Mathematics 9709 · Pure Mathematics 2

Logs and the laws: the toolkit you apply everywhere

A logarithm is just a power turned around: $a^x=b$ and $x=\log_a b$ mean the same thing.
One asks “what is the answer?”. The other asks “what was the power?”.
The whole chapter runs on that one swap plus the three laws below.

One idea runs the whole chapter: a logarithm and an exponential undo each other. Stuck because the unknown is in a power ( $a^x=b$ )? Take $\ln$ . Stuck in a messy $\ln$ equation? Raise $e$ to both sides. Got a curved data law? Take $\ln$ and it goes straight. That works because $\ln$ turns “multiply” into “add”, which is the shape of a straight line.

The swap: index form ⇄ log form

Quick recap from IGCSE. P3 marks come from using logs, not converting them.
The one thing to keep ready is the swap below.

$a^x=b \iff x=\log_a b$ : the log just means “the power you raise $a$ to”.

Two easy facts follow: $\log_a a=1$ (the power that gives $a$ is $1$ ) and $\log_a 1=0$ (the power that gives $1$ is $0$ ).
In P3 the base is nearly always $e$ , written $\ln$ . We cover it fully in the inverse-twins topic. (textbook §2.1–2.2, Key points 2.1–2.2)

The three laws (your everyday toolkit)

Multiply becomes add. Divide becomes subtract. A power drops to the front. That is the whole point: logs turn each hard job into an easier one.

Law	Rule	What it does
Product	$\log_a(xy)=\log_a x+\log_a y$	splits a product into a sum
Quotient	$\log_a\!\frac{x}{y}=\log_a x-\log_a y$	splits a quotient into a difference
Power	$\log_a(x^m)=m\,\log_a x$	drops an exponent to a coefficient

One more, which is really the power law again: $\log_a\!\frac{1}{x}=-\log_a x$ (because $\frac1x=x^{-1}$ ).
And the cancel both ways: $a^{\log_a x}=x$ . (textbook §2.3, Key points 2.3–2.4)

Common mistake

There is no law for

\log_a(x+y)

. You cannot split the log of a sum.

\ln(x+5)

stays as one piece. It is not

\ln x+\ln 5

. Get this wrong and your first line is already broken. Mark schemes give it M0.

The main move: combine into ONE log

Every solving and straightening question in this chapter starts the same way: squash several log terms into one log.

Worked example

Combine into one logarithm

$\log_2 3+\log_2 5=\log_2(3\times5)=\log_2 15$ (product law)
$2\log_5 2+\log_5 3=\log_5 2^2+\log_5 3=\log_5 12$ (power, then product)

In line 2, use the power law first ( $2\log_5 2=\log_5 4$ ) so every term is a plain log, then combine. (textbook WE 2.8)

To go the other way, a number in front of a log becomes a power: $m\log_a x=\log_a x^m$ . That clears the number stuck before a $\ln$ . It turns $2\ln x$ into $\ln x^2$ , so you can compare what is inside each log.

The log laws are really the index rules you already know

$x^m\cdot x^n=x^{m+n}$ says multiplying powers adds the indices. A log is just “the index”. So reading that same rule through a log gives $\log(xy)=\log x+\log y$ . You already knew the product law in Year 1, only written as indices, not logs. Same for quotient ( $x^m/x^n=x^{m-n}$ ) and power ( $(x^m)^n=x^{mn}$ ). Nothing new here, just a new way to look at it. (textbook §2.3 proofs)

Your turn— tap to reveal the worked answer (quick check)

Write $\log_4 p^5-\log_4 q^2$ using $x=\log_4 p$ and $y=\log_4 q$ . Use the power law on each term: $5\log_4 p-2\log_4 q$ .

Answer

5x-2y

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