A progressive collection — from basic numerical examples to parametric identities — designed to master the fundamental laws of exponents. Each exercise highlights the relevant rule, provides a step-by-step application, and concludes with the final result.
Exercise of 19/04/2026 - 09:00 — level ★☆☆☆☆
\[ 2^3 \cdot 2^4 \]
Result
\[ 128 \]
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 of 19/04/2026 - 09:10 — level ★☆☆☆☆
\[ 5^6 \div 5^4 \]
Result
\[ 25 \]
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 of 19/04/2026 - 09:20 — level ★☆☆☆☆
\[ \left(3^2\right)^3 \]
Result
\[ 729 \]
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 of 19/04/2026 - 09:30 — level ★☆☆☆☆
\[ (2 \cdot 5)^3 \]
Result
\[ 1000 \]
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 of 19/04/2026 - 09:40 — level ★☆☆☆☆
\[ \left(\frac{6}{3}\right)^4 \]
Result
\[ 16 \]
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 of 19/04/2026 - 09:50 — level ★★☆☆☆
\[ x^4 \cdot x^5 \]
Result
\[ x^9 \]
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 of 19/04/2026 - 10:00 — level ★★☆☆☆
\[ \frac{x^9}{x^4} \]
Result
\[ x^5 \]
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 of 19/04/2026 - 10:10 — level ★★☆☆☆
\[ \left(x^3\right)^5 \]
Result
\[ x^{15} \]
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 of 19/04/2026 - 10:20 — level ★★☆☆☆
\[ (3x)^3 \]
Result
\[ 27x^3 \]
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
\[ \boxed{27x^3} \]
Exercise of 19/04/2026 - 10:30 — level ★★☆☆☆
\[ \left(\frac{x}{2}\right)^4 \]
Result
\[ \dfrac{x^4}{16} \]
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 of 19/04/2026 - 10:40 — level ★★☆☆☆
\[ 4^0 \cdot 7^2 \]
Result
\[ 49 \]
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 of 19/04/2026 - 10:50 — level ★★☆☆☆
\[ 3^{-2} \]
Result
\[ \dfrac{1}{9} \]
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
\[ \boxed{\dfrac{1}{9}} \]
Exercise of 19/04/2026 - 11:00 — level ★★★☆☆
\[ x^{-3} \cdot x^7 \]
Result
\[ x^4 \]
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 of 19/04/2026 - 11:10 — level ★★★☆☆
\[ \left(x^2 y^3\right)^4 \]
Result
\[ x^8\, y^{12} \]
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 of 19/04/2026 - 11:20 — level ★★★☆☆
\[ \left(\frac{2}{x}\right)^{-3} \]
Result
\[ \dfrac{x^3}{8} \]
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 of 19/04/2026 - 11:30 — level ★★★☆☆
\[ 25^{\,1/2} \]
Result
\[ 5 \]
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 of 19/04/2026 - 11:40 — level ★★★☆☆
\[ 8^{\,1/3} \]
Result
\[ 2 \]
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 of 19/04/2026 - 11:50 — level ★★★☆☆
\[ x^{2/3} \cdot x^{1/3} \]
Result
\[ x \]
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 of 19/04/2026 - 12:00 — level ★★★☆☆
\[ 27^{2/3} \]
Result
\[ 9 \]
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 of 19/04/2026 - 12:10 — level ★★★☆☆
\[ \left(x^{-2}\right)^3 \cdot x^9 \]
Result
\[ x^3 \]
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 of 19/04/2026 - 12:20 — level ★★★★☆
\[ \frac{(4x^3)^2}{(2x^2)^3} \]
Result
\[ 2 \]
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 of 19/04/2026 - 12:30 — level ★★★★☆
\[ \frac{\left(a^2 b^3\right)^4}{a^5\, b^7} \]
Result
\[ a^3\, b^5 \]
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 of 19/04/2026 - 12:40 — level ★★★★☆
\[ \frac{(2x^3)^4}{(4x^2)^3} \]
Result
\[ \dfrac{x^6}{4} \]
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 of 19/04/2026 - 12:50 — level ★★★★☆
\[ \left(a^{1/2} \cdot b^{1/3}\right)^6 \]
Result
\[ a^3\, b^2 \]
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 of 19/04/2026 - 13:00 — level ★★★★☆
\[ \left(\frac{x^2}{y^3}\right)^{-2} \]
Result
\[ \dfrac{y^6}{x^4} \]
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 of 19/04/2026 - 13:10 — level ★★★★☆
\[ \frac{2^n \cdot 4^n}{8^n} \]
Result
\[ 1 \]
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 of 19/04/2026 - 13:20 — level ★★★★☆
\[ \frac{(3x^2)^3 \cdot (2x)^2}{(6x^4)^2} \]
Result
\[ 3 \]
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 of 19/04/2026 - 13:30 — level ★★★★★
\[ \frac{a^{m+n} \cdot a^{m-n}}{\left(a^m\right)^2} \]
Result
\[ 1 \]
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
\[ \boxed{1} \]
Exercise of 19/04/2026 - 13:45 — level ★★★★★
\[ \frac{3^{n+2} - 3^{n+1}}{2 \cdot 3^n} \]
Result
\[ 3 \]
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
\[ \boxed{3} \]
Exercise of 19/04/2026 - 14:00 — level ★★★★★
\[ \frac{x^{a+b} \cdot x^{b+c} \cdot x^{c+a}}{\left(x^a \cdot x^b \cdot x^c\right)^2} \]
Result
\[ 1 \]
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
\[ \boxed{1} \]