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Math Help - Logs

  1. #1
    Member M670's Avatar
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    Logs

    Solve for :
    I made it into

    log2x9/2=-3
    2-3=x9/2
    .125x-4.5=x
    -0.5625=x


    but I don't think it's right
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  2. #2
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    skeeter's Avatar
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    Re: Logs

    \frac{9}{2} \log_2{x} = -3

    \log_2{x} = -\frac{2}{3}

    x = 2^{-\frac{2}{3}} = \frac{1}{\sqrt[3]{4}}
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  3. #3
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    Re: Logs

    Skeeter's method, dividing both sides by 9/2 first, is simpler than yours but it can be done your way:

    \frac{9}{2}log_2(x)= -3
    log_2(x^{9/2})= -3

    x^{9/2}= 2^{-3}= \frac{1}{8}

    I don't know how you now got ".125x-4.5=x". Was that first "x" supposed to be a multiplication sign? Bad idea! Never use the same symbol to mean two different things!

    But even so, you get rid of the "9/2" power on the left by taking the 9/2 root- which is the same thing as the 2/9 power, on both sides: \left(x^{9/2}\right)^{2/9}= x= (1/2^3)^{2/9}= 1/2^{6/9}= 1/2^{2/3}= \frac{1}{\sqrt[3]{4}}

    If you want or are required to "rationalize the denominator", multiply both numerator and denominator by \left(\sqrt[3]{4}\right)^2= \sqrt[3]{16}: \frac{\sqrt[3]{16}}{4}.
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  4. #4
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    Re: Logs

    Hello, M670!

    Solve for x\!:\;\tfrac{9}{2}\log_2x \:=\:-3

    I made it into:

    \log_2\left(x^{\frac{9}{2}}\right) \:=\:-3 . . Right!

    2^{-3} \:=\:x^{\frac{9}{2}} . . Correct!

    0.125x-4.5\:=\:x . . I don't understand this step.

    -0.5625\:=\:x . . Nor this step.

    but I don't think it's right

    In your second step, you had: x^{\frac{9}{2}} \:=\:2^{-3}

    Raise both sides to the \tfrac{2}{9} power:
    . . . \left(x^{\frac{9}{2}}\right)^{\frac{2}{9}} \:=\:\left(2^{-3}\right)^{\frac{2}{9}} \quad\rightarrow\quad x \:=\:2^{-\frac{2}{3}}

    Therefore: . x\;=\;0.629960515
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  5. #5
    Member M670's Avatar
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    Re: Logs

    sorry guys I should of written my step .125x-4.5=x as follows .125(-4.5)=x
    but i understand what you have all shown me
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