Results 1 to 3 of 3

Math Help - Prove this log using...

  1. #1
    Junior Member
    Joined
    Nov 2010
    Posts
    58

    Prove this log using...

    Given that x2 + y2 = 11xy
    Show that log [ (x-y)/3 ] = 0.5 (logx +logy)

    Similarly,
    Give that log [ (x-y)/3 ] = 0.5 (logx +logy)
    Show that x2 + y2 = 11xy
    I really liked the log unit and understood it, but I was stumped on this question. It was for a test and it keeps bothering me.
    What I did was completely wrong, but I will be able to understand it if someone shows the solution.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Dec 2010
    Posts
    1
    This is a good question. Here's a proof one way...hopefully it'll help you do it in reverse too.

    We have x^2 + y^2 = 11xy.
    Therefore x^2 - 11xy + y^2 = 0.

    Note that (x - y)^2 = x^2 - 2xy + y^2.
    Therefore (x - y)^2 = 9xy.

    Take logs of both sides:
    log((x - y)^2) = log(9xy).
    Therefore 2log(x-y) = log(9) + log(x) + log(y).
    (*) Therefore log(x-y) = 0.5(log(9) + log(x) + log(y)).

    Now log((x-y)/3) = log(x-y) - log(3).
    Therefore log(x-y) = log((x-y)/3) + log 3.

    Putting this into (*):
    log((x-y)/3) = -log(3) + 0.5(log(9) + log(x) + log(y)).
    Therefore log((x-y)/3) = -log(3) + log(3) + 0.5(log(x) + log(y)).

    Done!

    Hope this helps.
    Last edited by mr fantastic; December 15th 2010 at 08:13 PM. Reason: Deleted advertising link.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,686
    Thanks
    617
    Hello, TN17!

    A variation of exambox's solution . . .


    \text{Given that: }\:x^2+ y^2 \:=\: 11xy

    \text{Show that: }\:\log\left(\dfrac{x-y}{3}\right) \:=\:\frac{1}{2}(\log x + \log y)

    We have: . x^2 + y^2 \:=\:11xy

    Subtract 2xy from both sides: . x^2 - 2xy + y^2 \:=\:9xy

    Then we have: . (x-y)^2 \:=\:(3\sqrt{xy})^2

    Take the positive square root: . x-y \;=\;3\sqrt{xy} \quad\Rightarrow\quad \dfrac{x-y}{3} \;=\;(xy)^{\frac{1}{2}}

    Take logs: . \log\left(\dfrac{x-y}{3}\right) \;=\;\log(xy)^{\frac{1}{2}} \;=\;\frac{1}{2}\log(xy)

    Therefore: . \log\left(\dfrac{x-y}{3}\right) \;=\;\frac{1}{2}(\log x + \log y)

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove that
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: May 21st 2010, 05:48 AM
  2. Prove n^2<= ......
    Posted in the Advanced Algebra Forum
    Replies: 12
    Last Post: November 17th 2009, 05:52 AM
  3. Replies: 2
    Last Post: August 28th 2009, 02:59 AM
  4. prove that
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 7th 2008, 05:14 PM
  5. prove
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 7th 2008, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum