[tex]\displaystyle\bf\\log\Bigg(\frac{12}{4}\Bigg)+log\Bigg(\frac{48}{4}\Bigg)-log\Bigg(\frac{36}{4}\Bigg)=\\\\=log\Big(3\Big)+log\Big(12\Big)-log\Big(9\Big)=\\\\=log\Bigg(\frac{3\times12}{9}\Bigg)=\\\\=log\Bigg(\frac{36}{9}\Bigg)=\\\\=\boxed{\bf log(4)}[/tex]