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Math Help - please help logarithms

  1. #1
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    please help logarithms

    solve simultaneously the equations
    log
    yx-4logxy=0
    log8x+log8y=1
    tks for ur help
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  2. #2
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    Try using the identities

    1.\;\log_y\,x\,=\,\frac{\ln\,x}{\ln\,y}

    2.\;a^{x\,+\,y}\,=\,a^xa^y

    3.\;a^{\log_a\,x}\,=\,x
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  3. #3
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    i tried using those identities but i did nt solve them
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  4. #4
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    \mbox{a.}\;\log_y\,x\,-\,4\log_x\,y\,=\,0

    \mbox{b.}\;\log_8\,x\,+\,\log_8\,y\,=\,1

    Try adding 4\log_x\,y to both sides of equation (a) and using identity (1). Then multiply both sides by ln x and ln y and take the square root of both sides.
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  5. #5
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    Quote Originally Posted by clutch View Post
    i tried using those identities but i did nt solve them
    In your 2nd equation, all the logarithms have the same base. So use the identity:

     log_{a}x + log_{a} y = log_{a} xy

    And then take the exponential of both sides!

    In the first question it's a bit trickier since they have different exponents. But you can use the following rule to write one of the logarithms in ANY base you like (although it would help if you put them into the same base as the other logarithm in the equation:

     log_{b}(x) = \frac{log_{k}(x)}{log_{k}(b)}

    And you simply get to choose what k is.

    I recommend you take  log_{y}(x) and turn it into \frac{log_{x}(x)}{log_{x}(y)}. Once you plug this back into the equation, it should be easy to solve using the basic laws of logarithms.
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  6. #6
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    Hello, clutch!

    We can use the identity: . \log_xy \:=\:\frac{1}{\log_yx}


    Solve the equations : . \begin{array}{cccc} \log_yx - 4\log_xy &=&0 & {\color{blue}[1]}\\ \log_8x + \log_8y &=& 1 & {\color{blue}[2]} \end{array}

    Use the identity on [1]: . \log_yx - \frac{4}{\log_yx} \:=\:0 \quad\Rightarrow\quad \bigg[\log_yx\bigg]^2 - 4 \:=\:0

    . . \bigg[\log_y\bigg]^2 =\:4 \quad\Rightarrow\quad \log_yx \:=\:\pm2 \quad\Rightarrow\quad x \:=\:y^{\pm2} .[3]


    From [2], we have: . \log_8(xy) \:=\:1 \quad\Rightarrow\quad xy \:=\:8^1 \quad\Rightarrow\quad x \:=\:\frac{8}{y} .[4]

    Equate [3] and [4]: . y^{\pm2} \:=\:\frac{8}{y} \quad\Rightarrow\quad y ^{1\pm2} \:=\:8 \quad\Rightarrow\quad\begin{array}{ccc}y^3 \:=\:8 & \Rightarrow & y \:=\:2 \\ y^{-1} \:=\:8 & \Rightarrow & y \:=\:\frac{1}{8}\end{array}<br />

    Substitute into [4] . \begin{array}{cccc}y = 2\!: & x = \frac{8}{2} & \Rightarrow & x = 4 \\ \\[-3mm] y = \frac{1}{8}\!: & x = \frac{8}{\frac{1}{8}} & \Rightarrow & x = 64 \end{array}


    Answers: . (4,2),\;\left(64,\,\tfrac{1}{8}\right)

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