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Homework Help: Exponential proof

  1. Feb 7, 2017 #1

    chwala

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    1. The problem statement, all variables and given/known data
    they say 1. ##e^{ln x}= x ## and 2.##e^-{ln(x+1)}= \frac 1 {x+1}## how can we prove this ##e^{ln x}= x ## and also ##e^-{ln(x+1)}= \frac 1 {x+1}##?

    2. Relevant equations


    3. The attempt at a solution
    let ## ln x = a## then
    ##e^a= x,
    ## a ln e= x,##
    →a= x, where
    ## ln x= x
     
    Last edited: Feb 7, 2017
  2. jcsd
  3. Feb 7, 2017 #2

    Delta²

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    Set ##y=ln(x+1)## so we ll have that ##e^y=x+1##.
     
  4. Feb 7, 2017 #3

    chwala

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    Delta what are you doing???????????????:smile: so?
     
  5. Feb 7, 2017 #4

    Delta²

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    Sorry I now see you have edited the original post.

    It is from the definition of the natural logarithm ##lnx## that ##e^{lnx}=x##. So not much to prove here, it follows directly from the definition.

    Now we want to look at ##e^{-ln(x+1)}##. This is equal to ##\frac{1}{e^{ln(x+1)}}## don't you agree? (it is like saying that ##e^{-y}=\frac{1}{e^y})##
     
    Last edited: Feb 7, 2017
  6. Feb 7, 2017 #5

    chwala

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    Thanks a lot, i was blind but now i see
    let ## ln x = y##............1
    it follows that ## e^y = x##
    taking logs on both sides,
    ## y ln.e = ln x##
    ##y = ln x## now substituting this in original equation 1,
    ##ln x = ln x##,
    implying that ##e^{ln x}= x##
    greetings from Africa, chikhabi
     
  7. Feb 7, 2017 #6

    Delta²

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    This is correct but what you doing is like driving in a circle, i.e you start from ##lnx=y## to end up with ##y=lnx## which is essentially the same thing.

    Substituting ##y=lnx## to ##e^y=x## is what proves (if we can call this a mini proof) that ##e^{lnx}=x##.

    Greetings from Athens, Greece.
     
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