gives:

    x^{\frac{2}{3}\times\frac{1}{1}} =2

    x^\frac{2}{3}=2

    (\sqrt[3]{x})^2=2

    \sqrt[3]{x}=\sqrt{2}

    x=(\sqrt{2})^3

    x=(2^\frac{1}{2})^3

    0.5 X 3=1.5, hence

    x=2^\frac{3}{2}
    Before sharing answers you should check whether they are correct or not.

    How can x = 2^{ \frac32} be a solution to x^{ \frac23} = 2 x :confused:

    How did you do divide both sides by x , i only see one side divided by it.

    You needed to factorize not divide it by x
    Last edited by raheem94; 07-07-2012 at 13:54.
  10. steve2005's Avatar
    10 07-07-2012  13:56
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    • Location: LONDON
    Re: Solving with indices
    (Original post by raheem94)
    x^{ \frac23} = 2 x

    Cube root both sides

    x^{ \frac23 \times 3 } = (2x)^3 \implies x^2 = 8x^3 \implies 8x^3 - x^2 = 0 \implies x^2 ( 8x -1 ) = 0
    You should have said " Cube both sides" rather than " Cube root both sides"
    Post rating:
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  11. raheem94's Avatar
    11 07-07-2012  14:04
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    • Posts: 5,512
    Re: Solving with indices
    (Original post by steve2005)
    You should have said " Cube both sides" rather than " Cube root both sides"
    Sorry, my bad Thanks for correcting.

    I didn't noticed my mistake.
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