Defining the Exponential Function

Idea

In everything you have met so far, the variable sits on the ground floor: $x^2$ , $x^3$ , $\sqrt{x}$ . The base changes and the power stays fixed.

An exponential function turns that around. The base is fixed and the variable is the power.

Rule

$f(x) = a^x, \quad a > 0, \quad a \ne 1$

Two conditions, and each rules out a real problem.

↳ $a > 0$ , or $a^x$ has no value at $x = \dfrac{1}{2}$

↳ $a \ne 1$ , or the function is the flat line $y = 1$

How it is used

1. Where the variable sits

Expression	Base	Power	Kind
$x^3$	$x$ , varies	$3$ , fixed	power function
$3^x$	$3$ , fixed	$x$ , varies	exponential

Swapping those two is the mistake this lesson exists to prevent. $2^3 = 8$ and $3^2 = 9$ are not the same number.

2. The rules of indices still hold

Nothing new is needed to work with $a^x$ . The rules you already have apply, with $x$ in the exponent.

a^x \cdot a^y = a^{x + y}

\frac{a^x}{a^y} = a^{x - y}

(a^x)^y = a^{xy}

$a^0 = 1$ for every allowed base

$a^{-x} = \dfrac{1}{a^x}$

3. It is never zero and never negative

$a$ is positive, so any power of it is positive. That is worth holding on to.

a^x > 0 \quad \text{for every } x

The graph therefore never touches the $x$ -axis, no matter how far left you go.

Exponential and logarithm

Try this. Compare bases 2 and 0.5: growth becomes decay. Move the input to 1; the logarithm is zero for either base.

BasePositive input x0.5

Base 2 · Blue: y = 2ˣ. At x = 0.5, y = 1.414.

4. Growth and decay

Base	Behaviour	As $x$ grows
$a > 1$	growth	climbs without limit
$0 < a < 1$	decay	falls towards $0$

Worked example

Solve $2^{x + 1} = 8$ .

Remember

$a^0 = 1$ and $a^1 = a$ for every allowed base $8 = 2^3$ , $16 = 2^4$ , $27 = 3^3$ equal bases means equal powers

Write both sides on the same base. Here $8$ is a power of $2$ .

2^{x + 1} = 2^3

The base is fixed and never $1$ , so it takes each value once. Equal outputs force equal powers.

x + 1 = 3

Solve.

x = 2

Check it: $2^{2 + 1} = 2^3 = 8$ .

Explore

Drag the curve to move the probe, and drag the slider to change the base $a$ .

Watch two things. The curve never crosses the $x$ -axis, whatever base you pick. And every curve passes through $(0, 1)$ , because $a^0 = 1$ for all of them.