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Logs question

  1. Dec 11, 2004 #1
    if log2 = x and log3 = y, solve for log(base5)36 in terms of x and y.

    can someone help me get started with this one? thanks.
     
    Last edited: Dec 11, 2004
  2. jcsd
  3. Dec 11, 2004 #2
    Try to use the translation of base formula.
    Or let log(base 5)10=z
    Try to think of how to convert log5 in terms of x.
    Notice that some special value you can get, such as log2=x , log3=y, log1=0, log 10=1,etc.
    Then you can express it in term of x.
     
  4. Dec 11, 2004 #3

    HallsofIvy

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    What is the base in the original log, 10?

    Assuming you mean log10(2)= x and log10(3)= y,

    log5(10)= 1/log10(5).

    5= 10/2 so log10(5)= log10(10/2)= log10(10)- log10(2)= 1- x.

    log10(3) doesn't enter into it.
     
  5. Dec 11, 2004 #4
    i'll post a little bit of what i've done, the teacher said its simple, and in the last questions we've converted the bases for x and y to the one for the final, not the other way around. heres what i've done, not sure if its right.
     

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    Last edited: Dec 11, 2004
  6. Dec 11, 2004 #5
    sorry, its supposed to be log(base5)36 to solve for
     
  7. Dec 11, 2004 #6

    Zurtex

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    36 = 3*3*2*2

    Now apply the fact that: log(ab) = log(a) + log(b)
     
  8. Dec 11, 2004 #7
    so in the end i get

    log(base5)36 = 2x + 2y / log(base10)5
    can it be simplified further?
     
  9. Dec 11, 2004 #8
    Yes.
    log 5=1-x
     
  10. Dec 11, 2004 #9

    Zurtex

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    You sure you have read the edits?
     
  11. Dec 14, 2004 #10
    What's edited?
     
  12. Dec 14, 2004 #11

    Zurtex

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    The original question.
     
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