﻿ Properties of Logarithms

# Properties of Logarithms

Before doing this exercise, you may want to review basic properties of logarithms:  Introduction to Logarithms

Recall one important viewpoint of logarithms:

In particular, what is the number $\,\log_b x\,$? Answer:  It is the power that $\,b\,$ must be raised to, in order to get $\,x\,.$

It follows that $\,\log_b x=y\,$ is equivalent to $\,b^y=x\,.$

The number $\,b\,$ in the expression $\,\log_b x\,$ is called the base of the logarithm. We need to pause and determine the allowable bases for logarithms. Then, we will study several important properties that logarithms exhibit.

## Allowable Bases for Logarithms

Since $\,1\,$ raised to any power is $\,1\,,$ and since $\,0\,$ raised to any positive power is $\,0\,,$ we certainly don't want to allow the base to be $\,1\,$ or $\,0\,.$

For example, consider these dilemmas:

• What is ‘$\,\log_1 5\,$’?  There isn't any power that $\,1\,$ can be raised to, to give $\,5\,.$
• Or, what is ‘$\,\log_1 1\,$’?  The number $\,1\,,$ raised to any power, is $\,1\,.$ So, which number would we ‘choose’ as our answer?

Negative bases present similar difficulties. We want $\,\log_b x = y\,$ to be equivalent to $\,b^y = x\,.$ But, consider an expression like $\,(-2)^y \,$ (where the base is $\,-2\,$). Since you can't take even roots of negative numbers, we run into lots of problems.

For example, $\,(-2)^1=-2\,,$ no problem. However, $$\cssId{s32}{(-2)^{1.01}} \cssId{s33}{=(-2)^{101/100}} \cssId{s34}{= \root{100}\of{(-2)^{101}}}$$ isn't even defined, since it involves an even root of a negative number. For these reasons, we only allow positive bases for logarithms, and we exclude the number $\,1\,.$

Now, a positive number, raised to any power, is always positive. (Think about this!) For this reason, logarithms cannot act on negative numbers. For example, what would ‘$\,\log_2 (-3)\,$’ be? There is no power that $\,2\,$ can be raised to, in order to get a negative answer.

We are now ready to make things precise:

Allowable Bases for Logarithms;
Allowable Inputs for Logarithms

Let $\,b\gt 0\,,$ $\,b\ne 1\,,$ and $\,x\gt 0\,.$

Then, $\,\log_b x\,$ is defined, and $$\cssId{s48}{\log_b x = y} \ \ \cssId{s49}{\text{is equivalent to}}\ \ \cssId{s50}{b^y = x}\ ,$$ for all real numbers $\,y\,.$

The equation ‘$\,\log_b x = y\,$’ is called the logarithmic form of the equation .

The equation ‘$\,b^y = x\,$’ is called the exponential form of the equation.

In particular, the only allowable bases for logarithms are positive numbers, not equal to $\,1\,.$ Also, the only allowable inputs to logarithms are positive numbers.

## Three Important Properties of Logarithms

Logarithms have some beautiful simplifying properties, which make them extremely valuable:

• They can take a multiplication problem, and ‘turn it into’ an addition problem (which is much simpler)!
• They can take a division problem, and ‘turn it into’ a subtraction problem (which is much simpler)!
• They can take an exponentiation problem, and ‘turn it into’ a multiplication problem (which is much simpler)!

Precisely, we have:

Laws of Logarithms

Let $\,b\gt 0\,,$ $\,b\ne 1\,,$ $\,x\gt 0\,,$ and $\,y\gt 0\,.$

$\displaystyle \log_b\,xy = \log_b x + \log_b y$
The log of a product is the sum of the logs.

$\displaystyle \log_b\frac{x}{y} = \log_b x - \log_b y$
The log of a quotient is the difference of the logs.

For this final property, $\,y\,$ can be any real number:

$\log_b\,x^y = y\,\log_b x$
You can bring exponents down.

The last descriptive phrase, you can bring exponents down, is of course a bit loose. It could be more correctly described as the log of a number raised to a power, is the power, times the log of the number. However, this is a bit long, and the shorter phrase seems to work well for students.

Here is a proof of the first property. The remaining two proofs are similar.

## Proof that $\,\log_b\,xy = \log_b x + \log_b y$

Let $\ \log_b x = u\$ and $\ \log_b y = v\,.$ Then, $\ b^u = x\$ and $\ b^v = y\ .$

Renaming, we have:

 $\log_b\,xy$ $=$ $\log_b\,b^u\,b^v$ substitution $=$ $\log_b\,b^{u+v}$ law of exponents $=$ $u+v$ definition of logarithm $=$ $\log_b x + \log_b y$ substitution