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Properties of Exponents: Worked Examples

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By Pimath, 20 April, 2026

A collection of worked exercises on the properties of exponents, designed to help you correctly apply the fundamental rules: product and quotient of powers with the same base, power of a power, power of a product and of a quotient, negative exponents, and zero exponents. Each exercise is solved step by step, with clear explanations of which property to use and why.

Exercise 1 — level ★☆☆☆☆

\[ 2^3 \cdot 2^4 \]

Answer

\[ 128 \]

Solution

Key idea

Both powers share the same base, \(2\). We apply the product rule for exponents: add the exponents and keep the base unchanged.

Rule applied

\[ a^m \cdot a^n = a^{m+n} \]

Identifying \(a\), \(m\), \(n\)

\[ a = 2 \qquad m = 3 \qquad n = 4 \]

Applying the rule

\[ 2^3 \cdot 2^4 = 2^{3+4} = 2^7 \]

Numerical computation

\[ 2^7 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 128 \]

Result

\[ \boxed{128} \]

Exercise 2 — level ★☆☆☆☆

\[ 5^6 \div 5^4 \]

Answer

\[ 25 \]

Solution

Key idea

Both powers share the same base, \(5\). We apply the quotient rule for exponents: subtract the exponent of the divisor from that of the dividend.

Rule applied

\[ \frac{a^m}{a^n} = a^{m-n} \qquad (a \neq 0) \]

Identifying \(a\), \(m\), \(n\)

\[ a = 5 \qquad m = 6 \qquad n = 4 \]

Applying the rule

\[ \frac{5^6}{5^4} = 5^{6-4} = 5^2 \]

Numerical computation

\[ 5^2 = 25 \]

Result

\[ \boxed{25} \]

Exercise 3 — level ★☆☆☆☆

\[ \left(3^2\right)^3 \]

Answer

\[ 729 \]

Solution

Key idea

A power is itself raised to an exponent. We apply the power of a power rule: multiply the two exponents.

Rule applied

\[ \left(a^m\right)^n = a^{m \cdot n} \]

Identifying \(a\), \(m\), \(n\)

\[ a = 3 \qquad m = 2 \qquad n = 3 \]

Applying the rule

\[ \left(3^2\right)^3 = 3^{2 \cdot 3} = 3^6 \]

Numerical computation

\[ 3^6 = 729 \]

Result

\[ \boxed{729} \]

Exercise 4 — level ★☆☆☆☆

\[ (2 \cdot 5)^3 \]

Answer

\[ 1000 \]

Solution

Key idea

A product of two factors is raised to an exponent. The power of a product rule allows us to distribute the exponent over each factor.

Rule applied

\[ (a \cdot b)^n = a^n \cdot b^n \]

Identifying \(a\), \(b\), \(n\)

\[ a = 2 \qquad b = 5 \qquad n = 3 \]

Applying the rule

\[ (2 \cdot 5)^3 = 2^3 \cdot 5^3 \]

Numerical computation

\[ 2^3 = 8 \qquad 5^3 = 125 \]

\[ 8 \cdot 125 = 1000 \]

Result

\[ \boxed{1000} \]

Exercise 5 — level ★☆☆☆☆

\[ \left(\frac{6}{3}\right)^4 \]

Answer

\[ 16 \]

Solution

Key idea

A quotient is raised to an exponent. We can apply the power of a quotient rule, or simplify the fraction first.

Method 1 — direct simplification

\[ \frac{6}{3} = 2 \qquad \Rightarrow \qquad \left(\frac{6}{3}\right)^4 = 2^4 = 16 \]

Method 2 — power of a quotient rule

\[ \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \]

\[ \left(\frac{6}{3}\right)^4 = \frac{6^4}{3^4} = \frac{1296}{81} = 16 \]

Result

\[ \boxed{16} \]

Exercise 6 — level ★★☆☆☆

\[ x^4 \cdot x^5 \]

Answer

\[ x^9 \]

Solution

Key idea

Same as Exercise 1, but with the literal base \(x\). Add the exponents.

