# Logarithm

## Introduction

Logarithms and exponents are very closely related. In fact, they are inverse functions. This means that logarithms can be used to reverse the result of exponentiation and vice versa, just as addition can be used to reverse the result of subtraction. Thus, if we have $a^x = b$, then taking the logarithm with base $a$ on both sides will give us $\displaystyle x=\log_a{b}$.

We would read this as "the logarithm of b, base a, is x". For example, we know that $3^4=81$. To express the same fact in logarithmic notation we would write $\log_3 81=4$.

Depending on the field, the symbol $\log$ without a base can have different meanings. Typically, in mathematics through the level of calculus, the symbol is used to refer to a base 10 logarithm. Thus, $\log(100)$ means $\log_{10}(100)=2$. Usually, the symbol $\ln$ (an abbreviation of the French "logarithme normal," meaning "natural logarithm") is introduced to refer to the logarithm base e. However, in higher mathematics such as complex analysis, the base 10 logarithm is typically disposed with entirely, the symbol $\log$ is taken to mean the logarithm base e and the symbol $\ln$ is not used at all. (This is an example of conflicting mathematical conventions.)

## Logarithmic Properties

We can use the properties of exponents to build a set of properties for logarithms.

We know that $a^x\cdot a^y=a^{x+y}$. We let $a^x=b$ and $a^y=c$. This also makes $\displaystyle a^{x+y}=bc$. From $a^x = b$, we have $x = \log_a{b}$, and from $a^y=c$, we have $y=\log_a{c}$. So, $x+y = \log_a{b}+\log_a{c}$. But we also have from $\displaystyle a^{x+y} = bc$ that $x+y = \log_a{bc}$. Thus, we have found two expressions for $x+y$ establishing the identity:

$\log_a{b} + \log_a{c} = \log_a{bc}.$

Using the laws of exponents, we can derive and prove the following identities:

• $\log_a b^n=n\log_a b$
• $\log_a b+ \log_a c=\log_a bc$
• $\log_a b-\log_a c=\log_a \frac{b}{c}$
• $(\log_a b)(\log_c d)= (\log_a d)(\log_c b)$
• $\frac{\log_a b}{\log_a c}=\log_c b$
• $\displaystyle \log_{a^n} b^n=\log_a b$

Try proving all of these as exercises.

## Problems

1. Evaluate $(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})$.
1. Evaluate $(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)$.
1. Simplify $\displaystyle \frac 1{\log_2 N}+\frac 1{\log_3 N}+\frac 1{\log_4 N}+\cdots+ \frac 1{\log_{100}N}$ where $N=(100!)^3$.