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

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1. Feb 28, 2016

### guifb99

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)

a∈ℝ
b∈ℝ
c∈ℝ / c>0

Last edited: Feb 28, 2016
2. Feb 28, 2016

### guifb99

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.

3. Feb 28, 2016

### Staff: Mentor

I don't see the point of writing this as a limit. Also, when you take a limit, use =, not ≈.
[QUOTE="guifb99]
a∈ℝ
b∈ℝ
[/quote]These won't do. You need to have a + b > 0 and ab > 0, not just arbitrary real numbers.
[QUOTE="guifb99]
c∈ℝ / c>0[/QUOTE]
Also, c cannot be 1.

4. Feb 29, 2016

### MrAnchovy

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.