Laws of Indices for Rational Exponents

Indices allow repeated multiplication, roots and reciprocals to be expressed using one consistent notation. In the TMUA, the main challenge is rarely remembering the laws themselves. It is recognising when they apply, using them in the correct direction and avoiding common algebraic traps.

However, based on patterns seen in actual TMUA past papers and worked solutions, there are certain recurring strategies and tricks that ExamAlly consistently emphasises. This article integrates both the core theory and these high-value exam insights.

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Indices appear frequently in TMUA questions, often disguised within algebraic expressions or equations. Success depends on:

• recognising when expressions can be simplified;
• rewriting numbers using a common base;
• avoiding traps designed to test conceptual understanding.

Students typically struggle not with the rules themselves, but with applying them efficiently under time pressure.

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For a positive integer $n$,

$$a^n=\underbrace{a\times a\times\cdots\times a}_{n\text{ factors}}.$$

For example,

$$3^4=3\times3\times3\times3=81.$$

This definition explains the first and most important index law:

$$a^m a^n=a^{m+n}.$$

When two powers with the same base are multiplied, the total number of factors is the sum of the two exponents.

For example,

$$2^3\times2^5 = (2\times2\times2)(2\times2\times2\times2\times2) = 2^8.$$

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For $a>0$, $b>0$, and rational numbers $m$ and $n$,

$$\boxed{a^m a^n=a^{m+n}}$$

$$\boxed{\frac{a^m}{a^n}=a^{m-n}}$$

$$\boxed{(a^m)^n=a^{mn}}$$

$$\boxed{(ab)^n=a^n b^n}$$

$$\boxed{\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}}$$

The quotient laws also require the relevant denominators to be non-zero.

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For $a\neq0$,

$$\boxed{a^0=1}.$$

This follows because

$$\frac{a^m}{a^m}=a^{m-m}=a^0,$$

while also

$$\frac{a^m}{a^m}=1.$$

Therefore,

$$a^0=1.$$

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For $a\neq0$,

$$\boxed{a^{-n}=\frac{1}{a^n}}.$$

For example,

$$2^{-3}=\frac{1}{2^3}=\frac{1}{8}.$$

A negative exponent does not make the value negative. It indicates a reciprocal.

For a non-zero fraction,

$$\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n.$$

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For $a>0$ and a positive integer $n$,

$$\boxed{a^{1/n}=\sqrt[n]{a}}.$$

More generally,

$$\boxed{a^{m/n}=\sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m}.$$

For example,

$$27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 =9.$$

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This is one of the most useful techniques for equations involving indices.

Example

Solve

$$9^{x+1}=27^{2x-1}.$$

Rewrite both sides using base $3$:

$$9=3^2, \qquad 27=3^3.$$

Therefore,

$$(3^2)^{x+1}=(3^3)^{2x-1}.$$

Using $(a^m)^n=a^{mn}$,

$$3^{2x+2}=3^{6x-3}.$$

Since the bases are equal and $3\neq1$, equate the exponents:

$$2x+2=6x-3.$$

Hence,

$$5=4x$$

and therefore

$$\boxed{x=\frac{5}{4}}.$$

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Trick 1: Look for Hidden Common Bases

Numbers are often disguised:

Number	Rewrite as
$4$	$2^2$
$8$	$2^3$
$16$	$2^4$
$27$	$3^3$
$32$	$2^5$
$64$	$2^6$
$81$	$3^4$

ExamAlly insight: If you see numbers such as $8$, $16$, $32$ or $64$, immediately consider rewriting them as powers of $2$.

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Trick 2: Fractional Powers Often Simplify Neatly

For example,

$$64^{2/3} = \left(\sqrt[3]{64}\right)^2 = 4^2 = 16.$$

ExamAlly insight: Taking the root first often avoids unnecessarily large calculations.

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Trick 3: Negative Powers Signal a Reciprocal

For example,

$$\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}.$$

ExamAlly insight: Reverse the fraction first, then apply the positive power.

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Trick 4: Simplify in Index Form Before Calculating

A less efficient approach is

$$(2^3)^4=8^4.$$

A better approach is

$$(2^3)^4=2^{12}.$$

ExamAlly insight: Stay in index form for as long as possible.

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Trick 5: Compare Positive Powers by Applying the Same Power

Compare $2^{3/2}$ and $3$.

