PROPERTIES OF LOGARITHMS

Recall one important viewpoint of logarithms:   a logarithm is an exponent.

In particular, what is the number [beautiful math coming... please be patient] $\,\log_b x\,$?
Answer:   It is the power that $\,b\,$ must be raised to, in order to get $\,x\,$.

It follows that [beautiful math coming... please be patient] $\,\log_b x=y\,$ is equivalent to $\,b^y=x\,$.

The number $\,b\,$ in the expression [beautiful math coming... please be patient] $\,\log_b x\,$ 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 [beautiful math coming... please be patient] $\,\log_b x = y\,$ 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, [beautiful math coming... please be patient] $\,(-2)^1=-2\,$, no problem.
However, [beautiful math coming... please be patient] $\,(-2)^{1.01}=(-2)^{101/100} = \root{100}\of{(-2)^{101}}\,$ 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 [beautiful math coming... please be patient] $\,b\gt 0\,$, $\,b\ne 1\,$, and $\,x\gt 0\,$.

Then, $\,\log_b x\,$ is defined, and $$ \log_b x = y \ \ \text{is equivalent to}\ \ b^y = x\ , $$ 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 [beautiful math coming... please be patient] $\,b\gt 0\,$, $\,b\ne 1\,$, $\,x\gt 0\,$, and $\,y\gt 0\,$.

[beautiful math coming... please be patient] $\log_b\,xy = \log_b x + \log_b y$
The log of a product is the sum of the logs.

[beautiful math coming... please be patient] $\displaystyle \log_b\frac{x}{y} = \log_b x - \log_b y$
The log of a quotient is the difference of the logs.

For this final property, $\,y\,$ can be any real number:
[beautiful math coming... please be patient] $\log_b\,x^y = y\,\log_b x$
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 [beautiful math coming... please be patient] $\ \log_b x = u\ $ and $\ \log_b y = v\,$.
Then, $\ b^u = x\ $ and $\ b^v = y\ $.
Renaming, we have:

[beautiful math coming... please be patient] $\log_b\,xy$ $=$ $\,\log_b\,b^u\,b^v$ (substitution)
  $=$ [beautiful math coming... please be patient] $\log_b\,b^{u+v}$ (law of exponents)
  $=$ $\,u+v\,$ (definition of logarithm)
$=$ [beautiful math coming... please be patient] $\log_b x + \log_b y$ (substitution)

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Change of Base Formula for Logarithms


On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
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(MAX is 21; there are 21 different problem types.)