Answer
$\log_{b}\frac{x}{y}={\log_{b}x}-{\log_{b}y}\\
proof:\\
let\,\,log_{b}\frac{x}{y} =f \,\,\,,\log_{b}x=g\,\,\,,\log_{b}y=h \\
log_{b}\frac{x}{y} =f \Leftrightarrow b^f=\frac{x}{y} \\
\log_{b}x=g \Leftrightarrow b^g=x \\
\log_{b}y=h \Leftrightarrow b^h=y \\
so\,\,b^f=\frac{x}{y}=\frac{b^g}{b^h}=b^{g-h}\\
\therefore f=g-h \\
\Rightarrow \log_{b}\frac{x}{y}={\log_{b}x}-{\log_{b}y}\\
$
Work Step by Step
$\log_{b}\frac{x}{y}={\log_{b}x}-{\log_{b}y}\\
proof:\\
let\,\,log_{b}\frac{x}{y} =f \,\,\,,\log_{b}x=g\,\,\,,\log_{b}y=h \\
log_{b}\frac{x}{y} =f \Leftrightarrow b^f=\frac{x}{y} \\
\log_{b}x=g \Leftrightarrow b^g=x \\
\log_{b}y=h \Leftrightarrow b^h=y \\
so\,\,b^f=\frac{x}{y}=\frac{b^g}{b^h}=b^{g-h}\\
\therefore f=g-h \\
\Rightarrow \log_{b}\frac{x}{y}={\log_{b}x}-{\log_{b}y}\\
$