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Simple question about logs

  1. Apr 11, 2012 #1
    if x^a=b (a,b are constants)
    then there are two ways of finding x: root and log
    so for example, x^2=4
    by root:
    (x^2)^(1/2)=(4)^(1/2)
    x=[itex]\pm[/itex]2
    by log:
    2 ln (x) = 2 ln 2
    x=2
    but it is yet impossible to obtain the negative x from logs. How are you supposed to do it? And heres a few questions:
    1.when a is rational how do you know x have 1 or 2 answers?
    2.what happens when a is not rational?
    3.what happens when a or(and) b is not real?


    thanks
    Victor Lu
    16
     
  2. jcsd
  3. Apr 11, 2012 #2
    x^a=b
    log[x](b)=a <--- [x] is base x
    log(b)/log(x)=a
    (1/a)*log(b)=log(x)
    log(b^(1/a))=log(x)
    b^(1/a)=x
     
  4. Apr 11, 2012 #3
    also i think i can say that there is no difference in whether you change a to rational or irrational, and there will always be two solutions, as it will always be some root
    but i dont know about non-real numbers, sorry.
     
  5. Apr 11, 2012 #4

    arildno

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    Hi, Victor!

    when you take the SQUARE root of x^2, the CORRECT answer is root(x^2)=|x|, not x.
    |x| is what we call "the absolute value" of the number x, i.e, its distance from 0 (irrespective of direction), which is always a non-negative number.

    Thus, solving x^2=4 with the square root operation gives you FIRST:
    |x|=2

    Then, when you wish to remove the absolute value sign, you get two solutions.

    root(4)=2 always, never -2
     
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