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Logarithm problem

  1. Jan 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Given 3log5 xy2 + log5x = 4+2log5y, prove xy = 5.

    2. Relevant equations



    3. The attempt at a solution

    Can i prove like this? Or i must use 3log5 xy2 + log5x = 4+2log5y to prove xy=5? Meaning my final result must be xy = 5? Please enlighten me..

    Mathematics 3.png
     
  2. jcsd
  3. Jan 10, 2012 #2

    ehild

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    It is a correct proof if you add the condition that both x and y are greater then 0, just to be strict.

    ehild
     
  4. Jan 10, 2012 #3

    micromass

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    I don't get it. You use xy=5, to prove that relationship.

    But you must do the opposite. You must use that relationship to prove xy=5. That means that you can't start from assuming xy=5. You must prove it.
     
  5. Jan 10, 2012 #4

    Curious3141

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    Technically, you've proven that xy = 5 satisfies the equation (that xy = 5 is sufficient). However, I believe the question is asking you to prove that xy = 5 is the only solution (that xy = 5 is necessary).

    You'd be better advised to just manipulate the equation with the laws of logs until you get the required result. It's quite simple.
     
  6. Jan 10, 2012 #5

    Curious3141

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    Beat me to the punch. :tongue:
     
  7. Jan 10, 2012 #6

    ehild

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    I have problems with the meaning of the sentence "prove xy=5" Can not it be understood as "prove xy=5 is solution of the equation"? When we solve an equation, we substitute the result back to see if it is really a solution.

    I think the other procedure, starting from the original equation would correspond to the task: "prove that xy=5 follows from equation
    3 log5 xy2 + log5 x = 4+2 log5 y ".

    ehild
     
  8. Jan 10, 2012 #7
    .............
     
  9. Jan 10, 2012 #8

    Curious3141

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    The way the question is phrased, viz. *Given* that equation (call this proposition [itex]P[/itex]), *prove* xy=5 (call this proposition [itex]Q[/itex]) indicates that one must establish [itex]P \Rightarrow Q[/itex].

    Substituting xy = 5 back into the equation only establishes [itex]Q \Rightarrow P[/itex]. I don't think this is sufficient.

    And, just to be pedantic, when one solves an equation, one tries to find all solutions (or at least all solutions within a defined range) that satisfy an equation. Substituting values back into an equation to see that LHS = RHS is essentially verifying that those values are solutions, not solving the equation. :smile:
     
  10. Jan 10, 2012 #9

    Curious3141

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    Sorry, your post is unclear, and I can't even tell if you're agreeing with me. But see my reply to ehild's post.
     
  11. Jan 10, 2012 #10
    Here is my take:


    To prove that xy = 5 is the only solution, does it suffice to show
    that xy = 5 ? No, but it is necessary that xy = 5.

    Suppose it is shown that xy = 5 satisfies the equation. It would
    be sufficient to show that no other solution satisfies the equation
    in order to prove that xy = 5 is the only solution.
     
  12. Jan 10, 2012 #11

    Curious3141

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    OK, so how would you show there are no other solutions?
     
  13. Jan 10, 2012 #12

    Curious3141

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    To illustrate my point, if I posed a question like this:

    Given [itex]\log_5{x} + \log_5{y} = \frac{1}{\log_5{x} + \log_5{y}}[/itex] prove that [itex]xy = 5[/itex],

    it would actually be an incorrect question. Because it doesn't allow for the possibility that [itex]xy = \frac{1}{5}[/itex], also a valid solution.

    However, if I posed the question like this:

    Verify that [itex]xy = 5[/itex] satisfies the equation [itex]\log_5{x} + \log_5{y} = \frac{1}{\log_5{x} + \log_5{y}}[/itex],

    that would be perfectly alright and the OP's method would be completely valid.

    It's not the best example, but it's the best argument I could come up with while suffering from a bad cold. :tongue:
     
  14. Jan 10, 2012 #13
    After using logarithmic and expontial rules, one of the steps
    I got to is:

    [itex](xy)^4 = 5^4[/itex]


    I see that xy = 5 works (with the acceptable ranges of x and y).



    Then I would show that all of the remaining cases

    xy = -5, xy = 5i, and xy = -5i fail to satisfy the equation,
    when taking into account the permissible real values for x and y.




    Note: I left this forum for the night.
     
  15. Jan 10, 2012 #14

    Curious3141

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    OK, I suppose that's fair enough. I guess any rigorous method one uses to solve this problem would involve observing that x and y should be restricted to positive values. It's probably easier to just state that at the start, as ehild suggests instead of seeking then excluding imaginary solutions after the fact.
     
    Last edited: Jan 11, 2012
  16. Jan 11, 2012 #15

    ehild

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    Thanks Curious. I know that you are right, I just quibble ... :smile: But it is usual in high-school Maths when showing that two relations P an G are equivalent that we say: we got Q from P through equivalence relations, so the proof works backwards, too. Applying the same relations in the opposite order we would get also Q→P.

    To admit the truth, it is not the case here. xy=5 follows from the equation supposed it is valid, that is, both x and y are positive, but not vice versa. The last step of the proof log x +log y →log(xy), is true, but log(xy)→log x+log y is not without excluding non-positive x, y values.

    To show that there is only one solution of the original equation it is enough to refer that it is linear in the logarithms and logarithm is a monotonous function:lg(x)=log(a)→x=a Proving that xy=5 is solution means also that it is the only solution.


    ehild
     
    Last edited: Jan 11, 2012
  17. Jan 11, 2012 #16
    This problem is not difficult :

    Just solve 3log5 xy2 + log5x = 4+2log5y using law : log a + log b = log ab

    You will surely get your way out easily rather than doing opposite...
    See post #2 also.
     
  18. Jan 11, 2012 #17

    Curious3141

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    Good point. Actually, if one wishes to consider logs of negative numbers, both ways can be iffy, depending on how one defines the log. Considering only principal values of the natural logs, [itex]\ln(-1) = i\pi[/itex] and [itex]\ln1 = 0[/itex] (the latter as usual).

    Even [itex]\log x + \log y = \log{xy}[/itex] doesn't work out as [itex]2\ln(-1) = 2{\pi}i \neq 0 = \ln1[/itex]. And going the other way (splitting it up) similarly doesn't work if we want to write [itex]0 = \ln1 \neq 2{\pi}i = 2\ln(-1)[/itex].

    I understand this, and I guess I was tacitly assuming that we were working with positive x,y (since this is at the elementary level). But it's better to state this explicitly, as you suggested. And the "resubstituting" (verifying) sort of solution is fine if (as you suggest), it's made rigorous by observing that logs are one-to-one.
     
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