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Math Log problems

  1. Apr 1, 2015 #1
    1. The problem statement, all variables and given/known data

    1. I have to prove that x+ln(1+e^(-x)) is equal to ln(1+e^(x))

    2. Relevant equations

    3. The attempt at a solution
    1. x+ln(1+e^(-x)) =ln(1+e^(x))

    Im blocked here.. :(

  2. jcsd
  3. Apr 1, 2015 #2


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    If ##e^a = e^b## what can you say about ##a## and ##b##?
  4. Apr 1, 2015 #3
    that a and b is equal, same number
  5. Apr 1, 2015 #4


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    Can you use that fact to help solve your problem?
  6. Apr 1, 2015 #5


    Staff: Mentor

    This is not the best way to start out, since this is what you're supposed to prove.
    A better way, IMO, is to start with this:
    ##x + ln(1 + e^{-x}) - ln(1 + e^x)##
    If you can show that this is equal to zero, then you will have proved that the equation is an identity.

    Hint: Write the first term as the ln of something, then combine all three terms using the properties of logs.
  7. Apr 2, 2015 #6


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    If [itex]y= x+ ln(1+ e^{-x})[/itex] then [itex]e^y= e^x(1+ e^{-x})[/itex]

    If [itex]z= ln(1+ e^x)[/itex] then [itex]e^z= ??[/itex].
  8. Apr 2, 2015 #7
    $$x=\ln (e^x)$$

  9. Apr 5, 2015 #8
    Sorry guys I still dont understand how to do this, been reading online tutorials on how to solve ln equation with no luck.
  10. Apr 6, 2015 #9


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    You need to know that ##e^{-x}=\frac{1}{e^x}## and ##\ln(\frac{a}{b})=\ln(a)-\ln(b)##.

    write 1+e-x in fraction form.
  11. Apr 6, 2015 #10
    so 1 + e^(-x) becomes 1 + 1/e^x which we can make one fraction with a common denominator of e^x. so it'll be (e^x/e^x) + (1/e^x)
    or (e^x + 1)/(e^x). we can use props of lns to change that into ln(e^x + 1) - ln(e^x)
    so it'd be x + ln(e^x + 1) - ln(e^x) = ln(1 + e^x)
    for ln(e^x) the ln and e cancel eachother, so only the x stays
    so it'd be x + ln(e^x + 1) - x = ln(e^x + 1)
    and the x and -x cancel eachother. so you have ln(e^x+1) = ln (e^x+1)

    is this right ?
  12. Apr 6, 2015 #11
    posted an attempt in my last post
  13. Apr 6, 2015 #12


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    It is right.
  14. Apr 6, 2015 #13
    awesome man thank you very much.
  15. Apr 6, 2015 #14
    I guess you saw no value in paying attention to the equation I wrote in post # 7. If you substitute this into your original equation, you get:
    $$x+\ln (1+e^{-x})=\ln e^x+\ln (1+e^{-x})=\ln [e^x(1+e^{-x})]$$
    What does that give you?

  16. Apr 6, 2015 #15
    yeah sorry I did not really understand what your post meant :/
    I am not very good with ln /e/log solver.
  17. Apr 6, 2015 #16
    The definition of lny is the power to which you have to raise e to get y. Now let's substitute ex for y in the previous statement:

    ##\ln (e^x)## is the power to which you have to raise e to get ##e^x##. But that's just x.

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