Results 1 to 15 of 15

Math Help - Prove the natural log can't equal 0

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    Prove the natural log can't equal 0

    I want to prove the natural log can't equal 0. Is this done via the limit definition or the fact the inverse of ln is e and e can't be 0?

    How would this proof be started as well thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by dwsmith View Post
    I want to prove the natural log can't equal 0. Is this done via the limit definition or the fact the inverse of ln is e and e can't be 0?

    How would this proof be started as well thanks.
    I don't understand. \ln(1)=0?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle ln(0)=DNE
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by dwsmith View Post
    \displaystyle ln(0)=DNE
    So the question isn't 'why can't the natural logarithm ever equal zero' as you stated, but instead 'why can't e^x ever equal zero'?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    I want to show that we can't take the natural log of 0. I wasn't sure if using the inverse function would be how to do it or not.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by dwsmith View Post
    I want to show that we can't take the natural log of 0. I wasn't sure if using the inverse function would be how to do it or not.
    I mean, this is all kind of semantical. I could call \ln(0)=e if I wanted to. If you're asking though, given the predefined function \exp:\mathbb{R}\to\mathbb{R}^+ then the inverse function \exp^{-1} has domain equal to the codomain of \exp, namely \mathbb{R}^+ and since 0\notin\mathbb{R}^+, we have that \exp^{-1} isn't (at least not with keeping it as an inverse of the exponential function) defined at zero.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    I don't necessarily want to use e. I was just throwing that out there. I want to prove that 0 isn't in the domain of ln without just saying. Hey the domain is (0, infinity).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by dwsmith View Post
    I don't necessarily want to use e. I was just throwing that out there. I want to prove that 0 isn't in the domain of ln without just saying. Hey the domain is (0, infinity).
    But that doesn't make sense. If you don't define \ln as being the inverse of \exp then I'd ask you 'what is it' and no matter how you've answered (assuming you've done so rigorously) you will, by necessity, have answered your own question. You can't define \ln without specifying its domain. A function can be thought of as the ordered triple \left(A,B,f\right) where A,B are sets and f\subseteq A\times B is a relation such that for each a\in A there is precisely one b\in B for which (a,b)\in f.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    I found a method using the derivative of the ln

    \displaystyle ln(|t|)=\int\frac{dt}{t}

    From this equation, we can easily see that \displaystyle t\neq 0 and \displaystyle |t|\in\mathbb{R}^+
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by dwsmith View Post
    I found a method using the derivative of the ln

    \displaystyle ln(|t|)=\int\frac{dt}{t}

    From this equation, we can easily see that \displaystyle t\neq 0 and \displaystyle |t|\in\mathbb{R}^+
    But that isn't really well-defined. Really what you may want to consider is that \displaystyle \ln(x)=\int_1^x \frac{dt}{t} and that if you take the domain to the largest for which that equation makes sense then surely 0 isn't included. But that's really quite silly.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by dwsmith View Post
    I found a method using the derivative of the ln

    \displaystyle ln(|t|)=\int\frac{dt}{t}

    From this equation, we can easily see that \displaystyle t\neq 0 and \displaystyle |t|\in\mathbb{R}^+
    When I see something like this it convinces me that indefinite integration should never be taught.

    That does not define the natural logarithm. It leaves it defined up to an additive constant. You need to define it as a definite integral when the non-convergence of that integral when the upper limit is zero shows that the natural log of zero does not exist.

    CB
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by dwsmith View Post
    I found a method using the derivative of the ln

    \displaystyle ln(|t|)=\int\frac{dt}{t}

    From this equation, we can easily see that \displaystyle t\neq 0 and \displaystyle |t|\in\mathbb{R}^+
    That is more exact to say that...

    \displaystyle \ln |t| = \left\{\begin{array}{ll}\int_{-1}^{t} \frac{d \tau}{\tau} ,\,\,t<0\\{}\\ \int_{1}^{t} \frac{d \tau}{\tau} ,\,\, t > 0\end{array}\right.

    In any case the function \ln |t| has a singularity in t=0...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by chisigma View Post
    That is more exact to say that...

    \displaystyle \ln |t| = \left\{\begin{array}{ll}\int_{-1}^{t} \frac{d \tau}{\tau} ,\,\,t<0\\{}\\ \int_{1}^{t} \frac{d \tau}{\tau} ,\,\, t > 0\end{array}\right.

    In any case the function \ln |t| has a singularity in t=0...

    Kind regards

    \chi \sigma
    Then:

    \displaystyle \ln(0.5)=\ln(|-0.5|)=\int_{-1}^{-0.5} \dfrac{1}{\tau}\ d\tau<0\ ?

    CB
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by dwsmith View Post
    I want to show that we can't take the natural log of 0. I wasn't sure if using the inverse function would be how to do it or not.

    Take the very basic, high-schoolish, definition of logarithm: \log_ab=x\Longleftrightarrow a^x=b , when we already

    know-define-demmand that a,b>0\,,\,a\neq 1 . From this, b=0\Longrightarrow a^x=0 for some

    x\in\mathbb{R} which, as Drexel already pointed out, is impossible by the very

    definition of power of positive real numbers (although he talked about inverse function and

    stuff, which isn't necessary in a first approach to this)

    Tonio
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,713
    Thanks
    1472
    Quote Originally Posted by Drexel28 View Post
    I mean, this is all kind of semantical. I could call \ln(0)=e if I wanted to. If you're asking though, given the predefined function \exp:\mathbb{R}\to\mathbb{R}^+ then the inverse function \exp^{-1} has domain equal to the codomain of \exp, namely \mathbb{R}^+ and since 0\notin\mathbb{R}^+, we have that \exp^{-1} isn't (at least not with keeping it as an inverse of the exponential function) defined at zero.
    In other words it makes no sense to try to proved anything about a function without saying how you have defined it! One way to show that 0 is not in the domain of ln(x) is to define ln(x) is as the inverse function of f(x)= e^x and then show that e^x= 0 is not true for any x. How you do that would, of course, depend upon how you defined e^x.

    Another way to define ln(x) is
    ln(x)= \int_1^x \frac{1}{t}dt

    You can then show that 0 is not in the domain of ln(x) by showing that \int_1^0 \frac{1}{t}dt does not exist.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove the divergence is equal to zero
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 5th 2012, 04:38 PM
  2. Prove me that pi is no equal 2
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 8th 2011, 09:40 AM
  3. Prove that g(x) greater than/equal to 0?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 24th 2009, 02:23 AM
  4. Prove that the determinants are equal
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 20th 2009, 03:27 AM
  5. prove H_T less than or equal to S_n
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 16th 2008, 02:59 PM

/mathhelpforum @mathhelpforum