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Easy and useful way to calculate Log(a+b)

  1. Feb 28, 2016 #1
    a>b ⇒ Lim(a–b)→0 Logc(a+b) ≈ Logc(a) + Logc(√a⋅b)

    b>a ⇒ Lim(b–a)→0 Logc(a+b) ≈ Logc(a) + Logc(√a⋅b)

    c∈ℝ / c>0
    Last edited: Feb 28, 2016
  2. jcsd
  3. Feb 28, 2016 #2
    Yes, this is the second post I do about it, but now I did it in a better format, the other one was too confusing because I didn't know how to use the mathematical symbols in the thread, I'm new here.
  4. Feb 28, 2016 #3


    Staff: Mentor

    I don't see the point of writing this as a limit. Also, when you take a limit, use =, not ≈.
    [/quote]These won't do. You need to have a + b > 0 and ab > 0, not just arbitrary real numbers.
    c∈ℝ / c>0[/QUOTE]
    Also, c cannot be 1.
  5. Feb 29, 2016 #4
    But this only calculates log(a+b) in the limit when (a-b) = 0 i.e. a=b. Then we have $$
    \begin{align*} \log(a+b) &= \log(2a) = \log 2 + \log a \\
    \log a + \log \sqrt{a \cdot a} &= 2 \log a \end{align*}
    $$ So your equation is only true when a=b=2: that doesn't look very useful to me.
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