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  1. #1
    Super Member dhiab's Avatar
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    log-system

    Solve the system

    $\displaystyle log_{a^{2}}(x)-log_{a^{4}}(y)=3$
    $\displaystyle log_{a^{6}}(x)+log_{a^{8}}(y)=4$
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by dhiab View Post
    Solve the system

    $\displaystyle log_{a^{2}}(x)-log_{a^{4}}(y)=3$
    $\displaystyle log_{a^{6}}(x)+log_{a^{8}}(y)=4$
    Is $\displaystyle a$ a constant?

    Here are some things you can do to the first equation.

    $\displaystyle log_{a^{2}}(x)-log_{a^{4}}(y)=3$

    $\displaystyle \Longrightarrow \frac{log_a(x)}{log_a(a^2)}-\frac{log_a(y)}{log_a(a^4)}=3$

    $\displaystyle \Longrightarrow \frac{log_a(x)}{2}-\frac{log_a(y)}{4}=3$

    $\displaystyle \Longrightarrow 2log_a(x)-log_a(y)=12$

    $\displaystyle \Longrightarrow log_a(x^2)-log_a(y)=12$

    $\displaystyle \Longrightarrow log_a\left(\frac{x^2}{y}\right)=12$

    Keep in mind that $\displaystyle x>0$ and $\displaystyle y>0$.
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  3. #3
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    Hello, dhiab!

    Edit: I've corrected my blunder . . .


    Solve the system:

    . . $\displaystyle \log_{a^2}(x)-\log_{a^4}(y) \;=\; 3 \;\;[1] $

    . . $\displaystyle \log_{a^6}(x)+\log_{a^8}(y)\:=\:4\;\;[2]$

    We have:

    . . $\displaystyle \frac{\ln x}{\ln a^2} \;{\color{red}-}\; \frac{\ln y}{\ln a^4} \;=\;3 \quad\Rightarrow\quad \frac{\ln x}{2\ln a} \;-\; \frac{\ln y}{4\ln a} \;=\;3 \quad\Rightarrow\quad 2\ln x \;-\; \ln y \;=\;12\ln a \;\;[1]$

    . . $\displaystyle \frac{\ln x}{\ln a^6} \;+\; \frac{\ln y}{\ln a^8} \;=\;4 \quad\Rightarrow\quad \frac{\ln x}{6\ln a} \;+\; \frac{\ln y}{8\ln a} \;=\;4 \quad\Rightarrow\quad 4\ln x \;+\; 3\ln y \;=\;96\ln a \;\;[2]$



    $\displaystyle \begin{array}{ccccccc}3\times [1]: & 6\ln x -3\ln y &=& 36\ln a \\
    \text{Add [2]:} & 4\ln x + 3\ln y &=& 96\ln a \end{array}$

    And we have: .$\displaystyle 10\ln x \:=\:132\ln a \quad\Rightarrow\quad \ln x \:=\:\tfrac{66}{5}\ln a \:=\:\ln\left(a^{\frac{66}{5}}\right)$

    . . Therefore: .$\displaystyle \boxed{x \;=\;a^{\frac{66}{5}}}$



    $\displaystyle \begin{array}{ccccc}\text{-}2\times [1]\!: & \text{-}4\ln x + 2\ln y &=& \text{-}24\ln a \\
    \text{Add [2]:} & 4\ln x + 3\ln y &=& 96\ln a \end{array}$

    And we have: .$\displaystyle 5\ln y \:=\:72\ln a \quad\Rightarrow\quad \ln y \:=\:\tfrac{72}{5}\ln a \:=\: \ln\left(a^{\frac{72}{5}}\right)$

    . . Therefore: .$\displaystyle \boxed{y \;=\;a^{\frac{72}{5}}}$

    Last edited by Soroban; May 28th 2010 at 12:03 PM.
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  4. #4
    Super Member dhiab's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, dhiab!

    It may be neater with natural logs . . . or maybe not.


    We have:

    . . $\displaystyle \frac{\ln x}{\ln a^2} + \frac{\ln y}{\ln a^4} \;=\;3 \quad\Rightarrow\quad \frac{\ln x}{2\ln a} + \frac{\ln y}{4\ln a} \;=\;3 \quad\Rightarrow\quad 2\ln x + \ln y \;=\;12\ln a \;\;[1]$

    . . $\displaystyle \frac{\ln x}{\ln a^6} + \frac{\ln y}{\ln a^8} \:=\:4 \quad\Rightarrow\quad \frac{\ln x}{6\ln a} + \frac{\ln y}{8\ln a} \:=\:4 \quad\Rightarrow\quad 4\ln x + 3\ln y \;=\;96\ln a \;\;[2]$


    $\displaystyle \begin{array}{ccccccc}\text{-}3\times [1]: & \text{-}6\ln x -3\ln y &=& \text{-}36\ln a \\$$\displaystyle
    \text{Add [2]:} & 4\ln x + 3\ln y &=& 96\ln a \end{array}$

    And we have: .$\displaystyle -2\ln x \:=\:60\ln a \quad\Rightarrow\quad \ln x \:=\:-30\ln a \:=\:\ln\left(a^{-30}\right)$

    . . Therefore: .$\displaystyle \boxed{x \;=\;a^{-30}}$


    $\displaystyle \begin{array}{ccccc}\text{-}2\times [1]\!: & \text{-}4\ln x - 2\ln y &=& \text{-}24\ln a \\$$\displaystyle
    \text{Add [2]:} & 4\ln x + 3\ln y &=& 96\ln a \end{array}$

    And we have: .$\displaystyle \ln y \:=\:72\ln a \quad\Rightarrow\quad \ln y \:=\:\ln\left(a^{72}\right)$

    . . Therefore: .$\displaystyle \boxed{y \;=\;a^{72}}$
    HELLO : thank you but in first equation you have - not +
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