Difference of Logarithms

From ProofWiki
Jump to navigation Jump to search

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

$\blacksquare$


Sources