Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Absolute value and log rules.

  1. Jul 24, 2011 #1
    1. The problem statement, all variables and given/known data

    y = ln | sin x |

    2. Relevant equations


    3. The attempt at a solution

    No attempt at a solution, because it's more of a "which part of this is incorrect" question.

    y = ln | sin x |

    Would everyone agree that this is equivalent?

    [itex]y = ln \sqrt{(sin x^2)}[/itex]

    Then by properties of logarithms:

    [itex]y = \frac{1}{2}ln (sin x^2)[/itex]

    And again,

    [itex]y = \frac{2}{2}ln sin x[/itex]


    [itex]y = ln sin x[/itex]

    So then, ln|sin x| is equivalent to ln sin x?

    Is this true? If so, is it just essentially a visual demonstration of the domain of the logarithm function?

    (note that I'm having trouble placing the 2 power on "sin" but I'm sure you all know what I mean, (sin x)^2)
  2. jcsd
  3. Jul 24, 2011 #2
    Go back to the beginning, think about the domain.
  4. Jul 24, 2011 #3
    As far as I can tell, the domain agrees with what I did..
  5. Jul 24, 2011 #4
    What is the domain?
  6. Jul 24, 2011 #5
    Greater than 0... so ln | x | has a range of all real numbers. So what's wrong with the steps I took above that says ln x = ln | x |?

    I'm not sure what you're trying to tell me, a little vague..
  7. Jul 25, 2011 #6
    Oh. Have you mentioned x > 0 earlier? My mistake if so ... otherwise ... how would we know?

    I thought the idea is to show what's wrong with the definition of y ... which is that the original function has a problem with its domain. But if you already know that x > 0 is the domain, then what are you supposed to do with this problem? What question is being asked?

    ps ... what about x < 0 ?
    Last edited: Jul 25, 2011
  8. Jul 25, 2011 #7

    I like Serena

    User Avatar
    Homework Helper

    Hi 1MileCrash! :smile:

    The problem does not put a limit on the domain.
    So for instance x = -pi/3 would be an acceptable value.

    If you substitute that in the original and also into the final expression, are the results then the same?

    Btw, the log function is only defined for positive numbers.
  9. Jul 25, 2011 #8


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    For [itex]y = \frac{1}{2}\ln \sqrt{(sin^2 x)}[/itex], the domain is all reals except fo the zeros of the sine function.

    For [itex]y = \frac{2}{2}\,\ln \sqrt{(sin\, x)}[/itex], the domain is all reals for which sin(x) > 0 .

    Therefore, these expressions are not equivalent.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook