Surds, Indices and Logarithms — CSEC Add Maths notes

The laws of indices

Law	Why
$a^m \times a^n = a^{m+n}$	the factors just pile up
$a^m \div a^n = a^{m-n}$	the shared factors cancel
$(a^m)^n = a^{mn}$	$n$ copies of $m$ factors
$a^0 = 1$	because $a^m \div a^m = 1$
$a^{-n} = \dfrac{1}{a^n}$	keeps the division law working past zero
$a^{1/n} = \sqrt[n]{a}$	because $\left(a^{1/n}\right)^n = a$

Surds

A surd is a root that cannot be written exactly as a fraction, like $\sqrt{2}$ or $\sqrt{75}$ .

To simplify one, look for the largest square factor:

$\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25}\,\sqrt{3} = 5\sqrt{3}$

Once every surd has been reduced to the same root, you can add and subtract them like ordinary like terms:

$\sqrt{50} - \sqrt{18} = 5\sqrt{2} - 3\sqrt{2} = 2\sqrt{2}$

Rationalising a denominator

An answer should not be left with a surd underneath. For a single surd, multiply top and bottom by that surd. For a two-term denominator, multiply by its conjugate — the same two terms with the middle sign reversed:

$\frac{6}{\sqrt5 - 2} = \frac{6}{\sqrt5-2} \times \frac{\sqrt5+2}{\sqrt5+2} = \frac{6(\sqrt5+2)}{5 - 4} = 6\sqrt5 + 12$

The conjugate works because $(a-b)(a+b) = a^2 - b^2$ , and squaring kills the root.

Logarithms

A logarithm is an index. That one sentence is most of the topic.

Read $\log_3 81$ as "what power of 3 gives 81?". The answer is 4, because $3^4 = 81$ .

Two consequences worth knowing on sight: $\log_a 1 = 0$ and $\log_a a = 1$ .

Note that $x$ must be positive. You cannot take the logarithm of zero or of a negative number, because no power of a positive base ever produces one. This is what makes the checking step at the end of a logarithmic equation compulsory rather than optional.

The laws of logarithms

They are the index laws in disguise: adding logs corresponds to multiplying powers, which is exactly $a^m \times a^n = a^{m+n}$ .

Solving exponential equations

If both sides can be written as powers of the same base, equate the indices:

$2^{x+1} = 32 = 2^5 \quad\Longrightarrow\quad x + 1 = 5 \quad\Longrightarrow\quad x = 4$

If they cannot, take logs of both sides and use the third law to bring the power down.

Some equations are quadratics in disguise. In $3^{2x} - 10\left(3^x\right) + 9 = 0$ , note that $3^{2x} = \left(3^x\right)^2$ . Substituting $y = 3^x$ turns it into $y^2 - 10y + 9 = 0$ , an ordinary quadratic — and once you have $y$ , you still have to get back to $x$ .

Solving logarithmic equations

Use the laws to combine everything into a single logarithm on each side.
Either equate the arguments, or rewrite in index form to remove the log.
Solve.
Check every solution in the original equation and reject any that asks for the log of a negative number or of zero.

That last step is not optional and it is not paranoia. Combining logs can create solutions that the original equation never had.