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Why is x^(1/x) = e^((1/x)lnx)?

  1. Dec 26, 2012 #1
    or why is 2gxjj15.jpg ?
    thx in advance!
     
  2. jcsd
  3. Dec 26, 2012 #2

    micromass

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    Do you understand why

    [tex]x=e^{ln(x)}[/tex]

    Think about the definition of the logarithm.
     
  4. Dec 26, 2012 #3
    I don't understand why.
     
  5. Dec 26, 2012 #4
    Try to make logarithm of the both sides. Regards.
     
  6. Dec 26, 2012 #5

    micromass

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    So what is your definition of the logarithm?
     
  7. Dec 26, 2012 #6
  8. Dec 26, 2012 #7
    I know it is a very basic definition but that is it for me.
     
  9. Dec 26, 2012 #8

    lurflurf

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    Logarithms can be used to define exponentiation as

    [itex]u(x)^{v(X)}=e^{v(x) \log(u(x))}[/itex]

    Otherwise it can be proved as a theorem.

    Note also that e^x is continuous is used in your example to justify moving the limit past e.
     
  10. Dec 26, 2012 #9

    micromass

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    OK, so you're saying that [itex]x=ln(b)[/itex] iff [itex]e^x=b[/itex].

    So take an arbitrary b. Then we can of course write [itex]ln(b)=ln(b)[/itex]. Define [itex]x=ln(b)[/itex]. The "iff" above yields directly that [itex]b=e^x = e^{ln(b)}[/itex].
     
  11. Dec 26, 2012 #10
    Do you know where can I find this theorem? I do not understand why one is equal to the otherI am doing some limits exercises and I came up with [tex]({\frac{n+3}{n+1}})^{n}[/tex] limit when n->infinity. And to calculate this limit wolphram alpha show this "formula" but I just can't understand. I have some basic knowledge of logarithm.
     
  12. Dec 26, 2012 #11
    I got it. But I fail to apply it for my case, when x has an exponential. Like [tex]{x}^{1/x}={e}^{(1/x)lnx}[/tex]
     
  13. Dec 26, 2012 #12

    micromass

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    So you understand why [itex]b=e^{ln(b)}[/itex]?? Good. Now apply it with [itex]b=x^{1/x}[/itex]. Then you get

    [tex]x^{1/x} = e^{ln\left(x^{1/x}\right)}[/tex]

    Do you agree with this? Now apply the rules of logarithms: what is [itex]ln(a^b)=...[/itex]. Can you apply this identity with a=x and b=1/x ?
     
  14. Dec 26, 2012 #13

    micromass

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    Alternative proof: apply [itex]e^{ln(b)}=b[/itex] on x=b. Then

    [tex]x=e^{ln(x)}[/tex]

    and thus

    [tex]x^{1/x} = \left(e^{ln(x)}\right)^{1/x}[/tex]
     
  15. Dec 26, 2012 #14
    [tex]x=ln({x}^{1/x})[/tex] so [tex]{e}^{x}={x}^{1/x}[/tex] and [tex]{e}^{ln({x}^{1/x})}={x}^{1/x}[/tex]
     
  16. Dec 26, 2012 #15

    micromass

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    Why is this true? This isn't correct for all x.
     
  17. Dec 26, 2012 #16
    Look:

    [tex]x = ln b[/tex]

    [tex]{e}^{x}=b[/tex]

    [tex]{e}^{ln b}=b[/tex]

    [tex]b={x}^{1/x}[/tex]

    [tex]y=ln{x}^{1/x}[/tex]

    [tex]{e}^{y} = {x}^{1/x}[/tex]

    [tex]{e}^{ln{x}^{1/x}}={x}^{1/x}[/tex]


    now
    a=x, b=1/x
    [tex]ln({a}^{b})=c[/tex]

    [tex]ln({x}^{1/x}) = c[/tex]

    [tex]{e}^{c}={x}^{1/x}[/tex]

    [tex]{x}^{1/x}={e}^{ln{x}^{1/x}}[/tex]
     
  18. Dec 26, 2012 #17
    because x=ln(x1/x)... that was my first statement, should be correct for all x I think
     
  19. Dec 27, 2012 #18

    symbolipoint

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    lucas7,

    The Natural logarithm function is the INVERSE of the exponential function (for base, e). That is why the method works.

    How do you start with a number, x, put into a function, and then put this function into another function, and the outcome be x? One function undoes the effect of the other function.

    y=e^x
    y is the value along the vertical number line, x is the value along the horizontal number line. What if you switch x and y?
    x=e^y
    What function is this? What is "y"? We call this the natural logarithm function,
    y=ln(x), and this is the inverse of y=e^x.

    Pick an x, any x. e^(ln(x))=x, and ln(e^x))=x
     
  20. Jan 3, 2013 #19
    [tex]e^{ln(x)}=x[/tex] because they are inverse functions. [tex]f(f^{-1}(x))=x[/tex] by definition
     
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