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

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

 

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

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

Theorem

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

Proof

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

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

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

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

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

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