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Thread: 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

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

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

    $\displaystyle x = 2^{-\frac{2}{3}} = \frac{1}{\sqrt[3]{4}}$
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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:

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

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

    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: $\displaystyle \left(x^{9/2}\right)^{2/9}= x= (1/2^3)^{2/9}= 1/2^{6/9}= 1/2^{2/3}= \frac{1}{\sqrt[3]{4}}$

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

    Hello, M670!

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

    I made it into:

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

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

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

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

    but I don't think it's right

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

    Raise both sides to the $\displaystyle \tfrac{2}{9}$ power:
    . . . $\displaystyle \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: .$\displaystyle x\;=\;0.629960515$
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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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