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!

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

Similar Discussions: Easy and useful way to calculate Log(a+b)
  1. Expanding log(a+b) (Replies: 8)

  2. Calculating logs (Replies: 8)