Answer
let
p = $\log_{b}xy$
q = $\log_{b}x$
s = $\log_{b}y$
then (By Definition of logarthim)
xy = $\ b^p$
x = $\ b^q $
y = $\ b^s $
xy = $\ b^q $ * $\ b^s $ =$ \ b^{q+s} $ also xy = $\ b^p $
$\ b^p $ = $ \ b^{q+s} $
hence , p = q + s then subsitube p,q and s with their original log's we get
$\log_{b}xy$ = $\log_{b}x$ + $\log_{b}y$
Work Step by Step
let
p = $\log_{b}xy$
q = $\log_{b}x$
s = $\log_{b}y$
then (By Definition of logarithm)
xy = $\ b^p$
x = $\ b^q $
y = $\ b^s $
xy = $\ b^q $ * $\ b^s $ =$ \ b^{q+s} $ also xy = $\ b^p $
$\ b^p $ = $ \ b^{q+s} $
hence , p = q + s then substitute p,q and s with their original log's we get
$\log_{b}xy$ = $\log_{b}x$ + $\log_{b}y$
