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Changing the bases of logs question

  1. Nov 17, 2012 #1
    1. The problem statement, all variables and given/known data
    If x= log2aa
    y= log3a2a
    z= log4a3a

    prove that xyz +1 = 2yz

    2. Relevant equations

    loga C = log10C/log10a


    3. The attempt at a solution

    log= log10

    x= log2aa = log a/ log2a


    y= log3a2a = log 2a/log 3a


    z= log4a3a = log 3a/log 4a


    xyz + 1 = (log a/log2a)x (log 2a/log 3a) x (log 3a/log4a) + log 10


    = log a/ log 4a + log 10

    Is this correct so far?
     
  2. jcsd
  3. Nov 17, 2012 #2

    Curious3141

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    Everything you've written there is correct. But instead of writing log 10 in the last step, I would suggest you write the "1" as [itex]\frac{\log 4a}{\log 4a}[/itex] and combine the fractions.
     
  4. Nov 17, 2012 #3
    So it's
    (loga/log4a) + (log4a/log4a) = (loga/log4a) x (log4a/log4a)

    = loga/log4a?
     
  5. Nov 17, 2012 #4

    Curious3141

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    No. How'd you get from the '+' to the 'x'?

    Are you getting confused by the log rule that says: [itex]\log m + \log n = \log mn[/itex]? Because that's something quite different.

    Just do the addition like normal fraction expressions. The numerator of the combined expression should then be simplified using that log rule I mentioned.
     
  6. Nov 17, 2012 #5
    so it's log4a^2/log4a?
     
  7. Nov 18, 2012 #6

    Curious3141

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    Yes. Now observe that [itex]4a^2 = (2a)^2[/itex].

    Use [itex]\log m^n = n\log m[/itex] here.

    It's already quite close to the form you require. You just need one more trivial trick to introduce [itex]\log 3a[/itex] into the expression. Remember that [itex]\frac{x}{y} = (\frac{x}{z})(\frac{z}{y})[/itex], where z can be anything because it just cancels out. A little bit more manipulation and you'll get the form you require.
     
  8. Nov 18, 2012 #7
    How about (2log2a/log3a) x (log3a/log4a)?
     
  9. Nov 18, 2012 #8
    Oh yes it is that because yz = (log3a/log4a) x (log2a/log3a)


    and xyz + 1 =( 2log2a/log3a) x (log3a/log4)

    and that is equal to 2yz!
     
  10. Nov 18, 2012 #9

    Curious3141

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    Yup, you got it. :smile:
     
  11. Nov 18, 2012 #10
    Thank you!!
     
  12. Nov 18, 2012 #11
    =/ My teacher said what I did was wrong... he said it must be done the way he did it.......... he changed base of x to base of y then he got xy I think. Then changed the base of xy to base of z.

    I don't remember exactly what he did i didn't get a chance to transcribe it... but it was something along the lines of what I said above. But I personally believe the question can be worked more than one way..
     
  13. Nov 18, 2012 #12

    haruspex

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    You have my sympathies.
     
  14. Nov 18, 2012 #13
    What does that mean? I have a bad teacher? =/
     
  15. Nov 18, 2012 #14

    Curious3141

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    Everything you did is mathematically valid. The only small caveat is that 'a' should be taken to be positive, but this is generally a given in this sort of question. I don't know what your teacher is on about.
     
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