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Thread: Logarithmic equations

  1. #1
    Senior Member pankaj's Avatar
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    Logarithmic equations

    If $\displaystyle (x_{1},y_{1})$ is solution of $\displaystyle log_{225}(x)+log_{64}(y)=4$ and $\displaystyle (x_{2},y_{2})$ is solution of
    $\displaystyle
    log_{x}(225)-log_{y}(64)=1
    $
    then find the value of $\displaystyle log_{30}(x_{1}x_{2}y_{1}y_{2})$

    I got the answer as 12.But i think that language of this question is not proper .It should have been
    If $\displaystyle (x_{1},y_{1})$ and$\displaystyle (x_{2},y_{2})$ are solutions of $\displaystyle log_{225}(x)+log_{64}(y)=4$ and
    $\displaystyle
    log_{x}(225)-log_{y}(64)=1
    $
    then find the value of $\displaystyle log_{30}(x_{1}x_{2}y_{1}y_{2})$.
    What do you guys think?
    According to the answer i obtained i treated the 2 equations to be simultaneous equations in x and y and obtained 2 solutions.But according to language of the original question if $\displaystyle (x_{1},y_{1})$ is solution of first equation then it need not be solution of the second equation.
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  2. #2
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    Hello, pankaj!

    If that's the original wording of the problem, it's criminally sloppy!

    I tried two different interpretations ... and one seems to work out.


    Here's one interpretation . . .

    If $\displaystyle (x_1,y_1)$ is any solution of $\displaystyle \log_{225}(x)+\log_{64}(y)\:=\:4\;\;{\color{blue}[1]}$

    and $\displaystyle (x_2,y_2)$ is any solution of $\displaystyle \log_{x}(225)-\log_{y}(64)\:=\:1\;\;{\color{blue}[2]} $

    then find the value of $\displaystyle \log_{30}(x_1x_2y_1y_2)$

    A solution of [1] is: .$\displaystyle x_1 = 225^2,\;y_1 = 64^2$

    A solution of [2] is: .$\displaystyle x_2 = 15,\;y_2 =64$

    Then: .$\displaystyle x_1x_2y_1y_2 \;=\;225^2\cdot15\cdot64^2\cdot64 \;=\;(3^4\cdot5^4)\cdot(3\cdot5)\cdot(2^{12})(2^6)$

    . . . . $\displaystyle = \;2^{18}\cdot3^5\cdot5^5 \;=\;2^{13}\cdot(2^5\cdot3^5\cdot5^5) \;-=\;2^{13}\cdot(2\cdot3\cdot5)^5 \;=\;2^{13}\cdot30^5$


    Therefore: .$\displaystyle \log_{30}(x_1x_2y_1y_2) \;=\;\log_{30}\left(2^{13}\cdot30^5\right) \;=\;\boxed{13\log_{30}(2) + 5}$


    But, of course, this result is not a constant.
    It varies, depending on the solutions we select: $\displaystyle (x_1,y_1),\;(x_2,y_2)$




    Here's another interpretation . . . and I got your answer!

    $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$ are solutions of the system: .$\displaystyle \begin{array}{cccc}\log_{225}(x)+\log_{64}(y) &=& 4\\ \log_{x}(225)-\log_{y}(64) &=& 1 \end{array}$
    Find the value of $\displaystyle \log_{30}(x_1x_2y_1y_2)$
    We have: . $\displaystyle \begin{array}{cccc}\log_{225}(x) + \log_{64}(y) &=& 4 \\ \\[-3mm]
    \dfrac{1}{\log_{225}(x)} + \dfrac{1}{\log_{64}(y)} &=& 1 \end{array}$


    Let: .$\displaystyle \begin{Bmatrix}X &=& \log_{225}(x) \\ Y &=& \log_{64}(y) \end{Bmatrix}\quad \text{ and we have: }\;\begin{array}{cccc} X + Y &=& 4 & {\color{blue}[1]} \\ \\[-4mm] \dfrac{1}{X} - \dfrac{1}{Y} &=& 1 & {\color{blue}[2]} \end{array}$

    $\displaystyle \begin{array}{cccc}\text{{\color{blue}[2]} becomes:} & Y - X &=& XY \\ \text{Add {\color{blue}[1]}:} & Y + X &=&4 \end{array}$

    And we get: .$\displaystyle 2Y \:=\:XY + 4 \quad\Rightarrow\quad Y \:=\:\frac{4}{2-X}$

    Substitute into [1]: .$\displaystyle X + \frac{4}{2-X} \:=\:4\quad\Rightarrow\quad X^2 - 6X + 4 \:=\:0$

    The Quadratic Formula gives us: .$\displaystyle X \:=\:3 \pm\sqrt{5}$

    Substitute into [1]: .$\displaystyle Y \:=\:1 \mp \sqrt{5}$


    Then we have: . $\displaystyle \begin{array}{ccccccc}x_1 &=&225^{3+\sqrt{5}} & \quad & y_1 &=& 64^{1-\sqrt{5}} \\
    x_2 &=& 225^{3-\sqrt{5}} & \quad & y_2 &=& 64^{1+\sqrt{5}} \end{array}$

    Then: .$\displaystyle x_1x_2y_1y_2 \;=\;\left(225^{3+\sqrt{5}}\right)\left(225^{3-\sqrt{5}}\right)\left(64^{1-\sqrt{5}}\right)\left(64^{1+\sqrt{5}}\right) \;=\;225^6\cdot64^2$

    . . . . . . $\displaystyle = \;3^{12}\cdot5^{12}\cdot2^{12} \;=\;(2\cdot3\cdot5)^{12} \;=\;30^{12}$


    Therefore: .$\displaystyle \log_{30}\left(30^{12}\right) \;=\;\boxed{12}$

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  3. #3
    Senior Member pankaj's Avatar
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    Thanks for the verification.
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