Sum of Logarithms/Natural Logarithm/Proof 1

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

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.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 14.2 \ \text{(i)}$