What is the difference between index laws and logarithm laws?

They are two sides of the same idea. Index laws tell you how to combine powers, such as adding the indices when you multiply. Logarithm laws are the mirror image, so multiplying inside a log becomes adding two logs. Because a logarithm undoes a power, every index law has a matching log law.

Why can you not take the logarithm of a negative number or zero?

A logarithm asks what power you raise the base to in order to get the number inside. A positive base raised to any power is always positive, so it can never produce a negative number or zero. That is why the inside of a log must be greater than zero, and why you must reject any solution that makes it zero or negative.

When do I need the change of base formula?

Use it whenever a logarithm has a base your calculator cannot handle directly, such as base 2 or base 5. The formula rewrites the log as a fraction of two natural logs, with the number you want on top and the original base underneath, so you can evaluate it on any calculator.

Index and Logarithm Laws | EduVerse Learn

Powers and logarithms are the same idea wearing two different costumes. A power asks “what do I get when I multiply this base together this many times”. A logarithm asks the reverse question, “how many times did I have to multiply”. Once you see that one undoes the other, the long list of index laws and log laws stops feeling like rote memory and starts feeling like common sense. This lesson is about reading both costumes fluently.

Index laws: shortcuts for stacking powers

An index (also called a power or exponent) tells you how many times to multiply a base by itself. So $2^{4}$ just means $2 \times 2 \times 2 \times 2$ . Every index law is simply a shortcut that saves you writing all that out.

The five you need are:

a^{m} \times a^{n} = a^{m+n}, \qquad \frac{a^{m}}{a^{n}} = a^{m-n}, \qquad (a^{m})^{n} = a^{mn}

a^{0} = 1, \qquad a^{-n} = \frac{1}{a^{n}}

The plain idea behind the first one is this. If you multiply $2^{3}$ by $2^{4}$ , you are multiplying three twos by four twos, which is seven twos in total. So you add the indices. Dividing means cancelling, so you subtract. A power of a power means repeating the multiplication, so you multiply the indices. The whole topic is bookkeeping for how many copies of the base you have.

Logarithms: the undo button for powers

A logarithm answers the question a power leaves hanging. The statement $\log_{b}(a) = c$ reads as “the power you raise $b$ to in order to get $a$ ”. So $\log_{2}(8) = 3$ because $2^{3} = 8$ . The base sits as a small subscript, and in Methods the most common base is $e$ , written $\log_{e}$ .

Because a log is the undo button for a power, the index laws flip into a matching set of log laws:

\log_{b}(mn) = \log_{b}(m) + \log_{b}(n)

\log_{b}\!\left(\frac{m}{n}\right) = \log_{b}(m) - \log_{b}(n)

\log_{b}(m^{p}) = p\,\log_{b}(m)

Notice the pattern. Multiplying inside a log turns into adding logs, because powers add when you multiply. Dividing turns into subtracting. A power inside the log jumps out the front as a coefficient. The single most common exam mistake is mixing these up, for example writing $\log(m) + \log(n)$ as $\log(m+n)$ . That is wrong. Adding logs multiplies the insides, it never adds them.

Because the log undoes the power, the two graphs are mirror images of each other across the line $y = x$ . The curve $y = e^{x}$ shoots upward and flattens onto the $x$ -axis on the left, while $y = \log_{e}(x)$ is its reflection, climbing slowly and hugging the $y$ -axis. Reflecting a point like $(0, 1)$ on the exponential gives $(1, 0)$ on the logarithm.

The exponential y = e^x (blue) and the logarithm y = ln(x) (red) are reflections of each other across the line y = x (grey dashed), showing that a logarithm undoes a power.

Converting between exponential and logarithmic form

These two sentences say exactly the same thing, just from opposite directions:

b^{c} = a \quad \Longleftrightarrow \quad \log_{b}(a) = c

To convert, find the base, keep it as the base, and swap which of the other two numbers is the answer. If you are told $\log_{e}(y) = 3x + 2$ , the base is $e$ , so raising $e$ to the right hand side gives $y = e^{3x+2}$ . Going the other way, $e^{k} = 7$ becomes $k = \log_{e}(7)$ .

This swap is the heart of solving exponential and logarithmic equations, so practise it until it is automatic.

Change of base

Your calculator and the standard rules are built around base $e$ , but exam questions often use other bases such as $2$ or $10$ . The change of base rule lets you rewrite any logarithm using a base you prefer:

\log_{b}(a) = \frac{\log_{e}(a)}{\log_{e}(b)}

The number you want the log of goes on top, and the original base goes underneath. So $\log_{5}(20) = \dfrac{\log_{e}(20)}{\log_{e}(5)}$ . A handy special case drops straight out of this: $\log_{b}(a) = \dfrac{1}{\log_{a}(b)}$ , which is just the rule with the roles swapped.

Learn it, then fade the help

1 · I do — fully worked

Condense to a single log Write $2\log_e(3) + \log_e(2) - \log_e(6)$ as one logarithm. The coefficient lifts to a power: $2\log_e(3) = \log_e(9)$ . Add by multiplying inside: $\log_e(9) + \log_e(2) = \log_e(18)$ . Subtract by dividing: $\log_e(18) - \log_e(6) = \log_e\!\left(\tfrac{18}{6}\right) = \log_e(3)$ .

2 · We do — your turn, guided

Same steps, your turn. Write $2\log_e(2) + \log_e(5) - \log_e(2)$ as a single logarithm. Start with $2\log_e(2) = \log_e(\;?\;)$ , then combine.

Check your step

$2\log_e(2) = \log_e(4)$ , then $\log_e(4) + \log_e(5) = \log_e(20)$ , and $\log_e(20) - \log_e(2) = \log_e(10)$ .

3 · You do — solo

No scaffolding now: write $3\log_e(2) + \log_e(3) - \log_e(4)$ as a single logarithm in simplest form.

Reveal solution

$3\log_e(2) = \log_e(8)$ ; then $\log_e(8) + \log_e(3) = \log_e(24)$ ; finally $\log_e(24) - \log_e(4) = \log_e(6)$ .

Two traps that cost real marks

First, always check the domain of a logarithm. You can only take the log of a positive number, so $\log_{e}(x)$ needs $x > 0$ . When solving log equations you often get two candidate answers, and examiner reports note that students lose marks by keeping a value that makes any inside expression zero or negative. Solve fully, then reject the invalid ones.

Second, when an exponential equation hides a quadratic, remember that $b^{x} > 0$ for every $x$ . After you substitute and factorise, discard any solution that asks $b^{x}$ to equal a negative number, then finish by solving for $x$ rather than stopping at the substitution variable.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

State the three core index laws for

a^m \times a^n

\dfrac{a^m}{a^n}

and

(a^m)^n

What does

\log_b(mn)

equal, and what is the common slip?

Convert between forms:

b^c = a \iff \;?

Write the change of base rule for

\log_b(a)

Why can you never take the log of a negative number or zero?

Recall · Solving Exponential and Logarithmic Equations

To solve

a^x = b

, what do you do and what is the exact answer?

Recall · Functional Notation

When can

f(x) = k

have no solutions?

Work through the recipe in the Worked Examples tab, then put both costumes to the test in Try It.

Index and Logarithm Laws

Index laws: shortcuts for stacking powers

Logarithms: the undo button for powers

Converting between exponential and logarithmic form

Change of base

Two traps that cost real marks

Lock it in with active recall

Try it yourself

Frequently asked questions

Index and Logarithm Laws

Learn

Index laws: shortcuts for stacking powers

Logarithms: the undo button for powers

Converting between exponential and logarithmic form

Change of base

Two traps that cost real marks

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions