# 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\$