If $\log (x) = a$ and $\log (y)=b$, then $x = 10^a$ and $y = 10^b$, and
$$ x \cdot y = 10^a \cdot 10^b = 10^{a+b}$$
so that $$ \log(xy) = a+b $$

\begin{xmpl}

\begin{eqnarray*}
log(100)&=& 2 \\
log(1000)&=& 3 \\
\end{eqnarray*}
\end{xmpl}
\begin{xmpl}
If $$\log(2) \approx 0.3$$
then $$10^y=2$$
\begin {notes}
Note that
$$2^{10}=1024 \approx 1000 = 10^3$$ therefore
$$2 \approx 10^{3/10}$$ so
$$\log (2) \approx 0.3$$
\end{notes}
\end{xmpl}