# The Product of Logarithm of A and Logarithm of B

Introduction

Recall that we define the logarithm of base 10 of x is the exponent needed to produce x. The equation

$\log_{10}x = c$ means $10^c = x$, where $x >0$.

Logarithms to the base 10 are called common logarithms. Most times, $\log_{10}x$ is written as $\log x$. In this post, we prove that

$\log AB = \log A + \log B$.

Theorem

$\log AB = \log A + \log B$.

Proof

Let $\log A = x$ and let $\log B = y$. Using the definition above, we have $AB = (10^x)(10^y)$.

By the law of exponents, $AB = 10^{x + y}$.

Getting the logarithm of each side, $\log AB = \log 10^{x+y}$.

By the definition we have mentioned above, $\log AB = x + y$.

Substituting $\log A$ to $x$ and substituting $\log B$ to $y$, we have

$\log AB = \log A + \log B.$ $\blacksquare$