# Difference of Logarithms

## Theorem

Let $x, y, b \in \R$ be strictly positive real numbers such that $b > 1$.

Then:

$\log_b x - \log_b y = \map {\log_b} {\dfrac x y}$

where $\log_b$ denotes the logarithm to base $b$.

## Proof 1

 $\displaystyle \log_b x - \log_b y$ $=$ $\displaystyle \map {\log_b} {b^{\log_b x - \log_b y} }$ Definition of General Logarithm $\displaystyle$ $=$ $\displaystyle \map {\log_b} {\frac {\paren {b^{\log_b x} } } {\paren {b^{\log_b y} } } }$ Quotient of Powers $\displaystyle$ $=$ $\displaystyle \map {\log_b} {\frac x y}$ Definition of General Logarithm

$\blacksquare$

## Proof 2

 $\displaystyle \log_b x - \log_b y$ $=$ $\displaystyle \frac {\log_e x} {\log_e b} - \frac {\log_e y} {\log_e b}$ Change of Base of Logarithm $\displaystyle$ $=$ $\displaystyle \frac {\log_e x - \log_e y} {\log_e b}$ $\displaystyle$ $=$ $\displaystyle \frac {\log_e \left({\frac x y}\right)} {\log_e b}$ Difference of Logarithms: Proof for Natural Logarithm $\displaystyle$ $=$ $\displaystyle \log_b \left({\frac x y}\right)$ Change of Base of Logarithm

$\blacksquare$

## Proof 3

 $\displaystyle \map {\log_b} {\frac x y} + \log_b y$ $=$ $\displaystyle \map {\log_b} {\frac x y \times y}$ Sum of Logarithms $\displaystyle$ $=$ $\displaystyle \log_b x$ $\displaystyle \leadsto \ \$ $\displaystyle \map {\log_b} {\frac x y}$ $=$ $\displaystyle \log_b x - \log_b y$ subtracting $\log_b y$ from both sides

$\blacksquare$