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Math Help - logarithms

  1. #1
    lrn24gve
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    logarithms

    I don't understand this problem:

    solve the following system for (x;y):
    log x base of 9 + log 8 base of y = 2
    log 9 base of x + log y base of 8 = 8/3
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  2. #2
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    Quote Originally Posted by lrn24gve View Post
    I don't understand this problem:

    solve the following system for (x;y):
    log x base of 9 + log 8 base of y = 2
    log 9 base of x + log y base of 8 = 8/3
    \log_9 x+\log_y 8=2
    \log_x 9+\log_8 y=8/3
    Multiply the two equations,
    (\log_9 x+\log_y 8)(\log_x 9+\log_8 y)=2(8/3)=16/3
    \log_9 x\log_x 9+\log_y 8\log_x 9+\log_8 y\log_9 x+\log_y 8\log_8 y=16/3
    2+\log_y8\log_x 9+\log_8 y\log_9 x=16/3
    \log_y 8\log_x 9+\log_8 y\log_9 x=10/3
    z=\log_8 y\log_9 x
    1/z+z=10/3
    3+3z^2=10z
    3z^2-10z+3=0
    z=\frac{10\pm \sqrt{100-36}}{6}=1/3,3
    The first equation,
    \log_9 x+\frac{1}{\log_8 y}=2
    A=\log_9 x, B=\log_8 y
    Thus,
    A+\frac{1}{B}=2
    z=AB=1/3,3
    This can be solved simply.
    From the A,B you can find x,y.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by lrn24gve View Post
    I don't understand this problem:

    solve the following system for (x;y):
    log x base of 9 + log 8 base of y = 2
    log 9 base of x + log y base of 8 = 8/3
    Convert all the logs to a common base ( e - natural logarithms will do) then we have:

    \frac{\ln(x)}{\ln(9)}+\frac{\ln(8)}{\ln(y)}=2
    \frac{\ln(9)}{\ln(x)}+\frac{\ln(y)}{\ln(8)}=\frac{  8}{3}

    Which is now a pair of simultaneous equations in \ln(x) and \ln(y) which have to be solved.

    RonL
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  4. #4
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    Hello, lrn24gve!

    My solution is not an improvement on the others . . .


    Solve the following system for (x,y):

    . . \begin{array}{cc}\log_9x + \log_y8\:=\:2 \\<br />
\log_x9+ \log_8y \:=\:\frac{8}{3}\end{array}

    Theorem: . \log_ba \:=\:\frac{1}{\log_ab}


    The equations become:

    . . \begin{array}{cc}\log_9x + \frac{1}{\log_8y} &=\:2 \\ \frac{1}{\log_9x} + \log_8y & =\:\frac{8}{3}\end{array}<br />
\begin{array}{cc}\bf{(1)}\\ \bf{(2)}\end{array}\;\;<br />
\begin{array}{cc}\Rightarrow \\ \Rightarrow\end{array}\;\;<br />
\begin{array}{cc}(\log_9x)(\log_8y) + 1 & =\:2(\log_8y) \\<br />
1 + (\log_9x)(\log_8y) & =\:\frac{8}{3}(\log_8x)\end{array}

    Hence: . 2(\log_8y) \:=\:\frac{8}{3}(\log_9x)\quad\Rightarrow\quad \log_8y \:=\:\frac{4}{3}(\log_9x)\;\;\bf{(3)}

    Subsitute into (1): . \log_9x + \frac{1}{\frac{4}{3}(\log_9x)} \:=\:2

    . . which simplifies to: . 4(\log_9x)^2 - 8(\log_9x) + 3 \:=\:0

    . . which factors: . \left[2\log_9x - 1\right]\left[2\log_9x - 3\right] \:=\:0


    And we have two equations to solve:

    . . 2\log_9x - 1 \:=\:0\quad\Rightarrow\quad \log_9x = \frac{1}{2}\quad\Rightarrow\quad x \,=\,9^{\frac{1}{2}}\quad\Rightarrow\quad\boxed{x \,=\,3}
    . . . . Substitute into (3): . \log_8y \,= \,\frac{4}{3}(\log_93) \,=\,\frac{4}{3}\cdot\frac{1}{2}\,=\,\frac{2}{3}
    . . . . . . Then: . y \,= \,8^{\frac{2}{3}}\quad\Rightarrow\quad\boxed{y\,=\  ,4}

    . . 2\log_9x - 3\:=\:0\quad\Rightarrow\quad \log_9x = \frac{3}{2}\quad\Rightarrow\quad x\,=\,9^{\frac{3}{2}}\quad\Rightarrow\quad\boxed{x  \,=\,27}
    . . . . Substitute into (3): . \log_8y \,=\,\frac{4}{3}(\log_927) \,=\,\frac{4}{3}\cdot\frac{3}{2} \,= \,2
    . . . . . . Then: . y \,=\,8^2\quad\Rightarrow\quad\boxed{y \,= \,64}


    Solutions: . (x,y) \:=\:(3,4),\:(27,64)

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