Exponent Laws — Master Quiz

Quick Reference

Core Exponent Laws — Memorise These First

\(a^m \cdot a^n = a^{m+n}\)

⚡ SAME BASE → ADD exponents

\(\dfrac{a^m}{a^n} = a^{m-n}\)

⚡ DIVIDE → SUBTRACT exponents

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

⚡ POWER of POWER → MULTIPLY

\((ab)^n = a^n b^n\)

⚡ DISTRIBUTE the power

\(a^0 = 1\)

⚡ ZERO power = ONE (always!)

\(a^{-n} = \dfrac{1}{a^n}\)

⚡ NEGATIVE → FLIP to denominator

\(a^{1/n} = \sqrt[n]{a}\)

⚡ FRACTION power = ROOT

\(6^n = 2^n \cdot 3^n\)

⚡ SPLIT composite bases

Part 1 — Factor & Simplify

Worked Example

Find \(a\) if \(2^{n+2}(3^n - 3^{n+1}) = a \times 6^n\).

Step 1 — Expand: \(2^{n+2} = 4 \cdot 2^n\)
Step 2 — Factor right bracket: \(3^n - 3^{n+1} = 3^n(1 - 3) = -2 \cdot 3^n\)
Step 3 — Multiply: \(4 \cdot 2^n \cdot (-2) \cdot 3^n = -8 \cdot (2 \cdot 3)^n = -8 \cdot 6^n\)
Answer: \(a = -8\)

Medium · Factor & Simplify

For all natural numbers \(n\), if \(2^{n+3}(3^n - 3^{n+1}) = a \times 6^n\), find \(a\).

SPLIT → FACTOR bracket → COMBINE bases

Solution

\(2^{n+3} = 8 \cdot 2^n\). Factor the bracket: \(3^n - 3^{n+1} = 3^n(1-3) = -2 \cdot 3^n\).
Product: \(8 \cdot 2^n \cdot (-2) \cdot 3^n = -16 \cdot 6^n\). So \(a = \mathbf{-16}\).

Medium · Factor & Simplify

For all natural numbers \(n\), if \(3^{n+2}(2^n - 2^{n+1}) = a \times 6^n\), find \(a\).

Factor \(2^n\) from bracket first!

Solution

\(3^{n+2} = 9 \cdot 3^n\). Bracket: \(2^n - 2^{n+1} = 2^n(1 - 2) = -1 \cdot 2^n\).
Product: \(9 \cdot 3^n \cdot (-1) \cdot 2^n = -9 \cdot 6^n\). So \(a = \mathbf{-9}\).

Hard · Factor & Simplify

For all natural numbers \(n\), if \(2^{n+1}(3^{n+1} - 3^{n+2}) = a \times 6^n\), find \(a\).

Factor 3^(n+1) first — don't skip!

Solution

\(2^{n+1} = 2 \cdot 2^n\). Bracket: \(3^{n+1} - 3^{n+2} = 3^{n+1}(1-3) = -2 \cdot 3^{n+1} = -6 \cdot 3^n\).
Product: \(2 \cdot 2^n \cdot (-6) \cdot 3^n = -12 \cdot 6^n\). So \(a = \mathbf{-12}\).

Easy · Basic Law

Simplify: \(\dfrac{(2^3)^4 \cdot 2^{-5}}{2^3}\)

POWER of POWER first, then ADD/SUBTRACT

Solution

\((2^3)^4 = 2^{12}\). Numerator: \(2^{12} \cdot 2^{-5} = 2^7\). Divide: \(2^7 / 2^3 = 2^4\). Answer: \(\mathbf{2^4 = 16}\).

Part 2 — Zero & Negative Exponents

Key Trap

Don't confuse: \((-2)^0 = 1\) but \(-2^0 = -1\).
The negative sign is NOT inside the base unless there are parentheses!

Hard · Zero Power Trap

Which of the following equals \(1\)?

Select the expression that is exactly equal to 1.

Parentheses MATTER for zero power!

Solution

Only \((-3)^0\) puts \(-3\) as the full base, giving \(1\).
\(-3^0 = -(3^0) = -1\). The negative sign is applied after the power when there are no outer parentheses.

Medium · Negative Exponent

Evaluate: \(2^{-3} + 4^{-1}\)

NEGATIVE power → flip to 1/base^n

Solution

\(2^{-3} = \frac{1}{8}\), \(4^{-1} = \frac{1}{4} = \frac{2}{8}\).
Sum: \(\frac{1}{8} + \frac{2}{8} = \mathbf{\frac{3}{8}}\).

Hard · Negative Exponent Trap

Simplify: \(\left(\dfrac{2}{3}\right)^{-2} \times \left(\dfrac{3}{4}\right)^{-1}\)

Flip the fraction, THEN raise to power

Solution

\(\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}\).
\(\left(\frac{3}{4}\right)^{-1} = \frac{4}{3}\).
Product: \(\frac{9}{4} \times \frac{4}{3} = \frac{9 \times 4}{4 \times 3} = \mathbf{3}\).

Medium · Negative Exponent

If \(x = 2\), find the value of \(x^{-2} - x^{-1}\).

Substitute AFTER simplifying powers

Solution

\(x^{-2} = \frac{1}{4}\), \(x^{-1} = \frac{1}{2} = \frac{2}{4}\).
Result: \(\frac{1}{4} - \frac{2}{4} = \mathbf{-\frac{1}{4}}\).

Part 3 — Rational & Radical Exponents

Key Formula

\(a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m\)

Common trap: \(27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9\). Don't try to compute \(27^2\) first!

