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Thread: x^4/(3)√x Not sure what I did wrong (i got to x^(8/2)/3x^(1/2) )

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    x^4/(3)√x Not sure what I did wrong (i got to x^(8/2)/3x^(1/2) )

    x^4/(3)√x Not sure what I did wrong

    i got to x^(8/2)/3x^(1/2)

    I tried 3x^(7/2) as the final answer but what was wrong. I'm pretty sure the problem isn't the exponent. Help is greatly appreciated!! thanks!
    Last edited by stapel; 12-01-2017 at 01:59 PM. Reason: Copying question from subject line into post.

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    How did the '3' get into the numerator?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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    Quote Originally Posted by tpace View Post
    i got to x^(8/2)/3x^(1/2)
    I tried 3x^(7/2) as the final answer but what was wrong. I'm pretty sure the problem isn't the exponent. Help is greatly appreciated!! thanks!
    If you intend for the problem to be equivalent to [tex] \ \ \dfrac{x^{\frac{8}{2}}}{3x^{\frac{1}{2}}}, \ \ [/tex] then you must put grouping symbols around the denominator,
    such as in x^(8/2)/[3x^(1/2)].

    Would that help you with where the "3" would end up?

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    Geeeezzzzz...why x^(8/2) instead of x^4
    Silly teacher
    I'm just an imagination of your figment !

  5. #5
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by tpace View Post
    x^4/(3)√x Not sure what I did wrong

    i got to x^(8/2)/3x^(1/2)
    Since we can't see your work, and because you haven't included the instructions, we cannot know what you were supposed to have done, nor where things might have gone sideways. Sorry!

    Quote Originally Posted by tpace View Post
    I tried 3x^(7/2) as the final answer but what was wrong.
    What do you mean by "trying this as the final answer"? What does that mean? Are you just guessing answer options, or are you doing something or other with the original expression?

    Quote Originally Posted by tpace View Post
    I'm pretty sure the problem isn't the exponent.
    What do you mean by this?

    Is the original expression either of the following?

    . . . . .[tex]\mbox{a. }\, \left(\dfrac{x^4}{3}\right)\, \sqrt{\strut x\,}[/tex]

    . . . . .[tex]\mbox{b. }\, \dfrac{x^4}{3\, \sqrt{\strut x\,}}[/tex]

    Or does "(3)√x" indicate "the cube root of x"? (This is often what it means, but I don't think this is the case for your posting.)

    What you "got to": Does this indicate that the instructions were to "Simplify the expression", perhaps by rationalizing the denominator? And that you've recently studied converting radicals to fractional-exponent notation? So your one step so far has been something along the lines of the following?

    . . . . .[tex]\dfrac{x^4}{3\, \sqrt{\strut x\,}}\, =\, \dfrac{x^{\frac{8}{2}}}{3\, x^{\frac{1}{2}}}[/tex]

    And now you're stuck with applying the exponent rules for cancelling off in this sort of situation? Or do you know the next step, but aren't sure how to subtract fractions? Or something else?

    Please be complete. Thank you!

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    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by Denis View Post
    Geeeezzzzz ... why x^(8/2) instead of x^4
    Silly teacher
    Because you misread the OP, perhaps?

    8/2 was an intermedate result, in tpace's first attempt.

    Quote Originally Posted by tpace View Post
    x^4/(3)√x

    i got to x^(8/2)/3x^(1/2)
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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