Indices, Surds and Exponential Functions | NIMCET

Quick answer: The laws of indices are $a^m \cdot a^n = a^{m + n}$ , $\frac{a^m}{a^n} = a^{m - n}$ , $(a^m)^n = a^{mn}$ , $a^0 = 1$ , $a^{-n} = \frac{1}{a^n}$ and $a^{\frac{p}{q}} = \sqrt[q]{a^p}$ . A surd is an irrational root such as $\sqrt{2}$ . To rationalise $\frac{1}{\sqrt{a} + \sqrt{b}}$ , multiply by $\sqrt{a} - \sqrt{b}$ to get $\frac{\sqrt{a} - \sqrt{b}}{a - b}$ . An exponential equation is solved by writing both sides with the same base.

Laws of indices

Law	Statement
Product	$a^m \cdot a^n = a^{m + n}$
Quotient	$\frac{a^m}{a^n} = a^{m - n}$
Power of a power	$(a^m)^n = a^{mn}$
Power of a product	$(ab)^n = a^nb^n$
Zero index	$a^0 = 1$ ( $a \ne 0$ )
Negative index	$a^{-n} = \frac{1}{a^n}$
Fractional index	$a^{\frac{p}{q}} = \left(\sqrt[q]{a}\right)^p$

Note that $a^{m^n}$ means $a^{(m^n)}$ , which is generally different from $(a^m)^n$ .

Surds

A surd $\sqrt[n]{a}$ is irrational, with a rational positive $a$ . For example $\sqrt{3}$ and $\sqrt[3]{5}$ are surds, but $\sqrt{9} = 3$ is not.

Rule	Example
$\sqrt{ab} = \sqrt{a}\sqrt{b}$	$\sqrt{50} = 5\sqrt{2}$
$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$	$\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$
Like surds add	$3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$
$\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$	$\sqrt{9} + \sqrt{16} = 7 \ne 5$

Rationalising the denominator

Denominator	Multiply by
$\sqrt{a}$	$\sqrt{a}$
$\sqrt{a} + \sqrt{b}$	$\sqrt{a} - \sqrt{b}$
$a + \sqrt{b}$	$a - \sqrt{b}$

Square root of a + 2√b

If $x + y = a$ and $xy = b$ , then

$\sqrt{a + 2\sqrt{b}} = \sqrt{x} + \sqrt{y}$

Example: $\sqrt{7 + 2\sqrt{10}}$ : numbers with sum 7 and product 10 are 5 and 2, so the answer is $\sqrt{5} + \sqrt{2}$ .

Comparing surds

Raise the surds to a common power. To compare $\sqrt{2}$ , $\sqrt[3]{3}$ and $\sqrt[6]{6}$ , raise each to the 6th power: $8$ , $9$ and $6$ . So $\sqrt[3]{3} > \sqrt{2} > \sqrt[6]{6}$ .

Exponential functions and equations

The function $f(x) = a^x$ ( $a > 0$ , $a \ne 1$ ) has domain ℝ and range $(0, \infty)$ . It is increasing if $a > 1$ and decreasing if $0 < a < 1$ .

Key rule: if $a^x = a^y$ with $a > 0$ , $a \ne 1$ , then $x = y$ .

Worked examples

Example 1: simplify

Simplify $\frac{2^{n + 3} - 2^{n + 1}}{2^{n + 2}}$ .

$\frac{2^{n + 1}(4 - 1)}{2^{n + 2}} = \frac{3}{2}$

Example 2: rationalise

Rationalise $\frac{1}{\sqrt{5} + \sqrt{3}}$ .

$\frac{\sqrt{5} - \sqrt{3}}{5 - 3} = \frac{\sqrt{5} - \sqrt{3}}{2}$

Example 3: exponential equation

Solve $4^x - 3 \cdot 2^{x + 1} + 8 = 0$ .

Step 1: Let $t = 2^x$ . Then $t^2 - 6t + 8 = 0$ , so $t = 2$ or $t = 4$ .

Step 2: $x = 1$ or $x = 2$ .

Example 4: same base

Solve $27^{x - 1} = 9^{x}$ .

$3^{3x - 3} = 3^{2x}$ , so $3x - 3 = 2x$ and $x = 3$ .

Summary table

Idea	Key result
Negative index	$a^{-n} = \frac{1}{a^n}$
Fractional index	$a^{\frac{p}{q}} = \sqrt[q]{a^p}$
Rationalising	Multiply by the conjugate
$\sqrt{a + 2\sqrt{b}}$	$\sqrt{x} + \sqrt{y}$ with $x + y = a$ , $xy = b$
Equal powers	Same base gives equal exponents

Exam tips

Convert every term to a common base (2, 3 or 5) before simplifying.
Substitute $t = a^x$ when an equation contains $a^{2x}$ and $a^x$ ; remember $t > 0$ .
To compare surds, take the LCM of the root orders.

Next, study logarithms and change of base, or browse the NIMCET Algebra chapter.

Common mistakes

Writing $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ .
Treating $a^{m^n}$ as $a^{mn}$ .
Keeping a negative value of $t = a^x$ in exponential equations.
Forgetting to change the sign when multiplying by the conjugate.

Frequently Asked Questions

What are the laws of indices?

aᵐ · aⁿ = aᵐ⁺ⁿ, aᵐ/aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁰ = 1 and a⁻ⁿ = 1/aⁿ.

What is a surd?

A surd is an irrational root of a rational number, such as √2 or ∛5.

How do you rationalise 1/(√a + √b)?

Multiply the numerator and denominator by √a − √b, giving (√a − √b)/(a − b).

How do you solve an exponential equation?

Write both sides with the same base and equate the exponents, or substitute t = aˣ.

What is the square root of 7 + 2√10?

It is √5 + √2, because 5 + 2 = 7 and 5 × 2 = 10.