Notes 28 Sep 2026

8. Logarithms

What a logarithm is, what conditions the base and the argument must meet, and the three types we will use.

Definition

Definition

The logarithm to base $a$ of $b$ is the number $c$ such that $a^{c} = b$:

$$\log_{\color{#e11d48}a} {\color{#059669}b} = {\color{#d97706}c} \iff {\color{#e11d48}a}^{\color{#d97706}c} = {\color{#059669}b}$$

${\color{#e11d48}a}$ is the base, ${\color{#059669}b}$ is the argument and ${\color{#d97706}c}$ is the result.

For it to make sense, the base and the argument must satisfy:

$$a \in (0,1) \cup (1,+\infty) \qquad b \in (0,+\infty)$$

The idea

Taking a logarithm means asking what power you have to raise the base to in order to get the argument.

Why these conditions: a negative base can't be raised to any exponent ($(-2)^{1/2}$ is not real), and base $1$ is useless because $1^{c} = 1$ always. And since a power with a positive base is always positive, the argument must be positive: the logarithm of $0$ or of a negative number does not exist.

Examples

Examples

LogarithmWhy
$\log_{2} 8 = 3$$2^{3} = 8$
$\log_{10} 100 = 2$$10^{2} = 100$
$\log_{4} 64 = 3$$4^{3} = 64$
$\log_{3} \frac{1}{9} = -2$$3^{-2} = \frac{1}{3^{2}} = \frac{1}{9}$
$\log_{\frac{2}{3}} \frac{3}{2} = -1$$\left(\frac{2}{3}\right)^{-1} = \frac{3}{2}$
$\log_{5} 5 = 1$$5^{1} = 5$
$\log_{5} 1 = 0$$5^{0} = 1$
$\log_{5} 0$does not exist: no power of $5$ equals $0$

Worked example

When the answer isn't obvious, rewrite it as a power and put everything in the same base:

$$\log_{\sqrt{2}} 8 = x \iff \left(\sqrt{2}\right)^{x} = 8$$
$$\left(2^{\frac{1}{2}}\right)^{x} = 2^{3} \;\Longrightarrow\; 2^{\frac{x}{2}} = 2^{3} \;\Longrightarrow\; \frac{x}{2} = 3 \;\Longrightarrow\; x = 6$$

If the bases are equal, the exponents must be equal too.

Types of logarithm

TypeNotationBase
General$\log_{a} b$$a$
Common (decimal)$\log b$$10$
Natural (Napierian)$\ln b$$e \approx 2.718$

When no base is written, it is $10$. The last two are the ones on the calculator, under the log and ln keys.

Examples: the unknown in the base or in the argument

If the unknown is the base:

$$\log_{x} 27 = 6 \iff x^{6} = 27 \;\Longrightarrow\; x = \sqrt[6]{27} = \sqrt[6]{3^{3}} = \sqrt{3}$$

If the unknown is the argument:

$$\log_{\frac{1}{3}} x = -2 \iff \left(\frac{1}{3}\right)^{-2} = x \;\Longrightarrow\; x = \left(\frac{3}{1}\right)^{2} = 9$$