Help Evaluating Logarithm

  1. 1. The problem statement, all variables and given/known data
    Evaluate without a calculator:
    (log34 + log29)2 - (log34 - log29)2


    2. Relevant equations



    3. The attempt at a solution
    (log34 + log29)2 - (log34 - log29)2

    (2log32 + 2log23)2 - (2log32 - 2log23)2

    And now I'm stuck....
     
    Last edited: Feb 22, 2012
  2. jcsd
  3. rock.freak667

    rock.freak667 6,220
    Homework Helper

    You went from a minus to a divide, I think you confused it with log(a/b)= loga-logb

    Try using this fact a2-b2 = (a+b)(a-b).
     
  4. [tex]a^2-b^2=(a-b)(a+b)[/tex]


    EDIT: how to delete message
     
  5. eumyang

    eumyang 1,347
    Homework Helper

    It also looks like you went from
    (log34 + log29)2
    to
    (log342 + log292),
    which is not true.
     
  6. Ohhhhh, I see. I shall re-attempt now. Thanks.
     
  7. Hmmm. I think it's simpler than that, guys -
    Setting A = log34 and B = log29

    (A+B)2 - (A-B)2 = (A2 + 2AB + B2) - (A2 - 2AB + B2)

    ...

    and later on using that lognxk = k.lognx
    and that logab [itex]\times[/itex] logbc = logac
     
  8. Erm...Nevermind. I'm still stuck. Can someone walk me through it please?

    Edit: I totally forgot about factoring. I'll try that, thanks!!
     
  9. jedishrfu

    Staff: Mentor

    try this (A+B)^2 - (A-B)^2 = X^2 - Y^2 = ( X + Y ) ( X - Y )

    where X=(A+B) and Y=(A-B)

    and you get (A+B)^2 - (A-B)^2 =(2A) (2B) = 4AB
     
  10. Okay one of my friends just told me to try changing the base. I did that, and now I have a giant mess on my hands. I have

    (ln2*ln4 + ln3*ln9 / ln3*ln2)^2 - (ln2*ln4 - ln3*ln9 / ln3*ln2)^2 and somehow I'm supposed to get 16 from all of that. I'm not really sure how...
     
  11. eumyang

    eumyang 1,347
    Homework Helper

    No, no, NO! Don't change the base at the beginning. Do what Joffan suggested. You can change the base much later if you really want to.
     
  12. Okie. Trying again now.

    Edit: I just can't seem to do it no matter what I try :(
     
    Last edited: Feb 22, 2012
  13. eumyang

    eumyang 1,347
    Homework Helper

    Show us what you have so far.
     
  14. Well I tried applying this:
    Setting A = log34 and B = log29
    (A+B)2 - (A-B)2 = (A2 + 2AB + B2) - (A2 - 2AB + B2)
    The A2's and B2's will cancel out leaving me with 4AB which would be

    4(log34 * log29)

    The answer in the book says I'm supposed to get 16. I tried changing the base to get
    4(ln4*ln9 / ln3*ln2)
    I think I could maybe do 4(2ln2*2ln3 / ln3*ln2) but I'm not sure if that's correct. I think then maybe the ln2's and ln3's would cancel out leaving me with 4(2*2) which would be 16. But I'm not sure if that's correct.
     
  15. eumyang

    eumyang 1,347
    Homework Helper

    That's correct.
     
  16. Yay!!! Thanks so much everyone!
     
  17. Also, if you didn't want to change base,
    [tex]
    \begin{align}
    log_34 \times log_29 & = log_32^2 \times log_23^2\\
    &= 2log_32 \times 2log_23\\
    &= 4(log_32.log_23)\\
    &= 4(log_33) \\
    &=4
    \end{align}
    [/tex]
     
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