TITLE: Sum of Logarithm Arguments
QUESTION [6 upvotes]: This is a very simple question I suspect but I just cannot seem to nail it...
I have values for $X,Y,Z $, where $X =\log (x)$,  $Y = \log (y)$ and $Z = \log (z)$ and I need to calculate $x + y + z$, well actually $\log(x + y + z)$ would suffice. Is there a clever way of doing this other than simply doing $e^X+e^Y+e^Z$? 
Its for an algorithm where I am trying to avoid underrun - $x=e^X,y=e^Y,z=e^Z$ likely to be very small.
Any pointers much appreciated.

With answer given by @response I ended up using $\log(x+y+z)=\log(e^R+e^S+e^T)-C$ where $R=X+C$, $S=Y+C$, $T=Z+C$

REPLY [6 votes]: Suppose, we add a constant $C$ to each one of the values: $X, Y, Z$ which would prevent underflow when they are raised to $e$. Then we get:
$X' = X + C$
$Y' = Y + C$
$Z' = Z + C$
Thus, we have:
$x' = e^{X'}$
$y' = e^{Y'}$
$z' = e^{Z'}$
Adding them, we have:
$x'+y'+z' = e^{X'} + e^{Y'} + e^{Z'}$
But, we know that:
$e^{X'} + e^{Y'} + e^{Z'} = e^{X+C} + e^{Y+C} + e^{Z+C} = e^C (e^{X} + e^{Y} + e^{Z})$
Thus, we have:
$x+y+z = (x'+y'+z') e^{-C}$