Add Exponents & Roots bundle: lesson + 7 questions

content/lessons/quant/quant-arithmetic-exponents-roots.md

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+---
+id: quant-arithmetic-exponents-roots
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+title: "Exponents & Roots"
+tags: [exponents, roots, radicals, powers]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Exponents and roots show up throughout GMAT Focus Quant — in pure arithmetic, in algebra, and
+hidden inside word problems. Almost every mistake comes from misapplying one of a small set of
+rules, so the payoff is learning the rules cold and knowing exactly which ones do **not** exist.
+
+## Core concepts
+
+**The exponent rules (same base).**
+
+- Multiply → **add** exponents: \(a^m \cdot a^n = a^{m+n}\).
+- Divide → **subtract** exponents: \(\dfrac{a^m}{a^n} = a^{m-n}\).
+- Power of a power → **multiply** exponents: \((a^m)^n = a^{mn}\).
+
+Notice multiplying powers *adds* the exponents while a power-of-a-power *multiplies* them. Swapping
+those two is the single most common exponent error.
+
+**Zero and negative exponents.**
+
+\[a^0 = 1 \quad (a \neq 0), \qquad a^{-n} = \frac{1}{a^n}\]
+
+A negative exponent means **reciprocal**, not a negative number: \(2^{-2} = \tfrac{1}{4}\), not \(-4\).
+
+**Products and quotients of different bases.** You can only combine bases that are equal. But a
+product raised to a power distributes: \((ab)^n = a^n b^n\). That lets you rewrite a mixed base like
+\(6^4 = (2\cdot 3)^4 = 2^4 \cdot 3^4\), which is often the key to a cancellation.
+
+**Roots are fractional exponents.**
+
+\[\sqrt[n]{a} = a^{1/n}, \qquad a^{m/n} = \left(a^{1/n}\right)^m = \left(\sqrt[n]{a}\right)^m\]
+
+So \(27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9\). Take the root first (smaller numbers), then the power.
+
+**Simplifying square roots.** Pull out perfect-square factors: \(\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}\).
+Radicals **add only when the radical part matches**: \(5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\), but
+\(\sqrt{50} + \sqrt{18}\) must first be simplified to \(5\sqrt{2} + 3\sqrt{2}\) before you can add.
+
+**Solving equations by matching bases.** If \(a^x = a^y\) then \(x = y\). To solve \(2^{x+3} = 8^{x-1}\),
+rewrite everything on one base: \(8 = 2^3\), so \(8^{x-1} = 2^{3(x-1)}\), then set the exponents equal.
+
+## Worked examples
+
+**Combine powers.** \(3^4 \cdot 3^2 = 3^{4+2} = 3^6 = 729\). (Not \(3^8\) — you add, not multiply.)
+
+**Fractional exponent.** \(27^{2/3}\): cube root of 27 is 3, then square it → \(9\).
+
+**Matching bases.** \(2^{x+3} = 8^{x-1} \Rightarrow 2^{x+3} = 2^{3(x-1)} = 2^{3x-3}\). Set
+\(x + 3 = 3x - 3 \Rightarrow 2x = 6 \Rightarrow x = 3\). Check: \(2^6 = 64 = 8^2\). ✓
+
+## Common traps
+
+- **Multiplying the exponents when you multiply powers.** \(3^4 \cdot 3^2 = 3^6\), not \(3^8\).
+- **Reading a negative exponent as a negative number.** \(2^{-2} = \tfrac14\), not \(-4\).
+- **Thinking \(a^0 = 0\).** Any nonzero base to the zero power is \(1\).
+- **Adding under the radical.** \(\sqrt{50} + \sqrt{18} \neq \sqrt{68}\). Simplify each radical first.
+- **Multiplying a base by an exponent.** \(27^{2/3}\) is a root-and-power, not \(27 \cdot \tfrac23 = 18\).
+- **Forgetting to distribute a converted exponent.** \(8^{x-1} = 2^{3(x-1)} = 2^{3x-3}\), not \(2^{3x-1}\).
+
+## Key takeaways
+
+- Multiply powers → add exponents; power of a power → multiply exponents. Don't swap them.
+- \(a^0 = 1\) and \(a^{-n} = 1/a^n\) (reciprocal, never negative).
+- \(a^{m/n} = (\sqrt[n]{a})^m\): take the root first, then the power.
+- Simplify radicals by pulling out perfect squares; add radicals only when the radical part matches.
+- To solve exponential equations, rewrite both sides on a common base and set the exponents equal.

