The logarithm function over $\mathbb{R}^{+}$ has the known property that:
$$\log(xy) = \log(x)+\log(y)$$
I wonder if we have an analogue for the case of $\log(x+y)$ where the RHS would be some approximation in terms of $\log(x),\log(y)$? If there was a way to approximate and calculate the error bounds that would be great. Any references would be appreciated.
Note: I am not quite sure what Tags are appropriate here so I chose some that are remotely related
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As @Dark Malthorp commented, without loss of generality assume $x \gg y$ to make $$\log(x+y)=\log(x)+\log \left(1+\frac{y}{x}\right)$$ Let $t=\frac{y}{x}$ and use the Taylor expansion $$\log(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}+O\left(t^4\right)$$
You could also define $1+t=\frac{1+u}{1-u}$ that is to say $u=\frac{t}{t+2}$ and $$\log \left(\frac{1+u}{1-u}\right)=2 u+\frac{2 u^3}{3}+O\left(u^5\right)$$ is much faster convergent.
Still better, using Padé approximants $$\log \left(\frac{1+u}{1-u}\right)=\frac{30 u-8 u^3}{15-9 u^2}$$ For illustration, consider $\log(110)$ that is to say $t=\frac 1{10}$ and $u=\frac 1{21}$. The last formula would give $$\log \left(1+\frac{1}{10}\right) \sim \frac{6611}{69363}=0.09531017978$$ while the exact value is $0.09531017980$.
I general for a concave function f we have the following:
$$\frac{f(x)+f(y)}{2} ≤ f\left(\frac{x+y}{2}\right)$$
Because $\ln$ is concave, we have the following:
$$\frac{\ln(x)+\ln(y)}{2} ≤ \ln\left(\frac{x+y}{2}\right)$$
which can be rearranged into the following lower bound for $\ln(x+y)$:
$$\frac{\ln(x)+\ln(y)}{2} + \ln(2) ≤ \ln(x+y)$$
Now for an upper bound, consider the following:
When $0<x,y$ we have $x^2+y^2 ≤ (x+y)^2$
If we replace $x$ with $\sqrt x$ and similarly for $y$ we have:
$x+y ≤ \left(\sqrt x+\sqrt y\right)^2$
Therefore $\ln(x+y) ≤ 2\ln\left(\sqrt x+\sqrt y\right)$
So in summary:
$$\frac{\ln(x)+\ln(y)}{2} + \ln(2) ≤ \ln(x+y) ≤ 2\ln\left(\sqrt x+\sqrt y\right)$$