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Log. functions - comparing solutions, which is best?

  1. Jul 3, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm not sure what my error is. Both solutions cannot be true.


    2. Relevant equations
    evaluating logarithms,
    power of logarithms?


    3. The attempt at a solution

    SOLUTION ONE

    =logxxn
    =n(logxx)
    =n(1)
    =n


    SOLUTION TWO

    =logxxn
    =(logxx)n
    =(1)n
    =1

    Please help me understand what I've done wrong. Thanks!
     
    Last edited: Jul 3, 2013
  2. jcsd
  3. Jul 3, 2013 #2

    jedishrfu

    Staff: Mentor

    the logx x^n means logx(x^n) not logx(x)^n so your solution two is wrong.

    the logx(x^n) is the same as n * logx(x)
     
  4. Jul 3, 2013 #3

    CAF123

    User Avatar
    Gold Member

    Let ##y = \log_x x^n##. Then ##x^y = x^n \Rightarrow y = n##, so your solution one is correct.
    ##x^n## is in the argument of the log, the log is not to the nth power.
     
  5. Jul 3, 2013 #4
    The other 2 posts are correct:

    Your solution 1 uses logx(xn)
    Your solution 2 uses (logx(x))n

    An easy way to avoid this is to use brackets whenever applying a function:
     
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