Daily run: Exponents & Roots content + growth playbook bootstrap

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 all over GMAT Focus Quant — inside number-properties questions, in
+disguised equations, and as the fast way to compare very large or very small quantities. Almost
+everything reduces to a short list of rules plus one habit: **rewrite every base as a power of the
+same prime before you do anything else.**
+
+## Core concepts
+
+**The exponent rules.** For any nonzero base and integer (or rational) exponents:
+
+- Product: \(a^m \cdot a^n = a^{m+n}\) — same base, *add* exponents.
+- Quotient: \(\dfrac{a^m}{a^n} = a^{m-n}\) — same base, *subtract* exponents.
+- Power of a power: \((a^m)^n = a^{mn}\) — *multiply* exponents.
+- Distributes over products: \((ab)^n = a^n b^n\).
+
+**Zero and negative exponents.**
+
+\[a^0 = 1 \quad (a \ne 0), \qquad a^{-n} = \frac{1}{a^n}\]
+
+A negative exponent means "reciprocal," not "negative number." So \(2^{-2} = \tfrac14\), which is
+positive.
+
+**Roots are fractional exponents.** This is the single most useful bridge:
+
+\[\sqrt[n]{a} = a^{1/n}, \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}\]
+
+So \(27^{2/3} = \left(27^{1/3}\right)^2 = 3^2 = 9\). Take the root first (smaller numbers), then the power.
+
+**Simplifying radicals.** Pull perfect-square factors out from under a square root, and combine only
+*like* radicals:
+
+\[\sqrt{ab} = \sqrt{a}\,\sqrt{b}, \qquad \sqrt{50} = \sqrt{25\cdot 2} = 5\sqrt{2}\]
+
+You can add \(5\sqrt2 + 3\sqrt2 = 8\sqrt2\), but you **cannot** turn \(\sqrt2 + \sqrt2\) into
+\(\sqrt4\).
+
+**Same-base strategy for equations.** When an equation has different bases that are all powers of one
+prime, rewrite them and set the exponents equal:
+
+\[4^{x+1} = 8^{x-1} \;\Rightarrow\; 2^{2(x+1)} = 2^{3(x-1)} \;\Rightarrow\; 2x+2 = 3x-3 \;\Rightarrow\; x = 5\]
+
+## Worked examples
+
+**Combine rules.** Simplify \((x^2)^3 \cdot x^4\). Power of a power gives \(x^{6}\); then the product
+rule adds exponents: \(x^{6} \cdot x^{4} = x^{10}\).
+
+**Fractional exponent.** Evaluate \(27^{2/3}\). Cube root of 27 is 3; square it: \(3^2 = 9\).
+
+**Radicals.** Simplify \(\sqrt{50} + \sqrt{18} = 5\sqrt2 + 3\sqrt2 = 8\sqrt2\).
+
+## Common traps
+
+- **Multiplying exponents when you should add.** \(2^5 \cdot 2^3 = 2^{8}\), not \(2^{15}\). You only
+  multiply exponents for a *power of a power*, \((2^5)^3 = 2^{15}\).
+- **Negative exponent → negative number.** \(3^{-1} = \tfrac13\), a positive fraction, not \(-3\).
+- **Adding under the radical.** \(\sqrt{a} + \sqrt{b} \ne \sqrt{a+b}\). \(\sqrt{9} + \sqrt{16} = 3+4 = 7\), not \(\sqrt{25} = 5\).
+- **Combining unlike bases.** \(2^3 \cdot 3^3\) is \((2\cdot3)^3 = 6^3\), but \(2^3 \cdot 3^2\) cannot be merged into one base.
+
+## Key takeaways
+
+- Same base: add exponents to multiply, subtract to divide, multiply to raise a power to a power.
+- \(a^0 = 1\) and \(a^{-n} = 1/a^n\); negative exponents give positive reciprocals.
+- Roots are fractional exponents: \(a^{m/n} = \sqrt[n]{a^m}\) — take the root first, then the power.
+- To solve exponential equations, rewrite every base as the same prime and equate the exponents.

