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On logarithms

  1. Mar 12, 2004 #1
    If (ln x) / x = (ln 2) / 2, does it mean that my x here is equal to 2?
  2. jcsd
  3. Mar 12, 2004 #2
    not just x=2. x could be 4, too. calculus can be used to prove those are the only two solutions, i think.
  4. Mar 12, 2004 #3

    matt grime

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    yep those are the anwsers - take hte derivative, it's monotone increasing for x less than e monotone decreasing for x greater than e.

    edit, the pronoun 'it' does not refer to the derivative, 'it' refers to the original function. Sorry if there was confusion there.
    Last edited: Mar 12, 2004
  5. Mar 12, 2004 #4


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    In general f(x)= f(y) leads to x= y IF f is "one to one" (in fact that's the definition of "one to one").

    f(x)= ln(x)/x is NOT one to one. If a function has derivative that is always positive or always negative, then it is "monotone" and so one to one. Here, f'(x)= (1-ln(x))/x2. That is continuous as long as x is positive and so can change from negative to positive only where it is 0: that is, 1- ln(x)= 0 which gives x= e.
    Now check x= 3> e: f'(3)= (1- ln(3))/32= -.011. That is negative so, as matt grime said, f(x) is decreasing, therefore monotone, for x> e.
    Check x= 2< e: f'(2)= (1- ln(2))/22= 0.0767: positive.
  6. Mar 12, 2004 #5


    how can I show that my "x" in my previous question " (ln x)/x =
    (ln 2) /2 has a different value aside from 2. Yes, I understand that it is not one-to -one , but how did phoenixthoth got "not just x=2. x could be 4, too. calculus can be used to prove those are the only two solutions"? In short, how did he got 4?
  7. Mar 12, 2004 #6


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    [ln(x)]/x = [ln(2)]/2 implies that if we take e to the power of each side, we get the equality exp{[ln(x)]/x} = exp{[ln(2)]/2}. This can be rewritten as {exp[ln(x)]}^(1/x) = {exp[ln(2)}^(1/2). This simplifies to x^(1/x) = 2^1/2. Taking both sides to the x power: x=2^(x/2). It is pretty clear that both x=2 and x=4 solve this.

    To reassure yourself that there are no other solutions, you could plot y=x (a straight line though the origin, with slope = 1) on the same graph as y=2^(x/2) which will be a curve concave upward, passing through (0,1). So even if you draw the curve real sloppy, you can reason that since it is monotonically increasing, and since its first derivative also increases monotonically, there are only three possibilities for the number of intersections of the two figures: 0, or 1 (osculating) or 2.
    Last edited: Mar 12, 2004
  8. Mar 12, 2004 #7
    i graphed it along with the horizontal line ln2/2 and saw two points of intersection. then i knew calc could be used to prove those are the only two. i cheated and used a computer to graph it but one could just use curve sketching techniques from calculus (or even precalc or plotting points) to graph it.

    actually, what happened is that i had mathematica numerically solve it. the thing is it kept giving me 2 as the only answer but i suspected that there must be some kind of trick to it and kept pushing (a bit) for a nother solution.
  9. Mar 13, 2004 #8
    Hello again

    Well, now I understand... thank you for the help that you show me. I really appreciate them. =)
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