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

[guifb99]

a>b ⇒ Lim_(a–b)→0 Log_c(a+b) ≈ Log_c(a) + Log_c(√a⋅b)

b>a ⇒ Lim_(b–a)→0 Log_c(a+b) ≈ Log_c(a) + Log_c(√a⋅b)

a∈ℝ
b∈ℝ
c∈ℝ / c>0

 

[guifb99]

a>b ⇔ Lim_(a–b)→0 Log_c(a+b) ≈ Log_c(a) + Log_c(√a⋅b)

b>a ⇔ Lim_(b–a)→0 Log_c(a+b) ≈ Log_c(a) + Log_c(√a⋅b)

a∈ℝ
b∈ℝ
c∈ℝ / c>0

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.

 

[Mark44]

a>b ⇒ Lim_(a–b)→0 Log_c(a+b) ≈ Log_c(a) + Log_c(√a⋅b)

b>a ⇒ Lim_(b–a)→0 Log_c(a+b) ≈ Log_c(a) + Log_c(√a⋅b)

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.

 

[pbuk]

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.