Answer
See proof
Work Step by Step
We have to prove the statement:
$\log_b \dfrac{x}{y}=\log_b x-\log_b y$
a) Let $x=b^p$
$y=b^q$
Solve these two equations for $p$ and $q$:
$\log_b x=\log_b b^p$
$\log_b x=p\log_b b$
$\log_b x=p$
$\log_b y=\log_b b^q$
$\log_b y=q\log_b b$
$\log_b y=q$
b) Use the property E2 for exponents to express $\dfrac{x}{y}$ in terms of $b,p,q$:
$\dfrac{x}{y}=\dfrac{b^p}{b^q}$
$\dfrac{x}{y}=b^{p-q}$
c) Compute $\log_b \dfrac{x}{y}$ and simplify:
$\log_b \dfrac{x}{y}=\log_b b^{p-q}$
$\log_b \dfrac{x}{y}=(p-q)\log_b b$
$\log_b \dfrac{x}{y}=p-q$
$\log_b \dfrac{x}{y}=\log_b x-\log_b y$