Both values are positive, so squaring preserves their order:

$$\left(2^{3/2}\right)^2=2^3=8, \qquad 3^2=9.$$

Since $8<9$,

$$\boxed{2^{3/2}<3}.$$

ExamAlly insight: Applying a suitable power can remove roots and make comparisons exact.

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Trick 6: Look for Cancellation

For example,

$$\frac{x^{5/2}}{x^{1/2}} = x^{5/2-1/2} = x^2.$$

ExamAlly insight: Complicated-looking fractions often simplify immediately when exponent laws are applied.

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Trick 7: Watch for Exponent Traps

In general,

$$a^m+a^n\neq a^{m+n},$$

and

$$(a+b)^n\neq a^n+b^n.$$

The index laws for adding exponents apply to multiplication, not addition.

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Trick 8: Use Structure, Not Brute Force

Consider

$$\left(\frac{8x^{-3/2}}{y^{1/2}}\right)^{-2/3}.$$

Rather than expanding numerically:

1. apply the outer power to each factor;
2. multiply the exponents;
3. simplify each resulting power;
4. combine the simplified factors.

ExamAlly insight: TMUA questions reward structural simplification more than brute-force calculation.

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Example

Simplify

$$\frac{12x^{5/2}y^{-1}}{3x^{1/2}y^{-4}}.$$

Simplify the numerical coefficient:

$$\frac{12}{3}=4.$$

For the powers of $x$,

$$\frac{x^{5/2}}{x^{1/2}} = x^{5/2-1/2} = x^2.$$

For the powers of $y$,

$$\frac{y^{-1}}{y^{-4}} = y^{-1-(-4)} = y^3.$$

Therefore,

$$\boxed{4x^2y^3}.$$

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To compare $2^{3/2}$ and $3$, square both positive quantities:

$$\left(2^{3/2}\right)^2=8$$

and

$$3^2=9.$$

Therefore,

$$\boxed{2^{3/2}<3}.$$

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Adding exponents across addition

The law

$$a^m a^n=a^{m+n}$$

does not imply

$$a^m+a^n=a^{m+n}.$$

Misusing a power of a power

The correct law is

$$(a^m)^n=a^{mn},$$

not $a^{m+n}$.

Forgetting the reciprocal

The correct interpretation is

$$a^{-n}=\frac{1}{a^n},$$

not $-a^n$.

Expanding too early

Expressions are often easier to simplify while they remain in index form.

Ignoring brackets

For example,

$$(a^3)^2=a^6,$$

whereas

$$a^{3^2}=a^9.$$

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For routine TMUA manipulations involving arbitrary rational exponents, the safest setting is

$$a>0.$$

This ensures that all required roots are real and that the index laws apply consistently.

There are important special cases:

• $a^0=1$ requires $a\neq0$;
• negative powers require a non-zero base;
• negative bases may work for some rational exponents but not for others;
• $0$ raised to a negative power is undefined.

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1. Rewrite numbers using common bases.
2. Stay in index form for as long as possible.
3. Simplify before calculating.
4. Convert fractional powers into roots.
5. Convert negative powers into reciprocals.
6. Look for cancellation.
7. Use brackets carefully.
8. Check that the base and denominator conditions are valid.
9. Avoid applying multiplication laws across addition.

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Simplify

$$\left(\frac{8x^{-3/2}}{y^{1/2}}\right)^{-2/3},$$

where $x>0$ and $y>0$.

Apply the outer exponent to each factor:

$$8^{-2/3} \left(x^{-3/2}\right)^{-2/3} \left(y^{-1/2}\right)^{-2/3}.$$

Now simplify each part:

$$8^{-2/3} = \frac{1}{8^{2/3}} = \frac{1}{\left(\sqrt[3]{8}\right)^2} = \frac{1}{4},$$

$$\left(x^{-3/2}\right)^{-2/3} = x^{(-3/2)(-2/3)} = x,$$

and

$$\left(y^{-1/2}\right)^{-2/3} = y^{1/3}.$$

Therefore,

$$\boxed{\frac{xy^{1/3}}{4}}.$$

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$$\boxed{a^m a^n=a^{m+n}}$$

$$\boxed{\frac{a^m}{a^n}=a^{m-n}}$$

$$\boxed{(a^m)^n=a^{mn}}$$

$$\boxed{a^0=1}$$

$$\boxed{a^{-n}=\frac{1}{a^n}}$$

$$\boxed{a^{1/n}=\sqrt[n]{a}}$$

$$\boxed{a^{m/n}=\sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m}$$

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