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Intermediate algebra , Natural logarithm question

  1. May 1, 2007 #1
    Hello everyone, I do not understand how:: e raised to ln of x = x ?
    this notation might make more sense:: e^ln x=x ?
    Thanks for the help.
     
  2. jcsd
  3. May 1, 2007 #2

    Office_Shredder

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    By definition, ln(x)=a means e^a=x (i.e. ln is the inverse of e^x).

    Obviously though, it seems like you're not using that definition (I hope...).

    How are you defining ln(x)?
     
  4. May 1, 2007 #3
    ln x = e ^ some number. we just use a calculator. we made a list in class of these ones, but they are simple.

    ln e^2=e
    ln e^x=x
     
  5. May 1, 2007 #4

    Office_Shredder

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    No.... it's the other way around

    ln(e^x) = x

    Note ln(e^2) = 2, not e.

    Anyway, the definition of natural log is ln(e^x) = x (well, the definition you should be working from)
     
  6. May 1, 2007 #5
    The natural log has a base of e, so if you raise e^ln(x), it will always just equal x.

    [tex]e^{ln_{e}(x)} = x[/tex]
     
    Last edited: May 1, 2007
  7. May 1, 2007 #6

    mathwonk

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    its for the same reason the father of the man whose father is john, is ....?
     
  8. May 2, 2007 #7

    VietDao29

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    If f(x) is a bijection (i.e, an onto, 1-to-1 function), then there will exists another function g(x), such that: g(f(x)) = x, and f(g(x)) = x, we denote that function g(x) by f-1(x), and call it the inverse of f(x) (i.e a function which does the reverse of f(x)).
    Say, if we have: f(a) = b, then we'll have g(b) = a. So we have: g(f(a)) = g(b) = a, and f(g(b)) = f(a) = b.

    ex has the inverse ln(x) (i.e, we define ln(x) to be the inverse of ex). So, if f(x) = ex, then f-1(x) = ln(x).
    So, we have: [tex]f(f ^ {-1} (x)) = e ^ {f ^ {-1} (x)} = e ^ {ln (x)} = x[/tex], as f(f-1(x)) = x
    And [tex]f ^ {-1} (f(x)) = \ln(f(x)) = \ln (e ^ x) = x[/tex], as f-1(f(x)) = x
     
  9. May 2, 2007 #8
    perhaps it is better to first understand what a log is. A log is an exponent. [tex]log_2{8}[/tex] means the exponent on 2 to get 8. (3)

    ln means log base e. So, ln(x) means "what is the exponent on e to get a value of x?" example ln(5) means "what is the exponent on e to get a value of 5?" In other words, e^? = 5.


    So, when you raise e^[ln(x)], that means e to the power: [the power on e to get x].

    So, e^[ln(5)] means raise e^(power that when e is raised to that power, the result is 5.)
     
  10. May 4, 2007 #9
    I just had to explain this to my calculus class.

    We know that e^(ln x) must be some number, call it q. So e^(ln x) = q.
    If we take the natural logarithm of both side of the equation we have
    ln (e^(ln x)) = ln q.

    Hopefully, we are convinced that ln (e^a) = a by the laws of logarithms. If not, here it is quickly.
    ln (e^a) = a ln e = a*1 = a

    So ln (e^(ln x)) = ln x.

    Thus we have ln x = ln q, which might convince you that q = x.

    If not, we continue
    ln x - ln q = 0, so ln (x/q) = 0. Then, by the definition of logarithm (as an inverse function), x/q = e^0 = 1, So x = q.

    Putting this back in the original equation, we have e^(ln x) = q = x.
    (We must include the caveat that x > 0).
     
  11. May 4, 2007 #10

    HallsofIvy

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    How, then, have you defined ln(x) in your class?
     
  12. May 4, 2007 #11
    I see your point. If ln x = a means e^a = x, then clearly, by substituting a into the second equation we get the result e^(ln x) = x.

    However, some students need a bit more convincing that the "complicated" formula e^(ln x) = x is right. They accept the definition, in principle, but do not always see connection to the results. By bringing it down to the situation where ln(x/q) = 0, they were then able to make the connection that x/q=1.

    BTW, as stated in my post, I use the inverse function definition of the natural logarithm. Sorry for the confusion.
     
  13. May 4, 2007 #12

    uart

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    This might seem a little silly but it really seems to work. When a student says that they cant understand why e^(ln x) = x is correct this is what I sometimes do. I say that I will explain it to them, but first they must explain something to me.

    I then tell them that I've just encounter [tex]\sqrt{\cdot}[/tex] for the first time and I don't understand it (they know I'm joking) and that they must explain to me why it is that for any positive number "x" that [tex]\sqrt{x^2} = x[/tex].

    It's quite interesting just how often in that in the process of explaining this (or thinking up an explaination) that they suddenly claim to now understand the e^(ln x) thing. I guess it's just a matter of putting it into a 1-1 relation with a problem that they already understand.
     
    Last edited: May 4, 2007
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