Results 1 to 4 of 4

Thread: logarithms

  1. #1
    lrn24gve
    Guest

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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
    $\displaystyle \log_9 x+\log_y 8=2$
    $\displaystyle \log_x 9+\log_8 y=8/3$
    Multiply the two equations,
    $\displaystyle (\log_9 x+\log_y 8)(\log_x 9+\log_8 y)=2(8/3)=16/3$
    $\displaystyle \log_9 x\log_x 9+\log_y 8\log_x 9+\log_8 y\log_9 x+\log_y 8\log_8 y=16/3$
    $\displaystyle 2+\log_y8\log_x 9+\log_8 y\log_9 x=16/3$
    $\displaystyle \log_y 8\log_x 9+\log_8 y\log_9 x=10/3$
    $\displaystyle z=\log_8 y\log_9 x$
    $\displaystyle 1/z+z=10/3$
    $\displaystyle 3+3z^2=10z$
    $\displaystyle 3z^2-10z+3=0$
    $\displaystyle z=\frac{10\pm \sqrt{100-36}}{6}=1/3,3$
    The first equation,
    $\displaystyle \log_9 x+\frac{1}{\log_8 y}=2$
    $\displaystyle A=\log_9 x, B=\log_8 y$
    Thus,
    $\displaystyle A+\frac{1}{B}=2$
    $\displaystyle z=AB=1/3,3$
    This can be solved simply.
    From the A,B you can find x,y.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    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 ($\displaystyle e$ - natural logarithms will do) then we have:

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

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

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, lrn24gve!

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


    Solve the following system for $\displaystyle (x,y):$

    . . $\displaystyle \begin{array}{cc}\log_9x + \log_y8\:=\:2 \\
    \log_x9+ \log_8y \:=\:\frac{8}{3}\end{array}$

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


    The equations become:

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

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

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

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

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


    And we have two equations to solve:

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

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


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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. logarithms help
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Apr 6th 2010, 06:29 PM
  2. Logarithms
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 6th 2010, 04:46 PM
  3. logarithms
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Mar 24th 2010, 04:26 AM
  4. Logarithms
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Mar 18th 2010, 02:52 PM
  5. logarithms
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Apr 16th 2008, 09:55 AM

Search Tags


/mathhelpforum @mathhelpforum