Results 1 to 2 of 2

Math Help - simplify log

  1. #1
    Junior Member
    Joined
    Aug 2007
    Posts
    28

    simplify log

    This isnt really homework, its a problem im attempting



    IS THIS SOLUTION CORRECT?

    for the first equation i get C_2 = Log((1-a)/(1+a)) I Do NOT see how they can get that answer

    then this gives the solution for the second bit as

    C_1 = ( log ((1+b)/(1-b) + log ((1-a)/(1+a)) )^-1

    HELP PLEASE, thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by BruceBronson View Post

    ...

    for the first equation i get C_2 = Log((1-a)/(1+a)) I Do NOT see how they can get that answer

    ...
    Hello,

    you are supposed to know that:

    -\ln\left(\frac ab \right) = \ln(1)-\ln\left(\frac ab \right) = \ln\left(\frac{1}{\frac ab} \right) = \ln\left( \frac ba \right)

    Now consider your first equation:


    1. Divide by C_1
    2. Subtract the ln-value:

    -\ln\left(\frac{1+a}{1-a}\right)=C_2 Use the property I mentioned above and you'll get

    \ln\left(\frac{1-a}{1+a}\right)=C_2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. simplify #2
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 21st 2009, 02:37 PM
  2. Please simplify
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 27th 2009, 04:32 AM
  3. Can you simplify this further?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 23rd 2009, 06:45 AM
  4. Simplify
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 12th 2009, 05:06 AM

Search Tags


/mathhelpforum @mathhelpforum