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Numerator log help

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex] log_4(5-x)-log_4(3-x) = 2 [/tex]



    3. The attempt at a solution

    [tex] log_4(5-x)-log_4(3-x) = 2 [/tex]
    [tex] log_4\frac{3-x}{5-x} = 2 [/tex]

    where do I go from here?
     
  2. jcsd
  3. Apr 29, 2009 #2

    Cyosis

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    Homework Helper

    Re: logarithms

    What do you know about [tex]a^\log_a x[/tex]?
     
  4. Apr 29, 2009 #3
    Re: logarithms

    Dont know, not come across that expression before. Could you please explain.

    Thanks.

    Are you using the power rule ?
     
  5. Apr 29, 2009 #4
    Re: logarithms

    It's actually much simpler than that cyosis.

    You're on the right track but what do you know about logs?

    Consider this;

    log 2 (4) = 2 Am I right? Isn't this just another way of saying, 2^2 = 4?

    Ok, let's apply this logic to your question.

    log 4 (3-x / 5-x ) = 2

    Now do the same
     
    Last edited: Apr 29, 2009
  6. Apr 29, 2009 #5

    HallsofIvy

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    Re: logarithms

    [itex]log_a(b)= c[/itex] is equivalent to [itex]b= a^c[/itex]. Here, a= 4, b= (3-x)/(5- x), and c= 2.
     
  7. Apr 29, 2009 #6
    Re: logarithms


    Oh right I get it now,
    thanks

    [tex] \frac{3-x}{5-x} = 4^2 [/tex]
     
  8. May 1, 2009 #7
    Re: logarithms

    Shouldn't the 5-x term be in the numerator and the 3-x term in the denominator?
    I'm not 100% certain, but I thought log b - log a = log (b/a), not log (a/b)?
     
  9. May 1, 2009 #8
    Re: logarithms

    Yeah you're right
     
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