A logarithmic function is a function of the form $\,f(x) = \log_b x\,$ for $\,b > 0\,$ and $\,b\ne 1\,$.
The expression ‘$\,\log_b x\,$’ is read aloud as ‘log base $\,b\,$ of $\,x\,$’;
‘$\,\log_b\,$’ (read aloud as ‘log base $\,b\,$’) is the name of the function, and it acts on the input $\,x\,$.
When there's no confusion about order of operations,
mathematical convention dictates that we can drop the parentheses used to ‘hold’ the input,
and write ‘$\,\log_b x\,$’ (without parentheses) instead of the more cumbersome ‘$\,\log_b(x)\,$’.
If order of operations becomes ambiguous, be sure to include parentheses for clarification.
Thus, for example, you should write either $\,\log_b(x + 1)\,$ or $\,(\log_b x) + 1\,$,
but never the ambiguous ‘$\,\log_b x + 1\,$’.
(Notice, however, that there is no ambiguity in the expression ‘$\,1 + \log_b x\,$’.)
The number $\,b\,$ in $\,\log_b x \,$ is called the base of the logarithm.
When the base is $\,\text{e}\,$, the function is called the natural logarithm,
and can be written as $\,y = \ln x\,$.
(Note: $\text{e}\approx 2.71828\,$)
When the base is $\,10\,$, the function is called the common logarithm,
and can be written as $\,y = \log x\,$.
Logarithmic functions and their graphs were introduced in the Algebra II curriculum:
When $\,b > 1\,$, $\,f(x) = \log_b x\,$ is an increasing function. That is, for all positive real numbers $\,x\,$ and $\,y\,$: $$ \cssId{s23}{x < y \ \ \Rightarrow\ \ \log_b x < \log_b y} $$ It is clear from the graph that the other direction is also true: $$ \cssId{s25}{\log_b x < \log_b y \ \ \Rightarrow\ \ x < y} $$ Together, we have that for all positive real numbers $\,x\,$ and $\,y\,$, and for $\,b > 1\,$: $$ \cssId{s27}{x < y\ \ \text{is equivalent to}\ \ \log_b x < \log_b y} \tag{1} $$ Notice that when the base of the logarithm is greater than one, the inequality symbols that compare the inputs ($\,x\,$ and $\,y\,$) and their corresponding outputs ($\,\log_b x\,$ and $\,\log_b y\,$) have the same direction: $$ \cssId{s31}{x \ {\bf\color{red}{<}}\ y\ \ \text{is equivalent to}\ \ \log_b x \ {\bf\color{red}{<}}\ \log_b y} \tag{1a} $$ This equivalence can be alternatively stated as: $$ \cssId{s33}{x \ {\bf\color{red}{>}}\ y\ \ \text{is equivalent to}\ \ \log_b x\ {\bf\color{red}{>}}\ \log_b y} \tag{1b} $$ 
$y = \log_b x\,$, for $\,b > 1\,$ an increasing logarithmic function $x < y \ \ \iff\ \ \log_b x < \log_b y$ 
When $\,0 < b < 1\,$, $\,f(x) = \log_b x\,$ is a decreasing function. That is, for all positive real numbers $\,x\,$ and $\,y\,$: $$ \cssId{s40}{x < y \ \ \Rightarrow\ \ \log_b x > \log_b y} $$ It is clear from the graph that the other direction is also true: $$ \cssId{s42}{\log_b x > \log_b y \ \ \Rightarrow\ \ x < y} $$ Together, we have that for all positive real numbers $\,x\,$ and $\,y\,$, and for $\,0 < b < 1\,$: $$ \cssId{s44}{x < y\ \ \text{is equivalent to}\ \ \log_b x > \log_b y} \tag{2} $$ Notice that when the base of the logarithm is between zero and one, the inequality symbols that compare the inputs ($\,x\,$ and $\,y\,$) and their corresponding outputs ($\,\log_b x\,$ and $\,\log_b y\,$) have different directions: $$ \cssId{s48}{x \ {\bf\color{red}{<}}\ y\ \ \text{is equivalent to}\ \ \log_b x \ {\bf\color{red}{>}}\ \log_b y} \tag{2a} $$ This equivalence can be alternatively stated as: $$ \cssId{s50}{x \ {\bf\color{red}{>}}\ y\ \ \text{is equivalent to}\ \ \log_b x \ {\bf\color{red}{<}}\ \log_b y} \tag{2b} $$ 
$y = \log_b x\,$, for $\,0 < b < 1\,$ a decreasing logarithmic function $x < y \ \ \iff\ \ \log_b x > \log_b y$ 
The graphs of logarithmic functions pass both vertical and horizontal lines tests,
so they are onetoone functions.
Thus, for all positive real numbers $\,x\,$ and $\,y\,$:
$$
\cssId{s59}{x = y \ \ \text{is equivalent to}\ \ \log_b x = \log_b y}
\tag{3}
$$
Consequently, logarithmic functions have inverses.
Logarithmic functions and exponential functions with the same bases are inverses—they ‘undo’ each other.
Roughly:
Drop $\,x\,$ in the ‘exponential function with base $\,a\,$’ box;
get $\,y\,$ out the bottom. (See the top box in the diagram at right.) This is expressed mathematically as: $$\cssId{s71}{y = a^x}$$ Drop $\,y\,$ in the ‘logarithmic function with base $\,a\,$’ box; get $\,x\,$ out the bottom. (See the bottom box in the diagram at right.) This is expressed mathematically as: $$\cssId{s76}{\log_a y = x}$$ Since the exponential and logarithmic functions with base $\,a\,$ are inverses, these two operations are equivalent: $$\cssId{s78}{y = a^x\ \ \iff\ \ \log_a y = x}$$ If one is true, so is the other. If one is false, so is the other. They are true and false at the same time. In other words:
‘$\log_a y = x\,$’ is called the logarithmic form of the equation. 
The exponential and logarithmic functions with the same bases ‘undo’ each other! When ‘EXP fct, base $\,a\,$’ takes $\,x\,$ to $\,y\,$, then ‘LOG fct, base $\,a\,$’ takes $\,y\,$ back to $\,x\,$. 
Going from logarithmic form ‘$\log_a y = x\,$’ to the equivalent exponential form ‘$\,y = a^x\,$’ is easy!
Start with the base ‘$\,a\,$’, circle counterclockwise to $\,x\,$, and then to $\,y\,$:
This circular motion actually gives ‘$\,a^x = y\,$’. Then, write it again from righttoleft to get ‘$\,y = a^x\,$’ ! 
Here's a thought process to go from exponential form ‘$\,y = a^x\,$’
to the equivalent logarithmic form ‘$\log_a y = x\,$’:
Let $\,a > 0\,$, $\,a \ne 1\,$.
These are the allowable bases for logarithmic and exponential functions.
For exponential functions, $\,y = a^x\,$:
Now, you should be comfortable with both important views of logarithms:
Equivalences (1), (2), and (3) can be used to solve certain types of mathematical sentences involving logarithms,
as illustrated in the following examples.
IMPORTANT:
Whenever you have an expression of the form ‘$\,\log_a (\text{stuff})\,$’, the ‘stuff’ must be positive!
EXAMPLE 1: Solve: $\log_2(3x1) < \log_2(5x)$ 
This is an inequality of the form
$$\,\cssId{sb49}{\log_b(\text{stuff1}) < \log_b(\text{stuff2})}\,$$
where the base of the logarithm, $\,b\,$, is $\,2\,$. Notice that the logarithmic functions on both sides use the same base. In this example, the base is greater than one, so we'll use equivalence (1): $$\cssId{sb53}{x < y \ \ \iff\ \ \log_b x < \log_b y}$$ 
SOLUTION:
$$\begin{alignat}{2}
\log_2(3x1) \ &< \ \log_2(5x)\qquad&&\text{original inequality}\cr\cr
3x1 &< \ 5x&&\text{use (1); inequality symbol doesn't change}\cr\cr
2x \ &< 1&&\text{addition property for inequalities}\cr\cr
x \ &> \frac12&&\text{divide by a negative #; reverse inequality}
\end{alignat}
$$
Since logarithms only act on positive numbers, we must also have:
$$
\begin{gather}
3x  1 > 0 \quad \text{and} \quad 5x > 0\cr
x > \frac 13 \quad \text{and} \quad x > 0
\end{gather}
$$
Putting everything together:
$$\cssId{sb59}{\bigl(x > \frac 12 \ \ \text{and} \ \ x > \frac 13 \ \ \text{and} \ \ x > 0\bigr) \iff x > \frac 13}$$
The solution set of this inequality is:
$$
\cssId{sb61}{\underbrace{\{x\ \ x > \frac 13\}}_{\text{setbuilder notation}} =
\underbrace{(\frac 13,\infty)}_{\text{interval notation}}}
$$




