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Exponential having ln exponent

  1. Jul 1, 2016 #1
    1. The problem statement, all variables and given/known data
    How is ## e^log√(1-x^2)## equal to ##√(1-x^2)?##


    2. Relevant equations


    3. The attempt at a solution

    taking ln on the function, ln√(1-x^2). lne⇒ ln√(1-x^2) .............
     
  2. jcsd
  3. Jul 1, 2016 #2

    Math_QED

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    Log a = b <=> e^b = a.
    Now notice on the right we have e^b = a, but we know b = log a. Therefor, e^log(a) = a. Apply this to your exercise.
     
  4. Jul 1, 2016 #3

    SammyS

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    Re: LaTeX.
    To have more than a single character in a superscript or subscript or either part of a fraction or ... ,
    place the desired string of characters inside a pair of braces: { ... } .

    For many well-known functions, place a \ in front of the function name: e.g.: \ln , \sin , \tan , \sqrt , ...​

    .
     
  5. Jul 1, 2016 #4

    Ray Vickson

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    It is good that you are trying to use LaTeX, but the next step is to learn to use it properly. Which of the following three expressions look best to you?
    (1) ##e^log√(1-x^2)##; (2) ##e^{log√(1-x^2)}##; or (3) ##e^{\log \sqrt{1-x^2}}##.
    The first is a copy of what you wrote; the second inserts the brackets { and } needed with a multi-character exponent (or subscript); the third uses '\log' instead of 'log' and uses '\sqrt{ ...}' instead of '√' ; that also allows you to write and print 1-x^2 instead of (1-x^2), producing cleaner formula that is easier to read. You can right-click on each of the expressions to see their TeX structure.

    Note added in edit: I see that SammyS has beaten me to it.
     
    Last edited: Jul 1, 2016
  6. Jul 2, 2016 #5
    Thanks a lot Ray Vickson and Sammy, next time i will type my work well in Latex. Noted.
     
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