An exponential function is a function of the form
[beautiful math coming... please be patient]$\,f(x) = b^x\,$ for
[beautiful math coming... please be patient]$\,b > 0\,$ and
[beautiful math coming... please be patient]$\,b\ne 1\,$.
Notice that the variable is in the exponent.
For example, $\,y = 2^x\,$, $\,y = (\frac 12)^x\,$, and $\,y = \text{e}^x\,$ are exponential functions.
Exponential functions and their graphs were introduced in the Algebra II curriculum:
When $\,b > 1\,$, $\,f(x) = b^x\,$ is an increasing function. That is, for all real numbers $\,x\,$ and $\,y\,$: $$x < y \ \ \Rightarrow\ \ b^x < b^y$$ It is clear from the graph that the other direction is also true: $$b^x < b^y \ \ \Rightarrow\ \ x < y$$ Together, we have: $$x < y\ \ \text{is equivalent to}\ \ b^x < b^y \tag{1} $$ Notice that when the base is greater than one, the inequality symbols that compare the inputs ($\,x\,$ and $\,y\,$) and their corresponding outputs ($\,b^x\,$ and $\,b^y\,$) have the same direction: $$ x \ {\bf\color{red}{<}}\ y\ \ \text{is equivalent to}\ \ b^x\ {\bf\color{red}{<}}\ b^y \tag{1a} $$ This equivalence can be alternately stated as: $$ x \ {\bf\color{red}{>}}\ y\ \ \text{is equivalent to}\ \ b^x\ {\bf\color{red}{>}}\ b^y \tag{1b} $$ 
$y = b^x\,$, for $\,b > 1\,$ an increasing exponential function $x < y \ \ \iff\ \ b^x < b^y$ 
When $\,0 < b < 1\,$, $\,f(x) = b^x\,$ is a decreasing function. That is, for all real numbers $\,x\,$ and $\,y\,$: $$x < y \ \ \Rightarrow\ \ b^x > b^y$$ It is clear from the graph that the other direction is also true: $$b^x > b^y \ \ \Rightarrow\ \ x < y$$ Together, we have: $$x < y\ \ \text{is equivalent to}\ \ b^x > b^y \tag{2}$$ Notice that when the base is between zero and one, the inequality symbols that compare the inputs ($\,x\,$ and $\,y\,$) and their corresponding outputs ($\,b^x\,$ and $\,b^y\,$) have different directions: $$ x \ {\bf\color{red}{<}}\ y\ \ \text{is equivalent to}\ \ b^x\ {\bf\color{red}{>}}\ b^y \tag{2a} $$ This equivalence can be alternately stated as: $$ x \ {\bf\color{red}{>}}\ y\ \ \text{is equivalent to}\ \ b^x\ {\bf\color{red}{<}}\ b^y \tag{2b} $$ 
$y = b^x\,$, for $\,0 < b < 1\,$ a decreasing exponential function $x < y \ \ \iff\ \ b^x > b^y$ 
The graphs of exponential functions pass both vertical and horizontal lines tests,
so they are onetoone functions.
Thus:
$$x = y \ \ \text{is equivalent to}\ \ b^x = b^y
\tag{3}$$
Consequently, exponential functions have inverses.
In a future section, we'll see that the class of logarithmic functions provide inverses
to the class of exponential functions.
Equivalences (1), (2), and (3) can be used to easily solve certain types of mathematical sentences, as illustrated in the following examples.
Solve: $2^{3x1} < 2^{5x}$ 
This is an inequality of the form
$$\,b^{\text{stuff1}} < b^{\text{stuff2}}\,$$
where the base, $\,b\,$, is $\,2\,$. Notice that the exponential functions on both sides use the same base. In this example, the base is greater than one, so we'll use equivalence (1): $$x < y \ \ \iff\ \ b^x < b^y$$ 
SOLUTION: $$\begin{alignat}{2} 2^{3x1} \ &<\ 2^{5x} \qquad&&\text{original inequality}\cr\cr 3x1 \ &<\ 5x&&\text{use (1); inequality symbol doesn't change}\cr\cr 2x\ \ &<\ 1&&\text{addition property of inequality}\cr\cr x\ \ &>\ \frac12&&\text{divide by a negative #; reverse inequality} \end{alignat} $$ 
red curve: $y = 2^{5x}$ blue curve lies below red curve for $\,x > \frac 12$ 
Solve: $\displaystyle\frac 1{3^{x^21}} > 1$ 
This inequality will be solved in two different ways. As long as you use correct tools in a correct way, there are often different ways you can proceed. 
SOLUTION #1: $$\begin{alignat}{2} \frac 1{3^{x^21}} \ &>\ 1 \qquad&&\text{original inequality}\cr\cr \left(\frac 13\right)^{x^21}\ &>\ \left(\frac 13\right)^0 \ \ &&\text{rename both sides using exponent laws}\cr\cr x^2  1\ \ &<\ 0&&\text{use (2); inequality changes direction}\cr\cr x^2\ \ &<\ 1&&\text{addition property of inequality}\cr\cr 1 < \ &x\ < 1&&\text{inspection (knowledge of $x^2$ curve)} \end{alignat} $$ SOLUTION #2: $$\begin{alignat}{2} \frac 1{3^{x^21}} \ &>\ 1 \qquad&&\text{original inequality}\cr\cr 1\ &>\ 3^{x^21} \ \ \ \ &&\text{multiply both sides by $3^{x^21}\overset{\text{always}}{>}0$}\cr\cr 3^0 \ &>\ 3^{x^21}&&\text{rename: $1 = 3^0$}\cr\cr 0\ \ &>\ x^2  1\ \ &&\text{use (1)}\cr\cr x^2 \ &<\ 1&&\text{addition property; rearrange}\cr\cr 1 < \ &x\ < 1&&\text{inspection (knowledge of $x^2$ curve)} \end{alignat} $$ 
red curve: $y = 1$ blue curve lies above red curve for $\,x\,$ between $\,1\,$ and $\,1\,$ 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
