ELF.APP.1 : $\log_{base}( y\cdot b ) = \log_{base}(y) + \log_{base}(b)$
Proof : We begin by using the inverse relation of  $\log_{base}$ and $\exp_{base}$ for $y$ and $b$.
Let $\log_{base}(y) = x$ and $\log_{base}(b)= a$ so  $\exp_{base}(x) = y$ and $\exp_{base}(a)= b$

Here's what that looks like on a mapping diagram:
Next we consider the product:
$y\cdot b = base^x \cdot base^a = base^{x+a}$
and thus we have the corresponding logarithmic equation:
$\log_{base} (y\cdot b) = \log_{base} (base^{x+a}) = x+a = \log_{base}(y) + \log_{base}(b)$

Here's what that looks like on a mapping diagram:

And here's a mapping diagram showing the sum of the logarithms visualized as well:

Finally here is a dynamic visualization of the proof using mapping diagrams:
 Proof of $\log_{base}( y\cdot b ) = \log_{base}(y) + \log_{base}(b)$

Martin Flashman, 20 Sept 2014, Created with GeoGebra