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Math Help - Exponent Question...

  1. #1
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    Exponent Question...

    (x)^4/3 = 16


    How do I find this?


    Oh and more importantly, how do you do this?

    (9)^x = 4th root (27) / 3
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    (x)^4/3 = 16
    x^{4/3} = 16 <-- Cube both sides

    x^4 = 16^3 <-- Take the fourth root of both sides

    x = 16^{3/4}

    What number is this? Well if you recall that 2^4 = 16...

    x = 16^{3/4} = \left ( 2^4 \right ) ^{3/4} = 2^{4 \cdot (3/4)} = 2^3 = 8

    Of course, this is the principle value of the solution. You might note that x = -8 also solves the equation. (The easiest way to see this is to take x^4 = 16^3 and take the square root of both sides twice.) There are also two complex solutions. I can give you the general solution if you like; it's not much harder than what I derived here.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    (9)^x = 4th root (27) / 3
    You want to get both sides of the equation to the same base. In this case, the easiest base to use is 3.

    9 = 3^2 and 27 = 3^3, and \sqrt[4]{x} = x^{1/4}, so...

    9^x = \frac{\sqrt[4]{27}}{3}

    (3^2)^x = \frac{(3^3)^{1/4}}{3}

    3^{2x} = \frac{3^{3/4}}{3} = 3^{(3/4) - 1} = 3^{-1/4}

    Now we have that
    2x = -\frac{1}{4}

    x = -\frac{1}{8}

    -Dan
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