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C2.4 Indices II: Positive, Zero and Negative Powers

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01 — Explanation

What Happens When I Multiply or Divide Powers of the Same Base?

Index laws avoid writing out repeated multiplication in full. Whenever the base is the same, multiplying combines the powers by addition, and dividing combines them by subtraction:

x^m \times x^n = x^{m+n} \qquad\qquad x^m \div x^n = x^{m-n}

5x^4 \times 3x^6 = (5\times3)\,x^{4+6} = 15x^{10} \qquad\qquad \frac{24a^8}{6a^3} = (24\div6)\,a^{8-3} = 4a^5

Coefficients multiply or divide as ordinary numbers — only the powers of $x$ get added or subtracted. It's worth checking these rules make sense with tiny numbers: $x^2 \times x^3$ really is $(x\times x)\times(x\times x\times x) = x^5$ , five $x$ 's multiplied together in total, which is exactly $x^{2+3}$ .

02 — Explanation

What Happens When a Power Is Raised to Another Power?

Raise the coefficient to the outer power as normal, and multiply the two powers together:

(x^m)^n = x^{mn} \qquad\qquad (2k^5)^3 = 2^3 \times k^{5\times3} = 8k^{15}

Exam Tip

Don't confuse this with the multiplication rule above — $(x^2)^3$ means $x^2$ used three times as a factor, so the powers multiply ( $x^6$ ), whereas $x^2 \times x^3$ combines two separate terms, so the powers add ( $x^5$ ). Reading the expression carefully before choosing a rule avoids mixing the two up.

03 — Explanation

What Does a Power of Zero Actually Mean?

Any non-zero number raised to the power $0$ is always exactly $1$ — never $0$ :

x^0 = 1 \quad (x\neq 0) \qquad\qquad 8^0 = 1

This follows directly from the division rule: $\dfrac{x^3}{x^3} = x^{3-3} = x^0$ , but any number divided by itself is also just $1$ — so $x^0$ must equal $1$ for the rule to stay consistent.

04 — Explanation

What Does a Negative Power Mean?

A negative index is an instruction to take the reciprocal of the positive power — it never produces a negative number:

x^{-n} = \frac{1}{x^n} \qquad\qquad 2^{-3} = \frac{1}{2^3} = \frac{1}{8}

Exam Tip

$2^{-3}$ is a common trap answer of $-8$ . Read a negative index as "flip it and make the power positive", not "make the answer negative".

The same idea works in reverse for a fraction base: $\left(\dfrac{2}{3}\right)^{-1} = \dfrac{3}{2}$ , since taking the reciprocal of a fraction simply swaps the numerator and denominator.

05 — Explanation

How Do I Solve an Equation With Indices on Both Sides?

If both sides can be written as powers of the same base, the powers themselves must be equal — this turns an index equation into a simple linear one.

Given $2^{2x} = 32$ , find $x$ .

Rewrite $32$ as a power of $2$ : $32 = 2^5$ .
The bases now match, so the powers must be equal: $2x = 5$ .
Solve: $x = 2.5$ .

06 — Explanation

Exam-Style Question 1

Simplify $\dfrac{y^3}{y^7}$ , giving your answer as a single power of $y$ .

Solution

Subtract the powers: $y^3 \div y^7 = y^{3-7} = y^{-4}$ .

The answer is allowed to have a negative power — "simplify" doesn't mean the power has to be positive, only that it's written as a single term.

07 — Explanation

Exam-Style Question 2

Simplify $\dfrac{(3x^2)^3}{9x^4}$ .

Solution

Apply the power to everything inside the bracket first: $(3x^2)^3 = 3^3 \times x^{2\times3} = 27x^6$ .
Divide the coefficients and subtract the powers of $x$ :

\frac{27x^6}{9x^4} = (27\div9)\,x^{6-4} = 3x^2

Working out $(3x^2)^3$ completely before dividing avoids trying to combine the bracket's power with the denominator's power in a single confused step.