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!

Manipulating natural log

  1. Dec 12, 2015 #1
    How is it true that:

    Log[L+(Z^2+L^2)^(1/2)] - Log[-L+(Z^2+L^2)^(1/2)] = 2{Log[L+(Z^2+L^2)^(1/2)] - Log[Z]}

  2. jcsd
  3. Dec 12, 2015 #2


    User Avatar

    Staff: Mentor

    Is this homework?

    The equality is not very complicated to check. Use the properties of logarithms to get to the form log(x) = log(y), and then check if x = y.
  4. Dec 12, 2015 #3
    Not homework, but part of a textbook problem that I'm using to study. I can get as far as the form log(x) = log(y) by turning the differences into quotients and the multiplicative prefactor on the right into an exponent. I guess I'm having trouble with the easiest part, seeing the algebraic identity.
  5. Dec 12, 2015 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Is your equation:

    ##log\; [L + \sqrt{(Z^2 + L^2)}] -log\; [-L + \sqrt{(Z^2 + L^2)}]=2log\; [L + \sqrt{(Z^2 + L^2)}] - log (Z)## ?

    The Laws of Logarithms are:

    ##log\; (a) - log\; (b) = log\;(\frac{a}{b}) ##

    ##log\; (a+b) - log\; (a-b) = log\;(\frac{a+b}{a-b}) ##

    ##log\;(a^b) = b\;log\;(a)##

  6. Dec 12, 2015 #5
    Thanks for your reply. Yes, I understand this much. So the problem becomes showing that

    (L + Sqrt[Z^2 + L^2])/(-L + Sqrt[Z^2 + L^2]) = [(L + Sqrt[Z^2 + L^2])/Z]^2

    Maybe this is me being dumb, but I don't know how to get from the left side to the right side.
  7. Dec 12, 2015 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you start from the left side, simply multiply by ##(L+\sqrt{Z^2+L^2})/(L+\sqrt{Z^2+L^2})##. This is the same trick as when we multiply by conjugates of complex numbers to get rid of a term (a+ib) in the denominator.
  8. Dec 12, 2015 #7
    My lord, that is simple. Thank you.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook