Guided exercises on powers and exponential equations: a collection of solved problems to review together the properties of bases, negative exponents, and roots. You will find simple examples and clear steps to learn how to solve exponential equations, with explanations designed to calmly accompany you toward more difficult levels.
Exercise of 06/04/2026 - 09:00 — level ★☆☆☆☆
\[ 2^5 \]
Result
\[ 32 \]
Solution
Solving idea
Calculating a power means multiplying the base by itself as many times as the exponent indicates.
Calculation
\[ 2^5 = 2\cdot2\cdot2\cdot2\cdot2 = 32 \]
Result
\[ \boxed{32} \]
Exercise of 06/04/2026 - 09:10 — level ★☆☆☆☆
\[ (-3)^3 \]
Result
\[ -27 \]
Solution
Solving idea
A power with a negative base and an odd exponent is negative.
Calculation
\[ (-3)^3 = (-3)\cdot(-3)\cdot(-3) = 9\cdot(-3) = -27 \]
Result
\[ \boxed{-27} \]
Exercise of 06/04/2026 - 09:20 — level ★☆☆☆☆
\[ 7^0 \]
Result
\[ 1 \]
Solution
Solving idea
Any non-zero base raised to the zero exponent is equal to \(1\), by convention.
Rule
\[ a^0 = 1 \quad \text{for every } a \neq 0 \]
Result
\[ \boxed{1} \]
Exercise of 06/04/2026 - 09:30 — level ★☆☆☆☆
\[ (-1)^4 \]
Result
\[ 1 \]
Solution
Solving idea
A power with a negative base and an even exponent is always positive.
Calculation
\[ (-1)^4 = (-1)^2\cdot(-1)^2 = 1\cdot1 = 1 \]
Result
\[ \boxed{1} \]
Exercise of 06/04/2026 - 09:40 — level ★★☆☆☆
\[ 2^{-4} \]
Result
\[ \dfrac{1}{16} \]
Solution
Solving idea
A negative exponent indicates the reciprocal of the power with a positive exponent: \(a^{-n}=\dfrac{1}{a^n}\).
Calculation
\[ 2^{-4} = \frac{1}{2^4} = \frac{1}{16} \]
Result
\[ \boxed{\dfrac{1}{16}} \]
Exercise of 06/04/2026 - 09:50 — level ★★☆☆☆
\[ \left(\frac{3}{2}\right)^3 \]
Result
\[ \dfrac{27}{8} \]
Solution
Solving idea
The power of a fraction is the fraction of the powers: \(\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}\).
Calculation
\[ \left(\frac{3}{2}\right)^3 = \frac{3^3}{2^3} = \frac{27}{8} \]
Result
\[ \boxed{\dfrac{27}{8}} \]
Exercise of 06/04/2026 - 10:00 — level ★★☆☆☆
\[ \left(\frac{1}{2}\right)^{-3} \]
Result
\[ 8 \]
Solution
Solving idea
Negative exponent on a fraction: invert the fraction and apply the positive exponent.
Calculation
\[ \left(\frac{1}{2}\right)^{-3} = \left(\frac{2}{1}\right)^3 = 2^3 = 8 \]
Result
\[ \boxed{8} \]
Exercise of 06/04/2026 - 10:10 — level ★★☆☆☆
\[ 2^3 \cdot 2^5 \]
Result
\[ 256 \]
Solution
Solving idea
The product of powers with the same base is calculated by adding the exponents: \(a^m\cdot a^n = a^{m+n}\).
Calculation
\[ 2^3\cdot2^5 = 2^{3+5} = 2^8 = 256 \]
Result
\[ \boxed{256} \]
Exercise of 06/04/2026 - 10:25 — level ★★☆☆☆
\[ \frac{5^6}{5^3} \]
Result
\[ 125 \]
Solution
Solving idea
The quotient of powers with the same base is calculated by subtracting the exponents: \(a^m / a^n = a^{m-n}\).
Calculation
\[ \frac{5^6}{5^3} = 5^{6-3} = 5^3 = 125 \]
Result
\[ \boxed{125} \]
Exercise of 06/04/2026 - 10:40 — level ★★☆☆☆
\[ (3^2)^3 \]
Result
\[ 729 \]
Solution
Solving idea
The power of a power is calculated by multiplying the exponents: \((a^m)^n = a^{m\cdot n}\).
Calculation
\[ (3^2)^3 = 3^{2\cdot3} = 3^6 = 729 \]
Result
\[ \boxed{729} \]
Exercise of 06/04/2026 - 10:55 — level ★★☆☆☆
\[ 9^{1/2} \]
Result
\[ 3 \]
Solution
Solving idea
A fractional exponent \(\tfrac{1}{n}\) corresponds to the \(n\)-th root: \(a^{1/n}=\sqrt[n]{a}\).
Calculation
\[ 9^{1/2} = \sqrt{9} = 3 \]
Result
\[ \boxed{3} \]
Exercise of 06/04/2026 - 11:10 — level ★★☆☆☆
\[ 8^{2/3} \]
Result
\[ 4 \]
Solution
Solving idea
\(a^{m/n} = (\sqrt[n]{a})^m\). First calculate the cube root, then square the result.
Calculation
\[ 8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4 \]
Result
\[ \boxed{4} \]
Exercise of 06/04/2026 - 11:25 — level ★★★☆☆
\[ 27^{2/3} \]
Result
\[ 9 \]
Solution
Solving idea
First calculate the cube root of \(27\), then square the result.
Calculation
\[ 27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9 \]
Result
\[ \boxed{9} \]
Exercise of 06/04/2026 - 11:40 — level ★★★☆☆
\[ \left(\frac{4}{9}\right)^{3/2} \]
Result
\[ \dfrac{8}{27} \]
Solution
Solving idea
Apply the fractional exponent separately to the numerator and denominator.
Calculation
\[ \left(\frac{4}{9}\right)^{3/2} = \frac{4^{3/2}}{9^{3/2}} = \frac{(\sqrt{4})^3}{(\sqrt{9})^3} = \frac{2^3}{3^3} = \frac{8}{27} \]
Result
\[ \boxed{\dfrac{8}{27}} \]
Exercise of 06/04/2026 - 11:55 — level ★★★☆☆
\[ 2^x \cdot 4 = 32 \]
Result
\[ x = 3 \]
Solution
Solving idea
Rewrite all powers with base \(2\), then equate the exponents.
Rewrite in base 2
\[ 2^x \cdot 2^2 = 2^5 \implies 2^{x+2} = 2^5 \]
Equation of the exponents
\[ x+2=5 \implies x=3 \]
Verification
\[ 2^3\cdot4=8\cdot4=32 \]
Result
\[ \boxed{x=3} \]
Exercise of 06/04/2026 - 12:10 — level ★★★☆☆
\[ 3^{x+1} = 27 \]
Result
\[ x = 2 \]
Solution
Rewrite in base 3
\[ 3^{x+1} = 3^3 \]
Equation of the exponents
\[ x+1=3 \implies x=2 \]
Verification
\[ 3^3=27 \]
Result
\[ \boxed{x=2} \]
Exercise of 06/04/2026 - 12:25 — level ★★★☆☆
\[ 4^x = 8 \]
Result
\[ x = \dfrac{3}{2} \]
Solution
Rewrite in base 2
\[ (2^2)^x = 2^3 \implies 2^{2x} = 2^3 \]
Equation of the exponents
\[ 2x=3 \implies x=\frac{3}{2} \]
Verification
\[ 4^{3/2}=(\sqrt{4})^3=2^3=8 \]
Result
\[ \boxed{x=\dfrac{3}{2}} \]
Exercise of 06/04/2026 - 12:40 — level ★★★☆☆
\[ a^3 \cdot a^{-2} \]
Result
\[ a \]
Solution
Applying the product rule
\[ a^3\cdot a^{-2} = a^{3+(-2)} = a^1 = a \]
Result
\[ \boxed{a} \]
Exercise of 06/04/2026 - 12:55 — level ★★★☆☆
\[ (x^2 y^3)^3 \]
Result
\[ x^6 y^9 \]
Solution
Solving idea
The power of a product is distributed over each factor.
Calculation
\[ (x^2y^3)^3 = (x^2)^3\cdot(y^3)^3 = x^6\cdot y^9 \]
Result
\[ \boxed{x^6 y^9} \]
Exercise of 06/04/2026 - 13:10 — level ★★★☆☆
\[ (2a^2b)^3 \]
Result
\[ 8a^6 b^3 \]
Solution
Calculation
\[ (2a^2b)^3 = 2^3\cdot(a^2)^3\cdot b^3 = 8a^6b^3 \]
Result
\[ \boxed{8a^6 b^3} \]
Exercise of 06/04/2026 - 13:25 — level ★★★☆☆
\[ (a^2 b^{-1})^2 \cdot (a^{-1}b)^3 \]
Result
\[ ab \]
Solution
Development of individual powers
\[ (a^2b^{-1})^2 = a^4b^{-2} \qquad (a^{-1}b)^3 = a^{-3}b^3 \]
Product
\[ a^4b^{-2}\cdot a^{-3}b^3 = a^{4-3}\cdot b^{-2+3} = a\cdot b = ab \]
Result
\[ \boxed{ab} \]
Exercise of 06/04/2026 - 13:40 — level ★★★☆☆
\[ \sqrt{2^8} \]
Result
\[ 16 \]
Solution
Solving idea
The square root is the power with exponent \(\tfrac{1}{2}\).
Calculation
\[ \sqrt{2^8} = (2^8)^{1/2} = 2^{8\cdot\frac{1}{2}} = 2^4 = 16 \]
Result
\[ \boxed{16} \]
Exercise of 06/04/2026 - 14:00 — level ★★★★☆
\[ 2^{x+1} + 2^x = 48 \]
Result
\[ x = 4 \]
Solution
Solving idea
Factor out \(2^x\) as a common factor in the first member.
Factoring
\[ 2\cdot2^x + 2^x = 2^x(2+1) = 3\cdot2^x = 48 \]
Resulting equation
\[ 2^x = 16 = 2^4 \implies x=4 \]
Verification
\[ 2^5+2^4=32+16=48 \]
Result
\[ \boxed{x=4} \]
Exercise of 06/04/2026 - 14:15 — level ★★★★☆
\[ 3^{2x} - 4\cdot3^x + 3 = 0 \]
Result
\[ x = 0 \quad \text{or} \quad x = 1 \]
Solution
Solving idea
Observe that \(3^{2x}=(3^x)^2\). The substitution \(t=3^x\) transforms the equation into a quadratic equation.
Substitution \(t=3^x\)
\[ t^2-4t+3=0 \implies (t-1)(t-3)=0 \]
Solution of the two equations
\(t=1\): \(3^x=1=3^0 \implies x=0\)
\(t=3\): \(3^x=3^1 \implies x=1\)
Verification
\(x=0\): \(1-4+3=0\) \(x=1\): \(9-12+3=0\)
Result
\[ \boxed{x=0 \quad \text{or} \quad x = 1} \]
Exercise of 06/04/2026 - 14:30 — level ★★★★☆
\[ 4^x - 2^{x+1} - 8 = 0 \]
Result
\[ x = 2 \]
Solution
Solving idea
Rewrite \(4^x=(2^x)^2\) and \(2^{x+1}=2\cdot2^x\), then let \(t=2^x\).
Equation in \(t\)
\[ t^2-2t-8=0 \implies (t-4)(t+2)=0 \]
Solutions
\(t=4\): \(2^x=4=2^2 \implies x=2\)
\(t=-2\): discarded since \(2^x>0\) always.
Verification
\[ 4^2-2^3-8=16-8-8=0 \]
Result
\[ \boxed{x=2} \]
Exercise of 06/04/2026 - 14:45 — level ★★★★☆
\[ \frac{2^n \cdot 4^{n+1}}{8^n} \]
Result
\[ 4 \]
Solution
Solving idea
Rewrite all powers with base \(2\).
Rewrite in base 2
\[ 4^{n+1}=2^{2(n+1)}=2^{2n+2} \qquad 8^n=2^{3n} \]
Simplification
\[ \frac{2^n\cdot2^{2n+2}}{2^{3n}} = \frac{2^{3n+2}}{2^{3n}} = 2^2 = 4 \]
Result
\[ \boxed{4} \]
Exercise of 06/04/2026 - 15:00 — level ★★★★☆
\[ (\sqrt{2}+1)^0 + (\sqrt{3})^4 \]
Result
\[ 10 \]
Solution
Calculation of the two terms
First term: any non-zero base raised to the zero exponent is equal to \(1\):
\[ (\sqrt{2}+1)^0 = 1 \]
Second term:
\[ (\sqrt{3})^4 = \left[(\sqrt{3})^2\right]^2 = 3^2 = 9 \]
Sum
\[ 1+9=10 \]
Result
\[ \boxed{10} \]
Exercise of 06/04/2026 - 15:15 — level ★★★★☆
\[ 2^{1/2} \cdot 2^{3/2} \]
Result
\[ 4 \]
Solution
Product rule
\[ 2^{1/2}\cdot2^{3/2}=2^{1/2+3/2}=2^{4/2}=2^2=4 \]
Result
\[ \boxed{4} \]
Exercise of 06/04/2026 - 15:30 — level ★★★★☆
\[ \left(3^{1/3}\right)^6 \]
Result
\[ 9 \]
Solution
Power of a power
\[ \left(3^{1/3}\right)^6 = 3^{(1/3)\cdot6} = 3^2 = 9 \]
Result
\[ \boxed{9} \]
Exercise of 06/04/2026 - 15:45 — level ★★★★☆
\[ 5^x \cdot 5^{x+1} = 5^4 \]
Result
\[ x = \dfrac{3}{2} \]
Solution
Product rule
\[ 5^x\cdot5^{x+1}=5^{x+(x+1)}=5^{2x+1}=5^4 \]
Equation of the exponents
\[ 2x+1=4 \implies x=\frac{3}{2} \]
Verification
\[ 5^{3/2}\cdot5^{5/2}=5^{3/2+5/2}=5^4 \]
Result
\[ \boxed{x=\dfrac{3}{2}} \]
Exercise of 06/04/2026 - 16:00 — level ★★★★☆
\[ a^{1/2} \cdot a^{1/3} \]
Result
\[ a^{5/6} \]
Solution
Product rule with fractional exponents
\[ a^{1/2}\cdot a^{1/3}=a^{1/2+1/3} \]
Sum of fractions
\[ \frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6} \]
Result
\[ \boxed{a^{5/6}} \]
Exercise of 06/04/2026 - 16:15 — level ★★★★☆
\[ \frac{4^{n+1}-4^n}{3\cdot4^{n-1}} \]
Result
\[ 4 \]
Solution
Factoring \(4^n\) in the numerator
\[ \frac{4^n(4-1)}{3\cdot4^{n-1}}=\frac{3\cdot4^n}{3\cdot4^{n-1}} \]
Simplification
\[ \frac{4^n}{4^{n-1}}=4^{n-(n-1)}=4^1=4 \]
Result
\[ \boxed{4} \]
Exercise of 06/04/2026 - 16:30 — level ★★★★☆
\[ 9^x = 3^{x+2} \]
Result
\[ x = 2 \]
Solution
Rewrite in base 3
\[ (3^2)^x=3^{x+2} \implies 3^{2x}=3^{x+2} \]
Equation of the exponents
\[ 2x=x+2 \implies x=2 \]
Verification
\[ 9^2=81=3^4=3^{2+2} \]
Result
\[ \boxed{x=2} \]
Exercise of 06/04/2026 - 16:45 — level ★★★★☆
\[ 2^{3x-1} = 4^{x+2} \]
Result
\[ x = 5 \]
Solution
Rewrite in base 2
\[ 4^{x+2}=2^{2(x+2)}=2^{2x+4} \]
Equation of the exponents
\[ 3x-1=2x+4 \implies x=5 \]
Verification
\[ 2^{14}=4^7=2^{14} \]
Result
\[ \boxed{x=5} \]
Exercise of 06/04/2026 - 17:00 — level ★★★★★
\[ \frac{2^{n+2}-2^n}{2^{n-1}} \]
Result
\[ 6 \]
Solution
Factoring \(2^n\) in the numerator
\[ \frac{2^n(2^2-1)}{2^{n-1}}=\frac{2^n\cdot3}{2^{n-1}} \]
Simplification
\[ 3\cdot\frac{2^n}{2^{n-1}}=3\cdot2^{n-(n-1)}=3\cdot2=6 \]
Result
\[ \boxed{6} \]
Exercise of 06/04/2026 - 17:15 — level ★★★★★
\[ 4^x + 2^{x+1} = 8 \]
Result
\[ x = 1 \]
Solution
Solving idea
Rewrite \(4^x=(2^x)^2\) and \(2^{x+1}=2\cdot2^x\), then let \(t=2^x\).
Equation in \(t\)
\[ t^2+2t-8=0 \implies (t+4)(t-2)=0 \]
Solution
\(t=-4\): discarded (\(2^x>0\)). \(t=2\): \(2^x=2 \implies x=1\).
Verification
\[ 4^1+2^2=4+4=8 \]
Result
\[ \boxed{x=1} \]
Exercise of 06/04/2026 - 17:30 — level ★★★★★
\[ (a^m \cdot a^n)^{\frac{1}{m+n}} \]
Result
\[ a \]
Solution
Solving idea
Simplify the exponent using the product rule for powers with the same base.
Product rule
\[ a^m\cdot a^n = a^{m+n} \]
Power of a power
\[ (a^{m+n})^{\frac{1}{m+n}} = a^{\frac{m+n}{m+n}} = a^1 = a \]
Result
\[ \boxed{a} \]
Exercise of 06/04/2026 - 17:45 — level ★★★★★
\[ 2^x + 2^{-x} = \frac{5}{2} \]
Result
\[ x = 1 \quad \text{or} \quad x = -1 \]
Solution
Solving idea
Let \(t=2^x>0\). Then \(2^{-x}=1/t\) and the equation becomes a rational equation.
Substitution \(t=2^x\)
\[ t+\frac{1}{t}=\frac{5}{2} \implies 2t^2-5t+2=0 \]
Solution of the quadratic equation
\[ t=\frac{5\pm\sqrt{25-16}}{4}=\frac{5\pm3}{4} \]
\(t=2\): \(2^x=2 \implies x=1\)
\(t=\tfrac{1}{2}\): \(2^x=2^{-1} \implies x=-1\)
Verification
\(x=1\): \(2+\tfrac{1}{2}=\tfrac{5}{2}\) \(x=-1\): \(\tfrac{1}{2}+2=\tfrac{5}{2}\)
Result
\[ \boxed{x=1 \quad \text{or} \quad x=-1} \]
Exercise of 06/04/2026 - 18:00 — level ★★★★★
\[ (a^{2/3}\cdot b^{1/2})^6 \]
Result
\[ a^4 b^3 \]
Solution
Power of a product
\[ (a^{2/3})^6\cdot(b^{1/2})^6 \]
Power of a power
\[ a^{(2/3)\cdot6}\cdot b^{(1/2)\cdot6}=a^4\cdot b^3 \]
Result
\[ \boxed{a^4 b^3} \]
Exercise of 06/04/2026 - 18:15 — level ★★★★★
\[ 9^x - 4\cdot3^{x+1} + 27 = 0 \]
Result
\[ x = 1 \quad \text{or} \quad x = 2 \]
Solution
Solving idea
Rewrite \(9^x=(3^x)^2\) and \(3^{x+1}=3\cdot3^x\), then let \(t=3^x\).
Equation in \(t\)
\[ t^2-12t+27=0 \]
Discriminant and solutions
\[ \Delta=144-108=36 \implies t=\frac{12\pm6}{2} \]
\(t=9\): \(3^x=3^2 \implies x=2\)
\(t=3\): \(3^x=3^1 \implies x=1\)
Verification
\(x=1\): \(9-36+27=0\) \(x=2\): \(81-108+27=0\)
Result
\[ \boxed{x=1 \quad \text{or} \quad x = 2} \]