Sum of Logarithms/Natural Logarithm

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

Then:
 * $\log_b x + \log_b y = \log_b \left({x y}\right)$

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

Proof for Natural Logarithm
First we demonstrate the result for the natural logarithm, i.e. when $b$ is Euler's number $e$.

Let $y \in \R, y > 0$ be fixed.

Consider the function:
 * $f \left({x}\right) = \ln xy - \ln x$.

where we use the notation $\ln$ to mean $\log_e$.

Then from the definition of the natural logarithm, the Fundamental Theorem of Calculus and the Chain Rule:


 * $\displaystyle \forall x > 0: f^{\prime} \left({x}\right) = \frac 1 {xy} y - \frac 1 x = \frac 1 x - \frac 1 x = 0$.

Thus from Zero Derivative means Constant Function, $f$ is constant: $\forall x > 0: \ln xy - \ln x = c$.

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.