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Math Help - help needed with exponents

  1. #1
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    help needed with exponents

    hi there i've been looking everywhere and just can't seem to work this out. any help would be greatly appreciated.

    (x + 1)^(3/2)
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  2. #2
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    Quote Originally Posted by ambienboy View Post
    hi there i've been looking everywhere and just can't seem to work this out. any help would be greatly appreciated.

    (x + 1)^(3/2)
    what exactly do you want to do? write (x + 1)^(3/2) in terms of a radical? or solve something involving (x + 1)^(3/2)?
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  3. #3
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    i'm sorry, I realise I was being vague.
    Here is the full question:

    help needed with exponents-problem.jpg
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ambienboy View Post
    i'm sorry, I realise I was being vague.
    Here is the full question:

    Click image for larger version. 

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    2(x + 1)^(3/2) = 27/4 ...........we deal with powers last, so begin by getting rid of the 2

    => (x + 1)^(3/2) = 27/8 ..........divided both sides by 2
    now we want to get rid of the power. that is, we want the power of x + 1 to be 1. how do we do that? note that 3/2 * 2/3 = 1. so if we raise both sides to the 2/3 power, we make the left side less complicated.

    => [(x + 1)^(3/2)]^(2/3) = (27/8)^(2/3)
    => x + 1 = (27/8)^(2/3)
    => x = (27/8)^(2/3) - 1

    how do we eveluate (27/8)^(2/3)?

    remember, when we have a fractional power, think of "Flower Power"
    you treat the fraction as a flower basically, yeah i know it's weird. the division sign represents the ground, the numerator on top represent the flower (which rhymes with power, hence flower-power), and the denominator under the division sign represents the root. so in short, when we have a fractional power, the numerator represents the power, and the denominator represents the root. so,

    x = (27/8)^(2/3) - 1
    => x = cuberoot[(27/8)^2]- 1
    we can find the root or the power first, it is easier to find the root:
    => x = (3/2)^2 - 1
    => x = 9/4 - 1
    => x = 5/4

    you could also split the powers up:


    (27/8)^(2/3) = (27/8)^[2 * 1/3] = [(27/8)^2]^(1/3) = [(27/8)^(1/3)]^2

    note that we can take either power first, i found it easier to take the 1/3 power first, which is the same as the cuberoot

    so (27/8)^(2/3) = [(27/8)^(1/3)]^2 = (3/2)^2 = 9/4
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  5. #5
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    thank you so much
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