Exponentials & logarithms

The graph of aˣ, the laws of logarithms, and solving equations of the form aˣ = b.

8 min read

A logarithm is the inverse of an exponential: it answers "what power gives this number?". Logs turn equations with the unknown in the exponent into ones you can solve.

Graph showing y = a to the x and its inverse y = log base a of x, mirror images in the dashed line y = x. — y = aˣ and y = logₐx are inverses: reflections of each other in the line y = x.

$y = a^{x}$ (for $a > 1$ ) passes through $(0, 1)$ and grows rapidly; its inverse $y = lo g_{a} x$ passes through $(1, 0)$ . They are reflections in the line $y = x$ .

The definition: $a^{b} = c ⟺ b = lo g_{a} c$ . The laws:

$lo g_{a} x + lo g_{a} y = lo g_{a} (x y)$
$lo g_{a} x - lo g_{a} y = lo g_{a} (\frac{x}{y})$
$k lo g_{a} x = lo g_{a} (x^{k})$
$lo g_{a} (\frac{1}{x}) = - lo g_{a} x$ and $lo g_{a} a = 1$ .

Worked example

Write $2 lo g_{a} 3 + lo g_{a} 5$ as a single logarithm.

$lo g_{a} 3^{2} + lo g_{a} 5 = lo g_{a} (9 \times 5) = lo g_{a} 45.$

Take logs of both sides to bring the power down: $a^{x} = b \Rightarrow x = \frac{lo g b}{lo g a}$ . Some equations reduce to this form after a substitution.

Worked example

Solve $2 5^{x} - 3 \times 5^{x} + 2 = 0$ .

Let $y = 5^{x}$ (so $2 5^{x} = y^{2}$ ): $y^{2} - 3 y + 2 = 0 \Rightarrow (y - 1) (y - 2) = 0$ .

$5^{x} = 1 \Rightarrow x = 0$ , or $5^{x} = 2 \Rightarrow x = \frac{lo g 2}{lo g 5} .$

Exam tip

When you see $a^{2 x}$ and $a^{x}$ together, substitute $y = a^{x}$ , turning the equation into a quadratic you already know how to solve.

Writing $lo g (x + y)$ as $lo g x + lo g y$ ; the law is for $lo g (x y)$ .
Forgetting $lo g_{a} 1 = 0$ and $lo g_{a} a = 1$ .
Losing the $x = 0$ solution when substituting $y = a^{x}$ .

Exponentials & logarithms

Overview

The exponential & log graphs

The laws of logarithms

Solving aˣ = b

Common mistakes