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More Logarithims. Is my first method correct?

  1. Nov 13, 2012 #1
    If log 12 = a and log 18 = b express log 2 in terms of a and b

    Is this method correct?

    log 12= 2log2 + log 3 = a
    log 2 + 2log 3 = b

    log 3= (b-log2)/2
    2log2 + (b-log2)/2 = a

    (4log2+b-log2)/2 = 2a/2

    3log2= 2a-b
    log2= (2a-b)/3


    and express log 13.5 in terms of a and b, i'm not sure how to attempt this one.

    One last one. If loga x =m and logx2a=n, find the relation between m and n.

    Could I take the log of the 2nd on in base a

    so logan= x^2 or something like that?
     
  2. jcsd
  3. Nov 13, 2012 #2

    symbolipoint

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    Your first method (finding log of 2 ) is correct.
     
  4. Nov 13, 2012 #3

    symbolipoint

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    Second part: log 13.5 = log(27/2) ?
    = log((3^3)/2)
    =log(3^3)-log2
    =3*log3 - log2
     
  5. Nov 13, 2012 #4

    HallsofIvy

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    Yes, that is correct.
     
  6. Nov 13, 2012 #5

    SammyS

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    For the last one, if you do as you suggest, you get [itex]\displaystyle \ \log_a(n)=\log_a\left( \log_{x^2}(a) \right)\,,\ [/itex] which isn't helpful.

    Instead, you can use the change of base formula to change the logarithm on the left hand side to base a.

    But I would be inclined to write the exponential version of [itex]\displaystyle \ \log_{x^2}(a)=n\,,\ [/itex] then take loga of both sides of that result.
     
  7. Nov 13, 2012 #6
    Ohhhhhh thank you guys!
     
  8. Nov 13, 2012 #7
    loga x = m and log x2 a = n
    How how about this?

    (x2)n = a

    log a x2 = log a x

    Wait no

    log a (x^2)n = log a a

    so 2nloga x = 1

    so in terms of m and n

    2mn = 1
     
    Last edited: Nov 13, 2012
  9. Nov 14, 2012 #8

    HallsofIvy

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    Do you mean [itex]log_{x^2} a= n[/itex]?

    If you meant [itex]log_{x^2} a= n[/itex], yes that is true. [itex]log_{x^2} a= n[/itex] is the same as [itex](x^2)^n= x^{2n}= a[/itex] and, taking the logarithm, with respect to a, of that, [itex]2n log_a(x)= 1[/itex].

    Yes, that is correct.

    By the way, it is better to post new questions in new threads. Many people, seeing that a question has been answered, will not respond to that thread again.
     
    Last edited: Nov 14, 2012
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