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Verifying logarithms

  1. Jun 30, 2009 #1
    1. The problem statement, all variables and given/known data
    log(1/2xy^2) = ln x + 3 ln y - ln 2


    2. Relevant equations



    3. The attempt at a solution
    i have been trying for hours on with next to no results even just an insight on how to go about it would be helpful
     
  2. jcsd
  3. Jun 30, 2009 #2

    Mark44

    Staff: Mentor

    Do you know how to combine the terms on the right side into one logarithm expression?

    Also, you have used log and ln. Do you intend for these to be different-based logs?
     
  4. Jun 30, 2009 #3
    vaguely yes

    and yes well from the criteria sheet that is what it reads
     
  5. Jun 30, 2009 #4
    this is what i got by combining the terms on the right ln(xy^3/2)
     
  6. Jun 30, 2009 #5

    Mark44

    Staff: Mentor

    Then you'll need to study the properties of logarithms, and particularly what ln a + ln b equals.

    After you do that, we'll help you with the rest.
     
  7. Jun 30, 2009 #6
    i think there is a typo on the original criteria sheet as it reads 3 ln y when combined this = y^3 so im thinking that it should be 2 ln y

    if so i think i have figured it out if not im still stumped
     
  8. Jun 30, 2009 #7

    Mark44

    Staff: Mentor

    I don't know anything about your problem sheet, but 3 ln y = ln y3
     
  9. Jun 30, 2009 #8
    so therefore the whole right hand side would = ln(xy^3/2) where as the left = log(1/2xy^2)

    so where it says 3 ln y it should read 2 ln y

    i think i have got this now thanks for your help mark :)
     
  10. Jun 30, 2009 #9

    Mark44

    Staff: Mentor

    If you are supposed to verify that the given equation is true for all reasonable values of x and y, the question is not well stated, for two reasons: the 3 ln y vs 2 ln y that you mentioned, and log on one side and ln on the other. These mean different things. If I understand the problem, it should be given with one or the other, but not with both.
     
  11. Jul 1, 2009 #10
    I still do not know what is your equation.
    Is it:
    [tex]log(\frac{1}{2xy^2}) = ln x + 3 ln y - ln 2[/tex]

    or is it
    [tex]log(\frac{1}{2}*xy^2})=ln x + 3lny - ln2 [/tex]

    You still need to use the fact that
    ln y + ln z = ln (y*z)
    and
    ln y - ln z = ln(y/z)
     
  12. Jul 1, 2009 #11
    The highlighted one is the equation i needed to verify
     
  13. Jul 1, 2009 #12

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [itex]ln((1/2)xy^2)= ln(1/2)+ ln(x)+ 2 ln(y)[/tex]. Is that the same as the right side?
     
  14. Jul 2, 2009 #13
    right i actually remembered how to do this alls i had to do was expand the left side to show it = the right side
     
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