Sum of Logarithms/Natural Logarithm
Theorem
Let $x, y \in \R$ be strictly positive real numbers.
Then:
- $\ln x + \ln y = \map \ln {x y}$
where $\ln$ denotes the natural logarithm.
Proof 1
Let $y \in \R_{>0}$ be fixed.
Consider the function:
- $\map f x = \ln x y - \ln x$
From the definition of the natural logarithm, the Fundamental Theorem of Calculus and the Chain Rule for Derivatives:
- $\forall x > 0: \map {f'} x = \dfrac 1 {x y} y - \dfrac 1 x = \dfrac 1 x - \dfrac 1 x = 0$
Thus from Zero Derivative implies Constant Function, $f$ is constant:
- $\forall x > 0: \ln x y - \ln x = c$
To determine the value of $c$, put $x = 1$.
From Logarithm of 1 is 0:
- $\ln 1 = 0$
Thus:
- $c = \ln y - \ln 1 = \ln y$
and hence the result.
$\blacksquare$
Proof 2
\(\ds \ln x + \ln y\) | \(=\) | \(\ds \int_1^x \dfrac {\d t} t + \int_1^y \dfrac {\d s} s\) | Definition of Natural Logarithm | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_1^x \dfrac {\d t} t + \int_x^{x y} \dfrac {\d t / x} {t / x}\) | Integration by Substitution: $s \mapsto t / x$, $\d s \mapsto \d t / x$, $1 \mapsto x$, $y \mapsto x y$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_1^x \dfrac {\d t} t + \int_x^{x y} \dfrac {\d t} t\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_1^{x y} \dfrac {\d t} t\) | Sum of Integrals on Adjacent Intervals for Continuous Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \ln x y\) | Definition of Natural Logarithm |
$\blacksquare$
Proof 3
Let $\sequence {f_n}$ be the sequence of mappings $f_n : \R_{>0} \to \R$ defined as:
- $\map {f_n} x = n \paren {\sqrt [n] x - 1}$
Let $\map M t = \max \set {\size {t - 1}, \size {\dfrac {t - 1} t} }$
From Bounds of Natural Logarithm:
- $\forall t \in \R_{>0}: \size {\map {f_n} t} \le \map M t$
Fix $x, y \in \R_{>0}$.
Then:
\(\ds \size {\map {f_n} {x y} - \paren {\map {f_n} x + \map {f_n} x} }\) | \(=\) | \(\ds \size {n \paren {\sqrt [n] {x y} - 1 - \sqrt [n] x + 1 - \sqrt [n] y + 1} }\) | Definition of $\map {f_n} x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds n \size {\sqrt [n] {x y} - \sqrt [n] x - \sqrt [n] y + 1}\) | Absolute Value Function is Completely Multiplicative | |||||||||||
\(\ds \) | \(=\) | \(\ds n \size {\sqrt [n] x - 1} \size {\sqrt [n] y - 1}\) | Absolute Value Function is Completely Multiplicative | |||||||||||
\(\ds \) | \(=\) | \(\ds n \frac {\size {\map {f_n} x} } n \frac {\size {\map {f_n} y} } n\) | Definition of $\map {f_n} x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 n \size {\map {f_n} x} \size {\map {f_n} y}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac {\map M x \map M y} n\) | Inequality of Product of Unequal Numbers | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size {\map \ln {x y} - \paren {\map \ln x + \map \ln y} }\) | \(=\) | \(\ds \size {\lim_{n \mathop \to \infty} \map {f_n} {x y} - \paren {\lim_{n \mathop \to \infty} \map {f_n} x + \lim_{n \mathop \to \infty} \map {f_n} y} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \size {\map {f_n} {x y} - \paren {\map {f_n} x + \map {f_n} y} }\) | Modulus of Limit | |||||||||||
\(\ds \) | \(\le\) | \(\ds \lim_{n \mathop \to \infty} \frac {\map M x \map M y} n\) | Limit of Bounded Convergent Sequence is Bounded | |||||||||||
\(\ds \) | \(=\) | \(\ds \map M x \map M y \lim_{n \mathop \to \infty} \frac 1 n\) | Multiple Rule for Real Sequences | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
Thus:
- $\ds \lim_{n \mathop \to \infty} \map {f_n} {x y} = \lim_{n \mathop \to \infty} \paren {\map {f_n} x + \map {f_n} y}$
Hence the result, from the definition of $\ln$.
$\blacksquare$
Proof 4
Recall the definition of the natural logarithm as the definite integral of the reciprocal function:
- $\ds \ln x := \int_1^x \frac {\d t} t$
Consider the diagram above.
The value of $\ln x$ is represented by the area between the points:
- $\tuple {1, 0}, \tuple {1, 1}, \tuple {x, \dfrac 1 x}, \tuple {x, 0}$
which is represented by the yellow region above.
Similarly, the value of $\ln y$ is represented by the area between the points:
- $\tuple {1, 0}, \tuple {1, 1}, \tuple {y, \dfrac 1 y}, \tuple {y, 0}$
Let the second of these areas be transformed by dividing its height by $x$ and multiplying its length by $x$.
This will preserve its area, while making it into the area between the points:
- $\tuple {x, 0}, \tuple {x, 1 / x}, \tuple {x y, \dfrac 1 {x y} }, \tuple {x y, 0}$
which is exactly the green area.
The total of the green and yellow areas represents the value of $\ln x y$.
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Morphisms