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Thread: e, ln and absolute values

  1. #1
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    e, ln and absolute values

    Hello,

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

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

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

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

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

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

    Thanks for your time.
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  2. #2
    Ted
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    Quote Originally Posted by sael View Post
    Hello,

    An example in a text book I am studying uses the line:
    $\displaystyle e^{\ln|sec(x)|}=sec(x)$
    But I can't see how it could be anything other than:
    $\displaystyle e^{\ln|sec(x)|}={|sec(x)|}$
    Because $\displaystyle sec(x)$ can sometimes be a negative value and I can't see how $\displaystyle e$ 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?
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  3. #3
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    Sorry, still getting used to the forum interface.
    Last edited by sael; Feb 22nd 2010 at 02:22 PM.
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  4. #4
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    There is no interval given in the example.
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  5. #5
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    Isn't it illegal to take a natural log of a negative value? Could that be why they use the absolute value?
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  6. #6
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    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 :-)
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  7. #7
    Member billa's Avatar
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    in your defense, against the textbook

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

    Perhaps you should post more of the example
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  8. #8
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    Quote Originally Posted by sael View Post
    Hello,

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

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

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

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

    Because $\displaystyle sec(x)$ can sometimes be a negative value and I can't see how $\displaystyle e$ 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.
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  9. #9
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    Quote Originally Posted by billa View Post
    in your defense, against the textbook

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

    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 $\displaystyle y(x)=c|\cos(x)|- 2\cos^2(x)$ 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
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  10. #10
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    Forgetting the example I posted. I am starting to see what I might be missing.

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

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

    So somehow I have to make the connection to $\displaystyle e^{\ln|\sec{(x)}|}=\sec{(x)}$
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  11. #11
    Member billa's Avatar
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    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
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  12. #12
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    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 $\displaystyle \ln{|x|}$ being defined for all real x where as $\displaystyle e^{x}$ is the inverse for $\displaystyle \ln{x}$ for all $\displaystyle x>0$. 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.
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  13. #13
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    The best I can come up with is that $\displaystyle e^x$ is not just a number with an exponent but is defined (in my texts at least) in terms of being the inverse of $\displaystyle \ln{x}$ that just happens to behave as though it obeys the laws of exponents from algebra.

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

    My brain hurts
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  14. #14
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    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 $\displaystyle e^h=\sec(x)$ assumes that the constant of integration from $\displaystyle h=\int\!p\,dx$ 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:

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

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

    The constant $\displaystyle c_2$ is ignored while evaluating (4) and hence the absolute value vanishes
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