content/questions/quant/quant-arithmetic-exponents-roots-0017.md

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+---
+id: quant-arithmetic-exponents-roots-0017
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: easy
+tags: [exponents, powers, product-rule]
+choices:
+  A: "6561"
+  B: "729"
+  C: "9"
+  D: "81"
+  E: "1296"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(3^4 \cdot 3^2\)?
+
+## Explanation
+
+When you multiply two powers of the **same base**, you **add** the exponents:
+
+\[3^4 \cdot 3^2 = 3^{4+2} = 3^6\]
+
+Now evaluate \(3^6 = 729\).
+
+**The trap:** choice **A (6561)** is \(3^8\) — the result of *multiplying* the exponents
+(\(4 \times 2 = 8\)) instead of adding them. Multiplying exponents is the rule for a power of a
+power, \((3^4)^2\), not for a product of powers. Choice **C (9)** is \(3^2\), from subtracting the
+exponents; **D (81)** is \(3^4\), from using only one factor; **E (1296)** is \(6^4\), from adding
+the bases \((3+3)\).
+
+## Hints
+
+- Same base, multiplied together — what do you do with the exponents?
+- Add the exponents first, then compute the single power.

content/questions/quant/quant-arithmetic-exponents-roots-0018.md

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+---
+id: quant-arithmetic-exponents-roots-0018
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: easy
+tags: [exponents, powers, power-of-a-power]
+choices:
+  A: "32"
+  B: "512"
+  C: "64"
+  D: "8"
+  E: "12"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \((2^3)^2\)?
+
+## Explanation
+
+A **power raised to a power** means you **multiply** the exponents:
+
+\[(2^3)^2 = 2^{3 \cdot 2} = 2^6 = 64\]
+
+**The trap:** choice **A (32)** is \(2^5\), from *adding* the exponents (\(3 + 2\)) — that's the
+rule for multiplying powers like \(2^3 \cdot 2^2\), not for a power of a power. Choice **B (512)**
+is \(2^9\), from reading the expression as \(2^{(3^2)}\). Choice **D (8)** ignores the outer
+exponent entirely (\(2^3\)), and **E (12)** multiplies the digits \(2 \cdot 3 \cdot 2\).
+
+## Hints
+
+- The whole quantity \(2^3\) is being squared.
+- For \((a^m)^n\), do you add or multiply \(m\) and \(n\)?

content/questions/quant/quant-arithmetic-exponents-roots-0019.md

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+---
+id: quant-arithmetic-exponents-roots-0019
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: easy
+tags: [roots, radicals, simplifying-radicals]
+choices:
+  A: '\(8\sqrt{3}\)'
+  B: '\(6\sqrt{3}\)'
+  C: '\(2\sqrt{6}\)'
+  D: '\(6\sqrt{2}\)'
+  E: '\(36\sqrt{2}\)'
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+Which of the following is equal to \(\sqrt{72}\)?
+
+## Explanation
+
+To simplify a square root, factor out the **largest perfect square**. The largest perfect square
+dividing 72 is 36:
+
+\[\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36}\,\sqrt{2} = 6\sqrt{2}\]
+
+You can sanity-check: \(6\sqrt{2} \approx 6 \times 1.414 = 8.49\), and \(\sqrt{72} \approx 8.49\). ✓
+
+**The trap:** choice **E (\(36\sqrt{2}\))** pulls the whole factor 36 *outside* the radical instead
+of its square root, 6. Choices **A** and **B** keep the wrong number under the radical (a mis-factor
+of 72), and **C (\(2\sqrt{6}\))** equals \(\sqrt{24}\), not \(\sqrt{72}\). Only the perfect-square
+factor comes out, and it comes out as its root.
+
+## Hints
+
+- Find the biggest perfect square that divides 72.
+- \(\sqrt{ab} = \sqrt{a}\,\sqrt{b}\); the perfect square leaves as its square root, not itself.

content/questions/quant/quant-arithmetic-exponents-roots-0020.md

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+---
+id: quant-arithmetic-exponents-roots-0020
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: medium
+tags: [exponents, zero-exponent, negative-exponent]
+choices:
+  A: '\(\tfrac{5}{4}\)'
+  B: '\(\tfrac{1}{4}\)'
+  C: '\(-3\)'
+  D: '\(\tfrac{17}{4}\)'
+  E: '\(\tfrac{3}{4}\)'
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(4^0 + 2^{-2}\)?
+
+## Explanation
+
+Handle each term with the exponent rules for zero and negative powers:
+
+- **Zero exponent:** any nonzero base to the zero power is 1, so \(4^0 = 1\).
+- **Negative exponent:** a negative exponent means reciprocal, so \(2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}\).
+
+Add them:
+
+\[4^0 + 2^{-2} = 1 + \frac{1}{4} = \frac{5}{4}\]
+
+**The trap:** choice **C (\(-3\))** treats \(2^{-2}\) as the negative number \(-4\) — but a negative
+exponent gives a *reciprocal*, never a negative value. Choice **B (\(\tfrac14\))** comes from
+\(4^0 = 0\) (wrong — it's 1). Choice **D (\(\tfrac{17}{4}\))** uses \(4^0 = 4\), and **E
+(\(\tfrac34\))** reads \(2^{-2}\) as \(-\tfrac14\).
+
+## Hints
+
+- What is any nonzero number raised to the power 0?
+- A negative exponent flips the base into a fraction; it does not make the value negative.

