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Logarithms help

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data

    I cant seem to get my head round this problem, I know how to use the 'change the base rule'

    [tex] log_2 x + log_4x = 2 [/tex]

    [tex] \frac{logx}{log2} + \frac{logx}{log4} = 2 [/tex]

    why is this not correct ??
     
  2. jcsd
  3. Mar 18, 2009 #2

    mgb_phys

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    It looks correct - what's the problem
    (of course everytime I stray into maths I make an dumb mistake!)
     
  4. Mar 18, 2009 #3
    well the answer book gives the solution like this,

    [tex] log_2x + \frac{log_2 x}{log_2 4} = 2 [/tex] dont understand why its base 2 here?

    [tex] log_2 x + \frac{log_2 x}{2} =2 [/tex]

    [tex] \frac{3}{2}log_2 x = 2 [/tex] dont understand were 3/2 came from ?

    [tex] log_2 x = \frac{4}{3} [/tex]

    [tex] x = 2^{\frac{4}{3} }[/tex]
     
  5. Mar 18, 2009 #4

    mgb_phys

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    They are just collapsing it all into the same base log.

    [tex] log_2 x + \frac{log_2 x}{2} =2 [/tex] is just

    [tex] 1.5 * log_2 x =2 [/tex] which is

    [tex] \frac{3}{2} * log_2 x =2 [/tex]

    It's the same answer as you get - but you can do it this way without needing to calculate log() of anything
     
  6. Mar 18, 2009 #5
    so is this method still correct? can I just add both logs up ?
     
  7. Mar 18, 2009 #6

    mgb_phys

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    Except you are trying to find X so first you have to multiply out log(2) and log(4)
     
  8. Mar 18, 2009 #7
    sorry, I still don't get where 1.5 comes from ? I thought you just multiply both sides by two to get rid of the fraction ?

    [tex] 2log_2 x + log_2 x = 4 [/tex]
     
  9. Mar 18, 2009 #8

    mgb_phys

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    [tex] log_2 x + \frac{log_2 x}{2} = 2 [/tex]

    Which if you ignore the logs for now is just; a + a/2 = 2

    a(1+1/2) = 2

    1.5a = 2

    3/2 a =2
     
  10. Mar 18, 2009 #9
    oh I get it now, thanks a lot for your help!!
     
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