Medium · Rational Exponent

Evaluate: \(8^{2/3}\)

Take ROOT first (denominator), THEN power (numerator)

Solution

\(8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = \mathbf{4}\).
Trick: always take the cube root first. \(\sqrt[3]{8} = 2\), then square it.

Hard · Rational Exponent

Simplify: \(\dfrac{16^{3/4}}{4^{3/2}}\)

Convert everything to base-2 first!

Solution

\(16^{3/4} = (2^4)^{3/4} = 2^3 = 8\).
\(4^{3/2} = (2^2)^{3/2} = 2^3 = 8\).
Result: \(8/8 = \mathbf{1}\).

Hard · Rational Exponent Trap

Evaluate: \((-8)^{1/3} + (-27)^{1/3}\)

Odd roots of negatives ARE defined! \(\sqrt[3]{-a} = -\sqrt[3]{a}\)

Solution

\((-8)^{1/3} = -2\) and \((-27)^{1/3} = -3\).
Sum: \(-2 + (-3) = \mathbf{-5}\).
Odd-index roots of negative numbers are real. Even-index roots are not (in ℝ).

Medium · Radical to Exponent

Simplify: \(\sqrt[4]{81} \times \sqrt{9}\)

Convert ALL radicals to exponents first

Solution

\(\sqrt[4]{81} = 81^{1/4} = (3^4)^{1/4} = 3^1 = 3\).
\(\sqrt{9} = 9^{1/2} = (3^2)^{1/2} = 3^1 = 3\).
Product: \(3 \times 3 = \mathbf{9}\).
Common trap: students confuse \(\sqrt[4]{81} \approx 81\) without simplifying. Always convert to base powers first.

Part 4 — Equations & Mixed Challenge

Strategy for Exponent Equations

To solve \(2^{x} = 8\): write both sides as same base → \(2^x = 2^3\) → \(x = 3\).
Key: both sides MUST have the same base before comparing exponents.

Easy · Same Base Equation

Solve for \(x\): \(4^x = 32\)

Write as powers of 2: \(4 = 2^2\), \(32 = 2^5\)

Solution

\(4^x = (2^2)^x = 2^{2x}\) and \(32 = 2^5\).
So \(2x = 5 \Rightarrow x = \mathbf{\frac{5}{2}}\).

Medium · Exponent Equation

Solve for \(x\): \(9^{x-1} = 27^x\)

Base 3 is your friend: \(9 = 3^2\), \(27 = 3^3\)

Solution

LHS: \((3^2)^{x-1} = 3^{2x-2}\). RHS: \((3^3)^x = 3^{3x}\).
Equation: \(2x - 2 = 3x \Rightarrow -2 = x\). So \(x = \mathbf{-2}\).

Hard · Substitution Trick

If \(2^x = 3\), find \(8^x\).

Rewrite \(8^x = (2^3)^x = (2^x)^3\) — substitute!

Solution

\(8^x = (2^3)^x = (2^x)^3 = 3^3 = \mathbf{27}\).
The key insight: rearrange so the given condition \(2^x = 3\) appears naturally.

Hard · Substitution Trick

If \(3^a = 5\), find \(\dfrac{1}{3^{-a}}\).

Negative exponent = reciprocal. Think step by step.

Solution

\(\frac{1}{3^{-a}} = 3^{a} = \mathbf{5}\).
Because \(\frac{1}{x^{-n}} = x^n\). Direct substitution of \(3^a = 5\).

Hard · Mixed Laws

Simplify: \(\dfrac{6^{n+1} - 6^n}{5 \times 6^{n-1}}\)

Factor 6^n from numerator first!

Solution

Numerator: \(6^{n+1} - 6^n = 6^n(6-1) = 5 \cdot 6^n\).
Denominator: \(5 \times 6^{n-1}\).
Result: \(\dfrac{5 \cdot 6^n}{5 \cdot 6^{n-1}} = \dfrac{6^n}{6^{n-1}} = 6^{n-(n-1)} = 6^1 = \mathbf{6}\).

Hard · Factor with Negative

For all natural numbers \(n\), if \(3^{n+2} - 3^n = a \times 3^n\), find \(a\).

Factor out the SMALLEST power of 3

Solution

\(3^{n+2} - 3^n = 3^n(3^2 - 1) = 3^n(9-1) = 8 \cdot 3^n\).
So \(a = \mathbf{8}\).

Hard · Multi-step Simplify

Simplify: \(\left(\dfrac{a^2 b^{-3}}{a^{-1} b^2}\right)^2\)

SUBTRACT exponents inside, MULTIPLY by outer power

Solution

Inside: \(\frac{a^2 b^{-3}}{a^{-1} b^2} = a^{2-(-1)} b^{-3-2} = a^3 b^{-5}\).
Squaring: \((a^3 b^{-5})^2 = a^6 b^{-10} = \dfrac{a^6}{b^{10}}\). So answer is \(\mathbf{\frac{a^6}{b^{10}}}\).

Hard · Boss Question

For all natural numbers \(n\), if \(2^{n+2}(3^n - 3^{n+1}) = a \times 6^n\), what is \(a\)? (Original problem — master it!)

SPLIT 2^(n+2) → FACTOR bracket → COMBINE as 6^n

Solution

\(2^{n+2} = 4 \cdot 2^n\). Bracket: \(3^n - 3^{n+1} = 3^n(1-3) = -2 \cdot 3^n\).
Product: \(4 \cdot 2^n \cdot (-2) \cdot 3^n = -8 \cdot (2 \cdot 3)^n = -8 \cdot 6^n\).
Therefore \(a = \mathbf{-8}\). ✓

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