Product Rule for Logarithms
Functions & Advanced AlgebraThe product rule for logarithms states that the logarithm of a product equals the sum of the logarithms of the factors.
Formula
\log_b(xy) = \log_b(x) + \log_b(y)
Definition
The product rule says: log of a product = sum of the logs. When you multiply two numbers inside a log, you can split it into addition of two separate logs.
Example
$\log_2(4 \times 8) = \log_2(4) + \log_2(8) = 2 + 3 = 5$. Check: $\log_2(32) = 5$ because $2^5 = 32$. The rule works.
Key Insight
This rule is what made logarithm tables useful for centuries: to multiply two large numbers, just look up their logs and add. Then look up the antilog of the sum.
Definition
For any valid logarithm base $b$ and positive $x, y$: $\log_b(xy) = \log_b(x) + \log_b(y)$. This follows because $b^m \cdot b^n = b^{m+n}$: if $\log_b(x) = m$ and $\log_b(y) = n$, then $xy = b^m \cdot b^n = b^{m+n}$, so $\log_b(xy) = m + n$.
Example
Solve $\ln(x) + \ln(x - 3) = \ln(10)$: $\ln(x(x-3)) = \ln(10)$, so $x(x-3) = 10$, $x^2 - 3x - 10 = 0$, $(x-5)(x+2) = 0$. Since $x > 3$ (domain), $x = 5$.
Key Insight
The product rule turns multiplicative equations into additive ones, which are easier to solve. This is exactly how slide rules work: physical addition of log-scale lengths computes multiplication.
Definition
The product rule $\log(xy) = \log(x) + \log(y)$ is the defining property of logarithms as group homomorphisms from $(\mathbb{R}^+, \times)$ to $(\mathbb{R}, +)$. Any continuous function satisfying this functional equation is a logarithm (Cauchy's functional equation result). It also underpins the definition of the Haar measure on locally compact abelian groups.
Example
The product rule yields $\log(n!) = \sum_{k=1}^{n} \log(k)$, enabling Stirling's approximation via the integral of log. This connects the discrete product (factorial) to continuous analysis (integration).
Key Insight
Cauchy's functional equation $f(xy) = f(x) + f(y)$ has logarithms as the only continuous (or even measurable) solutions. Without a regularity condition, there exist pathological solutions via Hamel bases of $\mathbb{R}$ over $\mathbb{Q}$.