Rule applied

\[ a^m \cdot a^n = a^{m+n} \]

Applying the rule

\[ x^4 \cdot x^5 = x^{4+5} = x^9 \]

Result

\[ \boxed{x^9} \]

Exercise 7 — level ★★☆☆☆

\[ \frac{x^9}{x^4} \]

Answer

\[ x^5 \]

Solution

Rule applied

\[ \frac{a^m}{a^n} = a^{m-n} \qquad (x \neq 0) \]

Applying the rule

\[ \frac{x^9}{x^4} = x^{9-4} = x^5 \]

Result

\[ \boxed{x^5} \]

Exercise 8 — level ★★☆☆☆

\[ \left(x^3\right)^5 \]

Answer

\[ x^{15} \]

Solution

Rule applied

\[ \left(a^m\right)^n = a^{m \cdot n} \]

Applying the rule

\[ \left(x^3\right)^5 = x^{3 \cdot 5} = x^{15} \]

Result

\[ \boxed{x^{15}} \]

Exercise 9 — level ★★☆☆☆

\[ (3x)^3 \]

Answer

\[ 27x^3 \]

Solution

Key idea

We apply the power of a product rule with \(a = 3\) and \(b = x\). Note: the exponent must be distributed over the numerical coefficient as well, not just the variable.

Rule applied

\[ (a \cdot b)^n = a^n \cdot b^n \]

Applying the rule

\[ (3x)^3 = 3^3 \cdot x^3 \]

Computation

\[ 3^3 = 27 \]

Result

\[ \boxed{27x^3} \]

⚠ Common mistake: writing \(3x^3\) instead of \(27x^3\), forgetting to raise the coefficient to the given power.

Exercise 10 — level ★★☆☆☆

\[ \left(\frac{x}{2}\right)^4 \]

Answer

\[ \dfrac{x^4}{16} \]

Solution

Rule applied

\[ \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \]

Applying the rule

\[ \left(\frac{x}{2}\right)^4 = \frac{x^4}{2^4} = \frac{x^4}{16} \]

Result

\[ \boxed{\dfrac{x^4}{16}} \]

Exercise 11 — level ★★☆☆☆

\[ 4^0 \cdot 7^2 \]

Answer

\[ 49 \]

Solution

Key idea

Any non-zero base raised to \(0\) equals \(1\). This follows from the quotient rule: \(a^m \div a^m = a^{m-m} = a^0\), and also \(a^m \div a^m = 1\).

Rule applied

\[ a^0 = 1 \qquad (a \neq 0) \]

Applying the rule

\[ 4^0 \cdot 7^2 = 1 \cdot 49 = 49 \]

Result

\[ \boxed{49} \]

Exercise 12 — level ★★☆☆☆

\[ 3^{-2} \]

Answer

\[ \dfrac{1}{9} \]

Solution

Key idea

A negative exponent indicates the reciprocal of the power with a positive exponent. It does not produce a negative result — the outcome is a fraction.

Rule applied

\[ a^{-n} = \frac{1}{a^n} \qquad (a \neq 0) \]

Applying the rule

\[ 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \]

Result

\[ \boxed{\dfrac{1}{9}} \]

⚠ Common mistake: thinking that \(3^{-2} = -9\). A negative exponent does not change the sign of the result.

Exercise 13 — level ★★★☆☆

\[ x^{-3} \cdot x^7 \]

Answer

\[ x^4 \]

Solution

Key idea

The product rule applies even when one exponent is negative: the rule is identical — add the exponents algebraically.

Rule applied

\[ a^m \cdot a^n = a^{m+n} \]

Applying the rule

\[ x^{-3} \cdot x^7 = x^{-3+7} = x^4 \]

Result

\[ \boxed{x^4} \]

Exercise 14 — level ★★★☆☆

\[ \left(x^2 y^3\right)^4 \]

Answer

\[ x^8\, y^{12} \]

Solution

Key idea

The product \(x^2 y^3\) is raised to the fourth power. Distribute the exponent over each factor, then apply the power of a power rule to each.

