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Math Help - logarithmic equation

  1. #1
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    logarithmic equation

    Hello MHF, please help me solving this equation =[
    {27x}^{\log_5 x}={x}^{4}
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  2. #2
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    Re: logarithmic equation

    Quote Originally Posted by Chipset3600 View Post
    Hello MHF, please help me solving this equation =[
    {27x}^{\log_5 x}={x}^{4}
    Let

    y=\log_{5}(x) \iff x=5^y Then you equation becomes

    27(5^y)^y=(5^y)^4

    27 \cdot 5^y^2=5^{4y} \iff 27 \cdot 5^{y^2-4y}=1

    Now if we take log base 5 of both sides we get

    y^2-4y+\log_{5}(27)=0

    Solving this quadratic for y gives

    y=\frac{4 \pm \sqrt{16-4\log_{5}(27)}}{2}=2 \pm \sqrt{4-\log_{5}(27)}

    Pluging this back in we get the solution for x is

    x=5^{2 \pm \sqrt{4-\log_{5}(27)}}
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  3. #3
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    Re: logarithmic equation

    Quote Originally Posted by Chipset3600 View Post
    Hello MHF, please help me solving this equation =[
    {27x}^{\log_5 x}={x}^{4}
    Here's an alternative method...

    \displaystyle \begin{align*} 27x^{\log_5{x}} &= x^4 \\ \log_x{\left( 27x^{\log_5{x}} \right)} &= \log_x{\left(x^4\right)} \\ \log_x{(27)} + \log_x{\left( x^{\log_5{x}}\right)} &= 4 \\ \log_x{(27)} + \log_5{x} &= 4 \\ \frac{\log_5{(27)}}{\log_5{x}} + \log_5{x} &= 4 \\ \log_5{(27)} + \log_5^2{x} &= 4\log_5{x} \end{align*}

    which is now a quadratic.
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  4. #4
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    Re: logarithmic equation

    Hello, Chipset3600!

    And yet another method . . .


    \displaystyle 27x^{\log_5 x}\:=\:x^4

    Take logs, base 5:

    . . . . . . . . . . . \log_5\left(27x^{\log_5x}\right) \;=\;\log_5(x^4)

    . . . . . . \log_527 + \log_5\left(x^{\log_5x}\right) \;=\;4\log_5x

    . . . . . . \log_527 + \log_5x\!\cdot\!\log_5x \;=\;4\log_5x

    . . \left(\log_5x\right)^2 - 4\log_5x + \log_527 \;=\;0

    which is the quadratic equation found by Prove It.
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  5. #5
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    Re: logarithmic equation

    I understood all steps, but how can i solve that \sqrt{4-\log_{5}(27)} without calculator?
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  6. #6
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    Re: logarithmic equation

    3^{3}x^{\log_3 x } = x^{4}

    x^{\log_3 x } = \frac{x^{4}}{27} doing \log_3 x = y so 3^{y} = x

    27(3^{y})^{y} = (3^{y})^{y}

    3^{3}.3^{y^{2}}= 3^{4y}
    3^{3+y^{2}}= 3^{4y}
    3+y^{2}=3^{4y}
    3+y^{2}=4y
    y^{2}-4y+3=0
    y_{1}=1 or y_{2}=3 \log_3 x =1 and \log_3 x = 3

    So x = 3 or x=27
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  7. #7
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    Re: logarithmic equation

    Sorry guys! i did a mistake, the base of the Log is 3 and not 5...Thanks
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