Sum of Logarithms

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

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.

$\blacksquare$

Proof for General Logarithm

Next we expand the proof for logarithms to the general base $b$.

Proof 1

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \log_b x + \log_b y\)	\(=\)	\(\displaystyle \log_b \left(b^{\log_b x + \log_b y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By the definition of the general logarithm
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \log_b \left(\left(b^{\log_b x}\right)\left(b^{\log_b y}\right)\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	From Exponent Combination Laws
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \log_b \left(xy\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By the definition of the general logarithm

$\blacksquare$

Proof 2

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \log_b x + \log_b y\)	\(=\)	\(\displaystyle \frac {\log_e x} {\log_e b} + \frac {\log_e y} {\log_e b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Change of Base of Logarithm
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\log_e x + \log_e y} {\log_e b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\log_e \left({x y}\right)} {\log_e b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Logarithms: Proof for Natural Logarithm
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \log_b \left({x y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Change of Base of Logarithm

$\blacksquare$

Comment

If one presupposes Exponent Combination Laws, the above proof for logarithms to the general base can be used to directly prove the laws for $\log_e$

Whether this would be circular is ultimately dependent on which definition of $e^x$ one chooses.

Sources

• Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (1968): $\S 1.2.2$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 14.2 \ \text{(i)}$
• For a video presentation of the contents of this page, visit the Khan Academy.