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Simplifying a log expression with identities

  1. Apr 20, 2016 #1

    ProfuselyQuarky

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    I was supposed to simplify the expression ##\ln |\cot {x}|+\ln |\tan {x}\cos {x}|## and apparently it’s wrong. Where’s the mistake? Is it not simplified enough or . . . ?

    ##\ln |\cot {x}|+\ln |\tan {x}\cos {x}|##

    ##=\ln |\frac {\cos {x}}{\sin {x}}|+\ln |\frac {\sin {x}}{\cos {x}}\cdot \cos {x}|##

    ##=\ln |\frac {\cos {x}}{\sin {x}}|+\ln |\sin {x}|##

    ##=\ln |\frac {\cos {x}\sin {x}}{\sin {x}}|##

    ##=\ln |\cos {x}|##

    If it’s really obvious, I’ll be happy with a hint. I thought that this was easy, but apparently I was mistaken?
     
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  3. Apr 20, 2016 #2

    Math_QED

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    Looks correct though. Note that Ln0 is undefined, so you might want to specify when the equality is correct.
     
  4. Apr 20, 2016 #3

    ProfuselyQuarky

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    Well, I was given the problem with no specifications and would it really affect the answer? I mean, this expression doesn’t deal with ##\ln {0}##.
     
  5. Apr 20, 2016 #4

    ProfuselyQuarky

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    So what is the error?
     
  6. Apr 20, 2016 #5

    SteamKing

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    You sure about that? You mean cos x is never equal to zero?
     
  7. Apr 20, 2016 #6

    ProfuselyQuarky

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    Oh, I never said cos x never is equal to zero. ##\cos {\frac {\pi}{2}}=0## and ##\cos {\frac {3\pi}{2}}=0## and so on.
     
  8. Apr 20, 2016 #7

    ProfuselyQuarky

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    But would that affect the answer? How did I simplify wrong?
     
  9. Apr 20, 2016 #8

    SteamKing

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    So what happens to ln |cos x| when x = (2k+1)π/2 ?
     
  10. Apr 20, 2016 #9

    ProfuselyQuarky

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    We get ln 0 which is undefined
     
  11. Apr 20, 2016 #10

    ProfuselyQuarky

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    Oh, wait, so I'd have to specify that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.

    That's great but there wasn't a single example of any of that in my book which is why I was thinking that my error had something to do with using the identities wrong.
     
  12. Apr 20, 2016 #11

    Math_QED

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    The answer is correct, as long as you specify for which values the expression is undefined.
     
  13. Apr 20, 2016 #12

    SammyS

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    Frankly. I don't think this is the problem. After all, the original expression isn't defined for values of x which make cos(x)=0, either.

    However, there are values of x for which the original expression is undefined, but are defined for ln(cos(x)) . Throw those out.
     
  14. Apr 20, 2016 #13

    Math_QED

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    Why do you think your expression is wrong in the first place?
     
  15. Apr 20, 2016 #14

    ProfuselyQuarky

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    It was marked wrong.
     
  16. Apr 20, 2016 #15

    SammyS

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    I'm not sure I was very clear in my previous post.

    I don't think the problem is:
    that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.​
    Those values of x at which ##\cos {x}=0## are not in the domain of ## \ \ln |\cot {x}|+\ln |\tan {x}\cos {x}|\ ## in the first place.

    I think that the problem is that ##\ln |\cos {x}|## is defined for some values of x for which ## \ \ln |\cot {x}|+\ln |\tan {x}\cos {x}|\ ## is not defined. I think that you must restrict the domain of the answer to eliminate those.
     
  17. Apr 20, 2016 #16

    Math_QED

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    Even then, it shouldn't be marked wrong since the expression itself is correct.
     
  18. Apr 20, 2016 #17

    ehild

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    SammyS was right. The initial and final expressions are not identical, as their ranges are different. The final one is not defined when cos(x)=0. The initial one is not defined when either sin(x) or cos(x) is zero.
     
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