Rules applied

\[ (ab)^n = a^n b^n \qquad \text{and} \qquad (a^m)^n = a^{mn} \]

Applying the rule

\[ \left(x^2 y^3\right)^4 = \left(x^2\right)^4 \cdot \left(y^3\right)^4 = x^{2 \cdot 4} \cdot y^{3 \cdot 4} = x^8 y^{12} \]

Result

\[ \boxed{x^8\, y^{12}} \]

Exercise 15 — level ★★★☆☆

\[ \left(\frac{2}{x}\right)^{-3} \]

Answer

\[ \dfrac{x^3}{8} \]

Solution

Key idea

A quotient raised to a negative exponent equals the reciprocal quotient raised to the corresponding positive exponent. Swap numerator and denominator, then raise both to \(3\).

Rule applied

\[ \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n} \]

Applying the rule

\[ \left(\frac{2}{x}\right)^{-3} = \left(\frac{x}{2}\right)^{3} = \frac{x^3}{2^3} = \frac{x^3}{8} \]

Result

\[ \boxed{\dfrac{x^3}{8}} \]

Exercise 16— level ★★★☆☆

\[ 25^{\,1/2} \]

Answer

\[ 5 \]

Solution

Key idea

An exponent of the form \(\tfrac{1}{q}\) denotes the \(q\)-th root. In particular, \(a^{1/2} = \sqrt{a}\).

Rule applied

\[ a^{1/q} = \sqrt[q]{a} \]

Applying the rule

\[ 25^{1/2} = \sqrt{25} = 5 \]

Verification

\[ 5^2 = 25 \checkmark \]

Result

\[ \boxed{5} \]

Exercise 17 — level ★★★☆☆

\[ 8^{\,1/3} \]

Answer

\[ 2 \]

Solution

Rule applied

\[ a^{1/3} = \sqrt[3]{a} \]

Applying the rule

\[ 8^{1/3} = \sqrt[3]{8} = 2 \]

Verification

\[ 2^3 = 8 \checkmark \]

Result

\[ \boxed{2} \]

Exercise 18 — level ★★★☆☆

\[ x^{2/3} \cdot x^{1/3} \]

Answer

\[ x \]

Solution

Key idea

The product rule applies to fractional exponents as well. Add the fractions using a common denominator.

Rule applied

\[ a^m \cdot a^n = a^{m+n} \]

Sum of the exponents

\[ \frac{2}{3} + \frac{1}{3} = \frac{3}{3} = 1 \]

Result

\[ x^{2/3} \cdot x^{1/3} = x^1 = x \]

\[ \boxed{x} \]

Exercise 19 — level ★★★☆☆

\[ 27^{2/3} \]

Answer

\[ 9 \]

Solution

Key idea

An exponent \(\tfrac{p}{q}\) denotes the \(q\)-th root of the base raised to the \(p\)-th power. It is advisable to extract the root first, then raise to the power: the numbers stay smaller and easier to manage.

Rule applied

\[ a^{p/q} = \left(\sqrt[q]{a}\right)^p = \sqrt[q]{a^p} \]

Application — root first, then power

\[ 27^{2/3} = \left(27^{1/3}\right)^2 = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9 \]

Verification — alternative method

\[ 27^{2/3} = \left(27^2\right)^{1/3} = 729^{1/3} = \sqrt[3]{729} = 9 \checkmark \]

Result

\[ \boxed{9} \]

Exercise 20 — level ★★★☆☆

\[ \left(x^{-2}\right)^3 \cdot x^9 \]

Answer

\[ x^3 \]

Solution

Key idea

We proceed in two steps: first simplify the power of a power, then multiply using the product rule for the same base.

Step 1 — power of a power

\[ \left(x^{-2}\right)^3 = x^{(-2)\cdot 3} = x^{-6} \]

Step 2 — product with the same base

\[ x^{-6} \cdot x^9 = x^{-6+9} = x^3 \]

Result

\[ \boxed{x^3} \]

Exercise 21 — level ★★★★☆

\[ \frac{(4x^3)^2}{(2x^2)^3} \]

Answer

\[ 2 \]

Solution

Key idea

Numerator and denominator are expanded separately by distributing the outer exponent; then the quotient is simplified.

