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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

    DrClaude

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    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

    SteamKing

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    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)##

    ##log\;[(a+b)^c]=c\;log\;(a+b)##
     
  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

    nrqed

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    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.
     
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