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Thread: Logarithmic Proof

  1. #1
    Senior Member Stroodle's Avatar
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    Logarithmic Proof

    Given $\displaystyle 3^x=4^y=12^z$ show that $\displaystyle z=\frac{xy}{x+y}$

    Thanks for you help!
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  2. #2
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    Quote Originally Posted by Stroodle View Post
    Given $\displaystyle 3^x=4^y=12^z$ show that $\displaystyle z=\frac{xy}{x+y}$

    Thanks for you help!
    What a horrid question...


    First note that $\displaystyle 12^z = (3 \cdot 4)^z = 3^z \cdot 4^z$

    Focus firstly on

    $\displaystyle 3^x = 12^z$

    $\displaystyle 3^x = 3^z \cdot 4^z$

    $\displaystyle 3^{x - z} = 4^z$

    $\displaystyle \log{3^{x - z}} = \log{4^z}$

    $\displaystyle (x - z)\log{3} = z\log{4}$

    $\displaystyle \frac{x - z}{z} = \frac{\log{4}}{\log{3}}$.


    Now focus on

    $\displaystyle 4^y = 3^z \cdot 4^z$

    $\displaystyle 4^{y - z} = 3^z$

    $\displaystyle \log{4^{y - z}} = \log{3^z}$

    $\displaystyle (y - z)\log{4} = z\log{3}$

    $\displaystyle \frac{\log{4}}{\log{3}} = \frac{z}{y - z}$.



    Therefore

    $\displaystyle \frac{x - z}{z} = \frac{z}{y - z}$

    $\displaystyle \frac{x}{z} - 1 = \frac{y}{y - z} - 1$

    $\displaystyle \frac{x}{z} = \frac{y}{y - z}$

    $\displaystyle \frac{y - z}{z} = \frac{y}{x}$

    $\displaystyle \frac{y}{z} - 1 = \frac{y}{x}$

    $\displaystyle \frac{y}{z} = \frac{y}{x} + 1$

    $\displaystyle \frac{y}{z} = \frac{x + y}{x}$

    $\displaystyle \frac{z}{y} = \frac{x}{x + y}$

    $\displaystyle z = \frac{xy}{x + y}$.
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  3. #3
    Senior Member Stroodle's Avatar
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    Awesome. I get it now.

    Thanks for your help Prove It.
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  4. #4
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    Hello Stroodle
    Quote Originally Posted by Stroodle View Post
    Given $\displaystyle 3^x=4^y=12^z$ show that $\displaystyle z=\frac{xy}{x+y}$

    Thanks for you help!
    Here's an alternative approach:

    $\displaystyle 3^x = 12^z$

    $\displaystyle \Rightarrow x\log3 = z\log12$

    $\displaystyle \Rightarrow x = \frac{z\log12}{\log 3}$

    Similarly $\displaystyle y = \frac{z\log12}{\log4}$

    $\displaystyle \Rightarrow \frac{xy}{x+y}= \frac{\dfrac{z\log12}{\log 3}\cdot\dfrac{z\log12}{\log 4}}{\dfrac{z\log12}{\log 3}+\dfrac{z\log12}{\log 4}}$

    $\displaystyle = \frac{\dfrac{z\log12}{\log 3}\cdot\dfrac{z\log12}{\log 4}}{\dfrac{z\log12}{\log 3}+\dfrac{z\log12}{\log 4}}\color{red}\times\frac{\log3\log4}{\log3\log4}$

    $\displaystyle = \frac{z^2(\log12)^2}{z\log12(\log4+\log3)}$

    $\displaystyle =\frac{z\log12}{\log12}$

    $\displaystyle =z$

    Grandad
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  5. #5
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    Quote Originally Posted by Stroodle View Post
    Given $\displaystyle 3^x=4^y=12^z$ show that $\displaystyle z=\frac{xy}{x+y}$

    Thanks for you help!
    HI

    Or another approach .

    $\displaystyle 3^x=4^y=12^z=k$

    $\displaystyle 3^x=k\Rightarrow 3=k^{\frac{1}{x}}$

    $\displaystyle 4^y=k\Rightarrow 4=k^{\frac{1}{y}}$

    $\displaystyle 12^z=k\Rightarrow 12=k^{\frac{1}{z}}$

    we know for the fact that 4 x 3 = 12

    so $\displaystyle k^{\frac{1}{x}}\times k^{\frac{1}{y}}=k^{\frac{1}{z}}$

    $\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{1}{z}$

    yz+xz=xy

    z(x+y)=xy

    z=xy/(x+y)
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