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A log proof

  1. Jun 16, 2015 #1
    1. The problem statement, all variables and given/known data
    I am having problem with number 16ii)

    1smhk0.jpg
    2. Relevant equations
    log([x+y]/sqrt5)

    3. The attempt at a solution

    Sorry still not used to latex
    So I tried this
    if x/y + y/x =3 then(

    (x^2 + y^2)/xy = 3xy
    x^2 + y^2 = 3xy
    (x-y)^2 = xy
    (x-y) = sqrt(xy)
    (x-y)(x+y) = sqrt(xy)(x+y)
    (x+y) = (x^2-y^2)/sqrt(xy)

    But subbing in the new expression for (x+y) did not help me at all I end up getting the same thing over and over.
     
  2. jcsd
  3. Jun 16, 2015 #2

    mfb

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    Good so far.
    This leads to (i), for (ii) you can use the opposite direction. Getting (x-y)^2 again does not help I think.
     
  4. Jun 16, 2015 #3
    By using the opposite direction you mean using the right side?
     
  5. Jun 16, 2015 #4

    mfb

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    Instead of (x-y)^2, go for (x+y)^2.
     
  6. Jun 16, 2015 #5

    SammyS

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    What allows you to go from the above line to that below?
     
  7. Jun 16, 2015 #6

    mfb

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    The uppermost line you quoted in the lower quote has a typo, the lines above and below are correct.
     
  8. Jun 16, 2015 #7

    SteamKing

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    You start out OK.
    Whoops! You've jumped ahead of yourself here. Check your algebra. Remember, just because you add the terms on one side of an equation together, this does not imply that the other side is changed.
    Now, you've corrected yourself.
    Whoops! Back in the weeds, again.

    What happened to the factor of 3 on the RHS?
    Is (x - y)2 = x2 + y2 ?

    You've got to carefully check your algebra.
     
  9. Jun 16, 2015 #8
    Okay I see that mistake and I get (x-y) = sqrt(xy) the correct way now. So how I can I get (x+y) from the x/y + y/x = 3 ?
     
  10. Jun 16, 2015 #9

    SammyS

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    Yes. How did you get that? In other words, what is it that you intended to do in taking that step ?
     
  11. Jun 16, 2015 #10
    I was hoping to get a new expression to sub in the log([x+y]/sqrt5) but the expression I get I think I showed it up the top and it doesn't help me
     
  12. Jun 16, 2015 #11

    SammyS

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    simple question:

    What is the algebraic step you took there ?
     
  13. Jun 16, 2015 #12
    x/y + y/x =3 then

    (x^2 + y^2)/xy = 3
    x^2 + y^2 = 3xy
    x^2 - 2xy + y^2 = xy
    (x-y)^2 = xy
    (x-y) = sqrt(xy)
    (x-y)(x+y) = sqrt(xy)(x+y) [ multiplying both sides by (x+y) ]
    (x+y) = (x^2-y^2)/sqrt(xy)
     
  14. Jun 16, 2015 #13

    SammyS

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    At this point, I see that you subtracted 2xy . Why not add 2xy instead?
     
  15. Jun 16, 2015 #14
    Oh because I'm an idiot sigh haha hmm wow . Okay thank you! Not sure why I couldn't see that...
     
  16. Jun 16, 2015 #15

    mfb

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    Two steps in one, but this is correct. Your two questions answer each other.
     
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