content/questions/quant/quant-arithmetic-exponents-roots-0021.md

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+---
+id: quant-arithmetic-exponents-roots-0021
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: medium
+tags: [exponents, fractional-exponents, roots]
+choices:
+  A: "18"
+  B: "3"
+  C: "729"
+  D: "6"
+  E: "9"
+answer: E
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(27^{2/3}\)?
+
+## Explanation
+
+A fractional exponent is a root and a power combined: \(a^{m/n} = \left(\sqrt[n]{a}\right)^m\). The
+denominator is the root, the numerator is the power. Take the **root first** to keep the numbers
+small:
+
+\[27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9\]
+
+**The trap:** choice **A (18)** multiplies the base by the exponent, \(27 \cdot \tfrac{2}{3} = 18\)
+— but an exponent is not multiplication. Choice **B (3)** takes the cube root and forgets to square
+it. Choice **C (729)** squares 27 and ignores the cube root (\(27^2\)), and **D (6)** takes the cube
+root (3) and then doubles it instead of squaring.
+
+## Hints
+
+- The denominator of the exponent tells you which root; the numerator tells you the power.
+- Cube-root 27 first, then apply the remaining power.

content/questions/quant/quant-arithmetic-exponents-roots-0022.md

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+---
+id: quant-arithmetic-exponents-roots-0022
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: hard
+tags: [exponents, equations, common-base]
+choices:
+  A: "3"
+  B: "2"
+  C: "6"
+  D: "0"
+  E: "-1"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(2^{\,x+3} = 8^{\,x-1}\), what is the value of \(x\)?
+
+## Explanation
+
+The two sides have different bases, so rewrite them on a **common base**. Since \(8 = 2^3\):
+
+\[8^{\,x-1} = \left(2^3\right)^{x-1} = 2^{\,3(x-1)} = 2^{\,3x-3}\]
+
+Now both sides are powers of 2, so the exponents must be equal:
+
+\[x + 3 = 3x - 3 \;\Rightarrow\; 6 = 2x \;\Rightarrow\; x = 3\]
+
+Check: \(2^{3+3} = 2^6 = 64\) and \(8^{3-1} = 8^2 = 64\). ✓
+
+**The trap:** choice **B (2)** comes from distributing the 3 only to the \(x\) and forgetting the
+\(-1\): writing \(8^{x-1} = 2^{3x-1}\) gives \(x + 3 = 3x - 1\), so \(x = 2\). You must apply the
+exponent 3 to the *entire* \((x-1)\). Choices **C**, **D**, and **E** come from arithmetic slips in
+solving the linear equation.
+
+## Hints
+
+- Write 8 as a power of 2 so both sides share a base.
+- \((2^3)^{x-1} = 2^{3(x-1)}\) — distribute the 3 across the whole exponent before setting them equal.

content/questions/quant/quant-arithmetic-exponents-roots-0023.md

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+---
+id: quant-arithmetic-exponents-roots-0023
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: medium
+tags: [roots, radicals, adding-radicals]
+choices:
+  A: '\(\sqrt{68}\)'
+  B: '\(15\sqrt{2}\)'
+  C: '\(8\sqrt{2}\)'
+  D: "8"
+  E: '\(2\sqrt{2}\)'
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is \(\sqrt{50} + \sqrt{18}\) in simplest form?
+
+## Explanation
+
+You cannot add radicals until each one is simplified to the **same radical part**. Pull out the
+perfect-square factors:
+
+\[\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}, \qquad \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\]
+
+Both are multiples of \(\sqrt{2}\), so add the coefficients:
+
+\[5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\]
+
+**The trap:** choice **A (\(\sqrt{68}\))** adds the numbers *under* the radical (\(50 + 18\)) — but
+\(\sqrt{a} + \sqrt{b} \neq \sqrt{a+b}\). Choice **B (\(15\sqrt{2}\))** multiplies the coefficients
+(\(5 \times 3\)) instead of adding them; **D (8)** drops the \(\sqrt{2}\) altogether; and **E
+(\(2\sqrt{2}\))** subtracts the coefficients. Add coefficients only when the radical parts already
+match.
+
+## Hints
+
+- Simplify each square root first — what perfect squares divide 50 and 18?
+- Once both terms are multiples of \(\sqrt{2}\), add their coefficients (don't touch the \(\sqrt{2}\)).