Head up to wolframalpha.com and type in: log_2 (3x 1) < log_2 (5x) Voila! 
EXAMPLE 2: Solve: $\log_{0.3}(7)  \log_{0.3}(1  6x) > 0$ 
This is an inequality that can be easily transformed to
$$\,\cssId{sb73}{\log_b(\text{stuff1}) > \log_b(\text{stuff2})}\,$$
where the base of the logarithm, $\,b\,$, is $\,0.3\,$. Notice that the logarithmic functions use the same base. In this example, the base is between zero and one, so we'll use equivalence (2): $$\cssId{sb77}{x < y \ \ \iff\ \ \log_b x > \log_b y}$$ Compare the solution at right to what you get at wolframalpha.com: log_(0.3) (7)  log_(0.3) (1  6x) > 0 
SOLUTION:
$$
\begin{gather}
\log_{0.3}(7)  \log_{0.3}(1  6x) > 0\cr\cr
\log_{0.3}(7) > \log_{0.3}(1  6x)\cr\cr
7 < 1  6x \ \ \ \text{and}\ \ \ 1  6x > 0\cr\cr
6x < 6 \ \ \ \text{and}\ \ \ 6x > 1\cr\cr
x < 1 \ \ \ \text{and}\ \ \ x < \frac 16\cr\cr
x < 1
\end{gather}
$$
Here are the reasons for each step:

EXAMPLE 3: Solve: $\ln(2x + 1)$ $\ \ = \ln(7x  3)$ 
Here's a screenshot from wolframalpha.com: 
Writing a list of equivalent mathematical sentences:
$$
\begin{gather}
\ln(2x + 1) = \ln(7x  3)\cr\cr
2x + 1 = 7x  3 \ \ \text{and}\ \ 2x + 1 > 0 \ \ \text{and}\ \ 7x3 > 0\cr\cr
5x = 4 \ \ \text{and}\ \ 2x > 1 \ \ \text{and}\ \ 7x > 3\cr\cr
x = \frac 45 \ \ \ \text{and}\ \ \ x > \frac 12 \ \ \ \text{and}\ \ \ x > \frac{3}{7}\cr\cr
x = \frac 45
\end{gather}
$$

On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
