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Log Law Question

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data
    This is not a homework question.

    Understanding how [tex] log_{\frac{1}{2}}\frac{1}{9} = log_2x^2 [/tex]


    3. The attempt at a solution

    Somehow, the base of the first logarith was turned in to 2^-1, no problem, but he was able to put the negative up in front of a log to equal:

    [tex] -log_2\frac{1}{9} [/tex] How do you do that?
     
  2. jcsd
  3. Oct 14, 2009 #2

    Mark44

    Staff: Mentor

    Let's look at the left side [tex] log_{\frac{1}{2}}\frac{1}{9} [/tex]

    What this log means is the exponent on 1/2 that produces 1/9. An equivalent equation is (1/2)y = 1/9. This is turn is equivalent to 1/(2y) = 1/9, or equivalently, 2y = 9.

    Your equation can be rewritten as log2 9 = log2 x2, and from this we see that x2 = 9, which has two solutions.
     
  4. Oct 14, 2009 #3
    Thanks! that.. makes things really easy to understand... If i don't know the proof I won't remember it.. lol. Thank you so much :) Test today.. :(

    *edit*

    just to make sure, the solution is +/- 3 correct?
     
  5. Oct 14, 2009 #4

    Mark44

    Staff: Mentor

    Yes.
     
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