# 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 $\map M t = \max \set {\size {t - 1}, \size {\dfrac {t - 1} t} }$

$\forall t \in \R_{>0}: \size {\map {f_n} t} \le \map M t$

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 of Product $\ds$ $=$ $\ds n \size {\sqrt [n] x - 1} \size {\sqrt [n] y - 1}$ Absolute Value of Product $\ds$ $=$ $\ds n \frac {\size {\map {f_n} x} } n \frac {\size {\map {f_n} y} } n$ 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:

$\displaystyle \ln x := \int_1^x \frac {\d t} t$

Consider the diagram above.

The value of $\ln x$ is represented by the area between the points:

$\left({1, 0}\right), \left({1, 1}\right), \left({x, \dfrac 1 x}\right), \left({x, 0}\right)$

which is represented by the yellow region above.

Similarly, the value of $\ln y$ is represented by the area between the points:

$\left({1, 0}\right), \left({1, 1}\right), \left({y, \dfrac 1 y}\right), \left({y, 0}\right)$

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:

$\left({x, 0}\right), \left({x, 1 / x}\right), \left({x y, \dfrac 1 {x y}}\right), \left({x y, 0}\right)$

which is exactly the green area.

The total of the green and yellow areas represents the value of $\ln x y$.

$\blacksquare$