Results 1 to 14 of 14

Math Help - e, ln and absolute values

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    31

    e, ln and absolute values

    Hello,

    An example in a text book I am studying uses the line:

    e^{\ln|sec(x)|}=sec(x)

    But I can't see how it could be anything other than:

    e^{\ln|sec(x)|}={|sec(x)|}

    Because sec(x) can sometimes be a negative value and I can't see how e to any exponent could be negative.

    Is my way of thinking correct or am I missing something?

    Thanks for your time.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Ted
    Ted is offline
    Member
    Joined
    Feb 2010
    From
    China
    Posts
    199
    Thanks
    1
    Quote Originally Posted by sael View Post
    Hello,

    An example in a text book I am studying uses the line:
    e^{\ln|sec(x)|}=sec(x)
    But I can't see how it could be anything other than:
    e^{\ln|sec(x)|}={|sec(x)|}
    Because sec(x) can sometimes be a negative value and I can't see how e to any exponent could be negative.

    Is my way of thinking correct or am I missing something?

    Thanks for your time.
    Is there any given interval for x in the example?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2008
    Posts
    31
    Sorry, still getting used to the forum interface.
    Last edited by sael; February 22nd 2010 at 02:22 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2008
    Posts
    31
    There is no interval given in the example.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2009
    Posts
    74
    Isn't it illegal to take a natural log of a negative value? Could that be why they use the absolute value?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Apr 2008
    Posts
    31
    Quote Originally Posted by satis View Post
    Isn't it illegal to take a natural log of a negative value? Could that be why they use the absolute value?
    Yes ln(x) is only for x> 0. My problem is not with that but that the absolute value is not preserved in the text book example and I think it should be :-)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member billa's Avatar
    Joined
    Oct 2008
    Posts
    100
    in your defense, against the textbook

     e^{\ln|sec(PI)|}=1
     sec(PI)=-1

    Perhaps you should post more of the example
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jan 2010
    Posts
    65
    Quote Originally Posted by sael View Post
    Hello,

    An example in a text book I am studying uses the line:

    e^{\ln|sec(x)|}=sec(x)

    But I can't see how it could be anything other than:

    e^{\ln|sec(x)|}={|sec(x)|}

    Because sec(x) can sometimes be a negative value and I can't see how e to any exponent could be negative.

    Is my way of thinking correct or am I missing something?

    Thanks for your time.
    You are correct, but I suspect that there is a restriction on the domain in the example.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Apr 2008
    Posts
    31
    Quote Originally Posted by billa View Post
    in your defense, against the textbook

     e^{\ln|sec(PI)|}=1
     sec(PI)=-1

    Perhaps you should post more of the example
    The example doesn't belong in this forum but here it is anyway

    I get the result y(x)=c|\cos(x)|- 2\cos^2(x) which is discontinuous and their result isn't which makes me wonder what I am missing.

    e, ln and absolute values-fo_ode_ivp.png
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Apr 2008
    Posts
    31
    Forgetting the example I posted. I am starting to see what I might be missing.

    We don't write:
    (x^2)^\frac{1}{2}=\pm{x}

    Instead:
    (x^2)^\frac{1}{2}=x

    So somehow I have to make the connection to e^{\ln|\sec{(x)}|}=\sec{(x)}
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member billa's Avatar
    Joined
    Oct 2008
    Posts
    100
    I recommend you re-post this in the differential equations section... I am currently in a DE and I want to know the answer now... if you get please post it here lol
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Apr 2008
    Posts
    31
    Quote Originally Posted by billa View Post
    I recommend you re-post this in the differential equations section... I am currently in a DE and I want to know the answer now... if you get please post it here lol
    From what I understand, at the moment, the problem isn't related to the particular ODE. I think it has more to do with \ln{|x|} being defined for all real x where as e^{x} is the inverse for \ln{x} for all x>0. So somehow I have to get my mind around functions that are not one-to-one. This is new ground for me and it doesn't quite make sense yet.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Apr 2008
    Posts
    31
    The best I can come up with is that e^x is not just a number with an exponent but is defined (in my texts at least) in terms of being the inverse of \ln{x} that just happens to behave as though it obeys the laws of exponents from algebra.

    The example appears to be treating e^{\ln|\sec{(x)}|} as though it means "inverse of" \ln{|\sec{(x)}|}, so in general what I really need is an understanding of the inverse of \ln{|x|} which is not a one-to-one function of x.

    My brain hurts
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Apr 2008
    Posts
    31
    Okay this really doesn't belong in this forum but I thought I had better relay what I have found anyway.

    It appears that within the example problem the part e^h=\sec(x) assumes that the constant of integration from h=\int\!p\,dx is ignored while evaluating (4). This comes from the derivation of (4) which I shouldn't post here because it is ODE related.

    Adding in the constant of integration:

    h=\int\!p\,dx=\int\!\tan (x)\,dx=\ln{|\sec (x)| + c_1}

    e^h=e^{\ln{|\sec (x)|}+c_1}=e^{c_1}|\sec(x)|
    e^{c_1}|\sec(x)|=c_2\sec(x)\quad where\,(c_2=\pm{e^{c_1}})

    The constant c_2 is ignored while evaluating (4) and hence the absolute value vanishes
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: September 7th 2011, 07:11 AM
  2. Replies: 1
    Last Post: June 1st 2011, 01:47 AM
  3. Replies: 8
    Last Post: May 23rd 2010, 10:59 PM
  4. Replies: 2
    Last Post: November 8th 2009, 01:52 PM
  5. Absolute Values
    Posted in the Math Topics Forum
    Replies: 7
    Last Post: February 8th 2008, 02:53 PM

Search Tags


/mathhelpforum @mathhelpforum