1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Re-arranging logarithms

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that ln[ (N + M - 1)! /M! (N-1)! ] is equal to N ln((N+M) / N) + M ln((N+M) /M).

    2. Relevant equations

    Using stirling's formula ln N! ~ N lnN - N

    3. The attempt at a solution

    ln[ (N + M - 1)! /M! (N-1)! ] (a)
    = (N+M -1) ln(N+M -1)- (N+M -1) - M lnM + M - (N-1) ln(N-1) + (N-1) (b)
    =(N+M) ln(N+M) - M lnM - N lnN (c)
    = N ln( (N+M)/N) + M ln ( (N+M) / M) (d)


    I understand how to get from (a) to (b), and (c) to (d). But I don't understand what happens from (b) to (c). What has happened to the -1 values?

    Thanks.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Re-arranging logarithms
  1. Re-Post: Subgroups (Replies: 0)

Loading...