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Solving x for ln and e

  1. Apr 7, 2015 #1
    1. The problem statement, all variables and given/known data
    I need to solve x:
    ln (1+e^-x)=-x+2


    2. Relevant equations


    3. The attempt at a solution
    ln (1+e^-x)=-x+2
    x+ln (1+1/e^x)=2
    x+ln (e^x/e^x+1/e^x)=2
    x+ln ((e^x+1)/e^x)=2
    x+ln (e^x+1)-ln(e^x)=2
    x+ln (e^x+1)-x=2
    ln (e^x+1)=2


    im stuck here.
    thank you
     
  2. jcsd
  3. Apr 7, 2015 #2

    BiGyElLoWhAt

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    Perhaps try raising e to both sides of the equation? ##e^{lhs}=e^{rhs}##
     
  4. Apr 7, 2015 #3
    sorry I dont really understand.
     
  5. Apr 7, 2015 #4

    BiGyElLoWhAt

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    ln (e^x+1)=2...
    ##e^{ln(e^x+1)}=e^2##
    ...
     
  6. Apr 7, 2015 #5

    BiGyElLoWhAt

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    In my previous post, lhs was left hand side and rhs was right hand side. That'll come up from time to time, so it wouldn't hurt to keep that in mind.
     
  7. Apr 7, 2015 #6
    where can I read about the rules/ methods to solve ln /e equation because this is confusing me I dont know what to do. :'(
     
  8. Apr 7, 2015 #7
    How would you define the natural log of say a variable called y?
     
  9. Apr 7, 2015 #8

    BiGyElLoWhAt

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    To add to chet, define it in words, not maths, but of course using some math termonology, such as exponents and whatnot.
     
  10. Apr 7, 2015 #9
    ln (y)?
     
  11. Apr 7, 2015 #10

    BiGyElLoWhAt

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    yes
    define it in words.
     
  12. Apr 7, 2015 #11
    the exponential form of y?
     
  13. Apr 7, 2015 #12

    BiGyElLoWhAt

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    What is the natural log of y? If I punch ##ln(1)## in my calculator, it returns a 0, why is the answer 0? what does the 0 represent?
     
  14. Apr 7, 2015 #13
    Fill in the blanks: The natural log of y is the power to which you have to raise ____ to get ___.

    Chet
     
  15. Apr 7, 2015 #14
    to raise e to get y

    ln (e^y)=yln e=y
     
  16. Apr 7, 2015 #15

    BiGyElLoWhAt

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    Ahhh, so ln(y) is the exponent that I can put on e to get y? So what happens if I take that exponent and stick it on e? what do I get? ##e^{ln(y)}=?##
     
  17. Apr 7, 2015 #16
    Good. Now going back to post #4, fill in the blanks: ##\ln(1+e^x)## is the power to which you have to raise ____ to get _____.

    Chet
     
  18. Apr 8, 2015 #17
    raise e to get ln (1+e^x)
    e^(ln (1+e^x))
    then e and ln can cancels? to get 1+e^x
     
  19. Apr 8, 2015 #18
    OK. Good. Now, if you combine this result with the equation in post #4, what do you get for x?

    Chet
     
  20. Apr 8, 2015 #19
    e^x+1=e^2
    e^x=e^2-1
    lne^x=ln (e^2-1)
    x=ln (e^2-1)
     
  21. Apr 8, 2015 #20
    Excellent!!!
     
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