Expanding the numerator

\[ (4x^3)^2 = 4^2 \cdot (x^3)^2 = 16x^6 \]

Expanding the denominator

\[ (2x^2)^3 = 2^3 \cdot (x^2)^3 = 8x^6 \]

Quotient

\[ \frac{16x^6}{8x^6} = \frac{16}{8} \cdot \frac{x^6}{x^6} = 2 \cdot 1 = 2 \]

Result

\[ \boxed{2} \]

Exercise 22 — level ★★★★☆

\[ \frac{\left(a^2 b^3\right)^4}{a^5\, b^7} \]

Answer

\[ a^3\, b^5 \]

Solution

Expanding the numerator

\[ \left(a^2 b^3\right)^4 = a^{2 \cdot 4} \cdot b^{3 \cdot 4} = a^8 b^{12} \]

Quotient — subtract the exponents for each base

\[ \frac{a^8 b^{12}}{a^5 b^7} = a^{8-5} \cdot b^{12-7} = a^3 b^5 \]

Result

\[ \boxed{a^3\, b^5} \]

Exercise 23 — level ★★★★☆

\[ \frac{(2x^3)^4}{(4x^2)^3} \]

Answer

\[ \dfrac{x^6}{4} \]

Solution

Expanding the numerator

\[ (2x^3)^4 = 2^4 \cdot x^{12} = 16x^{12} \]

Expanding the denominator

\[ (4x^2)^3 = 4^3 \cdot x^6 = 64x^6 \]

Quotient

\[ \frac{16x^{12}}{64x^6} = \frac{16}{64} \cdot x^{12-6} = \frac{1}{4}\, x^6 = \frac{x^6}{4} \]

Result

\[ \boxed{\dfrac{x^6}{4}} \]

Exercise 24 — level ★★★★☆

\[ \left(a^{1/2} \cdot b^{1/3}\right)^6 \]

Answer

\[ a^3\, b^2 \]

Solution

Rules applied

\[ (ab)^n = a^n b^n \qquad \text{and} \qquad (a^m)^n = a^{mn} \]

Distributing the exponent 6

\[ \left(a^{1/2} \cdot b^{1/3}\right)^6 = \left(a^{1/2}\right)^6 \cdot \left(b^{1/3}\right)^6 \]

Power of a power

\[ \left(a^{1/2}\right)^6 = a^{\,\frac{1}{2} \cdot 6} = a^3 \]

\[ \left(b^{1/3}\right)^6 = b^{\,\frac{1}{3} \cdot 6} = b^2 \]

Result

\[ \boxed{a^3\, b^2} \]

Exercise 25 — level ★★★★☆

\[ \left(\frac{x^2}{y^3}\right)^{-2} \]

Answer

\[ \dfrac{y^6}{x^4} \]

Solution

Key idea

A quotient raised to a negative exponent equals the reciprocal quotient raised to the corresponding positive exponent. Swap numerator and denominator, then raise both to \(3\).

Inverting for the negative exponent

\[ \left(\frac{x^2}{y^3}\right)^{-2} = \left(\frac{y^3}{x^2}\right)^{2} \]

Power of a quotient

\[ \left(\frac{y^3}{x^2}\right)^{2} = \frac{(y^3)^2}{(x^2)^2} = \frac{y^6}{x^4} \]

Result

\[ \boxed{\dfrac{y^6}{x^4}} \]

Exercise 26 — level ★★★★☆

\[ \frac{2^n \cdot 4^n}{8^n} \]

Answer

\[ 1 \]

Solution

Key idea

All bases (\(2\), \(4\), \(8\)) are powers of \(2\). Rewrite everything in base \(2\), then apply the product and quotient rules.

Rewriting in base 2

\[ 4^n = (2^2)^n = 2^{2n} \qquad 8^n = (2^3)^n = 2^{3n} \]

Substitution

\[ \frac{2^n \cdot 2^{2n}}{2^{3n}} = \frac{2^{n + 2n}}{2^{3n}} = \frac{2^{3n}}{2^{3n}} = 2^0 = 1 \]

Result

\[ \boxed{1} \]

Exercise 27 — level ★★★★☆

\[ \frac{(3x^2)^3 \cdot (2x)^2}{(6x^4)^2} \]

Answer

\[ 3 \]

Solution

Expanding the numerator — first factor

\[ (3x^2)^3 = 27x^6 \]

Expanding the numerator — second factor

\[ (2x)^2 = 4x^2 \]

Product in the numerator

\[ 27x^6 \cdot 4x^2 = 108\, x^8 \]

Expanding the denominator

\[ (6x^4)^2 = 36x^8 \]

Final quotient

\[ \frac{108\, x^8}{36\, x^8} = \frac{108}{36} \cdot x^0 = 3 \cdot 1 = 3 \]

Result

\[ \boxed{3} \]

Exercise 28 — level ★★★★★

\[ \frac{a^{m+n} \cdot a^{m-n}}{\left(a^m\right)^2} \]

Answer

\[ 1 \]

Solution

Key idea

Numerator and denominator both reduce to the same power of \(a\) via the product rule, the power of a power rule, and the quotient rule. The identity holds for any values of \(m\) and \(n\).

Simplifying the numerator

\[ a^{m+n} \cdot a^{m-n} = a^{(m+n)+(m-n)} = a^{2m} \]

Simplifying the denominator

\[ \left(a^m\right)^2 = a^{2m} \]

Quotient

\[ \frac{a^{2m}}{a^{2m}} = a^{2m - 2m} = a^0 = 1 \]

Result

\[ \boxed{1} \]

This identity reveals that \((m{+}n)\) and \((m{-}n)\) are complementary with respect to \(2m\): their sum is always \(2m\), regardless of \(n\).

Exercise 29 — level ★★★★★

\[ \frac{3^{n+2} - 3^{n+1}}{2 \cdot 3^n} \]

Answer

\[ 3 \]

Solution

Key idea

The numerator contains two powers of \(3\) with consecutive parametric exponents. Factor out \(3^n\) from the numerator, then simplify with the denominator.

Rewriting the exponents in the numerator

\[ 3^{n+2} = 3^n \cdot 3^2 = 9 \cdot 3^n \]

\[ 3^{n+1} = 3^n \cdot 3^1 = 3 \cdot 3^n \]

Factoring out \(3^n\)

\[ 3^{n+2} - 3^{n+1} = 9 \cdot 3^n - 3 \cdot 3^n = 3^n(9 - 3) = 6 \cdot 3^n \]

Quotient

\[ \frac{6 \cdot 3^n}{2 \cdot 3^n} = \frac{6}{2} = 3 \]

Result

\[ \boxed{3} \]

Factoring out \(3^n\) in the numerator is the key step: without it, the expression would seem impossible to simplify.

Exercise 30 — level ★★★★★

\[ \frac{x^{a+b} \cdot x^{b+c} \cdot x^{c+a}}{\left(x^a \cdot x^b \cdot x^c\right)^2} \]

Answer

\[ 1 \]

Solution

Key idea

Both numerator and denominator reduce to a single power of \(x\) with an exponent in terms of \(a\), \(b\), \(c\). The identity holds for any real values of \(a\), \(b\), \(c\) (with \(x \neq 0\)).

Simplifying the numerator

We use the product rule, summing all exponents:

\[ x^{a+b} \cdot x^{b+c} \cdot x^{c+a} = x^{(a+b)+(b+c)+(c+a)} \]

Sum of the exponents:

\[ (a+b)+(b+c)+(c+a) = 2a + 2b + 2c = 2(a+b+c) \]

Therefore the numerator equals \(x^{2(a+b+c)}\).

Simplifying the denominator

First reduce the inner product, then square:

\[ x^a \cdot x^b \cdot x^c = x^{a+b+c} \]

\[ \left(x^{a+b+c}\right)^2 = x^{2(a+b+c)} \]

Quotient

\[ \frac{x^{2(a+b+c)}}{x^{2(a+b+c)}} = x^0 = 1 \]

Result

\[ \boxed{1} \]

The symmetry of the expression is the key: the pairs \((a{+}b)\), \((b{+}c)\), \((c{+}a)\) in the numerator each cover every variable exactly twice — exactly as the square of \(x^{a+b+c}\) in the denominator.

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