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, quotient-rule]
+choices:
+  A: "5"
+  B: "25"
+  C: "125"
+  D: "625"
+  E: "5^{11}"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(\dfrac{5^{7}}{5^{4}}\)?
+
+## Explanation
+
+Same base, division — **subtract** the exponents:
+
+\[\frac{5^{7}}{5^{4}} = 5^{7-4} = 5^{3} = 125\]
+
+The answer is **125**.
+
+**The traps:** choice **E** (\(5^{11}\)) *adds* the exponents, which is the rule for
+*multiplication*, not division. Choices that divide the exponents (\(7 \div 4\)) or cancel bases
+incorrectly land on the smaller powers of 5.
+
+## Hints
+
+- Same base with division: what do you do to the exponents?
+- \(5^{7} / 5^{4} = 5^{7-4}\). Now just evaluate the small 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, zero-exponent, negative-exponent]
+choices:
+  A: "0"
+  B: "1"
+  C: "1.25"
+  D: "1.5"
+  E: "9"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(7^{0} + 2^{-2}\)?
+
+## Explanation
+
+Handle each term with the definitions of the zero and negative exponents:
+
+\[7^{0} = 1, \qquad 2^{-2} = \frac{1}{2^{2}} = \frac{1}{4}\]
+
+So the sum is
+
+\[1 + \frac{1}{4} = \frac{5}{4} = 1.25\]
+
+The answer is **1.25**.
+
+**The traps:** choice **A (0)** assumes \(7^0 = 0\), but any nonzero base to the zero power is 1.
+Choice **E (9)** reads \(2^{-2}\) as \(-2\times 2\) or \(2^2\) plus a sign slip — a negative exponent
+gives a positive *reciprocal*, not a negative number.
+
+## Hints
+
+- Any nonzero number raised to the power 0 equals 1 — not 0.
+- A negative exponent means take the reciprocal: \(2^{-2} = 1/2^{2}\), a positive fraction.

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: medium
+tags: [exponents, power-of-a-power, product-rule, variables]
+choices:
+  A: "x^{9}"
+  B: "x^{10}"
+  C: "x^{12}"
+  D: "x^{14}"
+  E: "x^{24}"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+For \(x \neq 0\), which of the following is equal to \(\left(x^{2}\right)^{3} \cdot x^{4}\)?
+
+## Explanation
+
+Apply the rules in order. First, **power of a power** — multiply the exponents:
+
+\[\left(x^{2}\right)^{3} = x^{2\cdot 3} = x^{6}\]
+
+Then the **product rule** — same base, add the exponents:
+
+\[x^{6} \cdot x^{4} = x^{6+4} = x^{10}\]
+
+The answer is **\(x^{10}\)**.
+
+**The traps:** choice **E** (\(x^{24}\)) multiplies all three exponents (\(2\cdot 3\cdot 4\)) as if
+everything were a power of a power. Choice **A** (\(x^{9}\)) adds \(2+3+4\), mixing up which
+operation adds and which multiplies. Multiply for a power raised to a power; add when multiplying
+like bases.
+
+## Hints
+
+- Do the inner operation first: \((x^2)^3\) is a power raised to a power.
+- Once you have a single power of \(x\) times another, add those two exponents.

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: [roots, radicals, simplifying-radicals]
+choices:
+  A: "\\sqrt{68}"
+  B: "2\\sqrt{17}"
+  C: "8\\sqrt{2}"
+  D: "15\\sqrt{2}"
+  E: "34"
+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
+
+Pull the perfect-square factor out of each radical, then combine the *like* radicals:
+
+\[\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}, \qquad \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\]
+
+Both are multiples of \(\sqrt2\), so add the coefficients:
+
+\[5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\]
+
+The answer is **\(8\sqrt{2}\)**.
+
+**The traps:** choice **A** (\(\sqrt{68}\)) adds *under* the radical (\(50+18\)), but
+\(\sqrt{a} + \sqrt{b} \neq \sqrt{a+b}\). Choice **D** (\(15\sqrt2\)) multiplies the coefficients
+instead of adding them. You may add radicals only when the parts under the root already match.
+
+## Hints
+
+- Factor out the largest perfect square from each radical first.
+- After simplifying, both terms are multiples of \(\sqrt2\). Add only the numbers in front.

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: "6"
+  B: "9"
+  C: "18"
+  D: "81"
+  E: "729"
+answer: B
+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 combines a root and a power: the denominator is the root, the numerator is the
+power. Take the root **first** to keep the numbers small:
+
+\[27^{2/3} = \left(27^{1/3}\right)^{2} = \left(\sqrt[3]{27}\right)^{2} = 3^{2} = 9\]
+
+The answer is **9**.
+
+**The traps:** choice **E** (\(729 = 27^2\)) ignores the cube root entirely. Choice **C (18)**
+multiplies the base by the exponent (\(27 \times \tfrac23\)). Choice **A (6)** takes the cube root
+but then multiplies by 2 instead of squaring.
+
+## Hints
+
+- Rewrite the exponent as a root and a power: the bottom of the fraction is the root.
+- Cube root of 27 is a small number — find it 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, exponential-equations, same-base]
+choices:
+  A: "2"
+  B: "3"
+  C: "4"
+  D: "5"
+  E: "7"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(4^{x+1} = 8^{x-1}\), what is the value of \(x\)?
+
+## Explanation
+
+The bases 4 and 8 are both powers of 2, so rewrite everything with base 2:
+
+\[4^{x+1} = \left(2^{2}\right)^{x+1} = 2^{2(x+1)}, \qquad 8^{x-1} = \left(2^{3}\right)^{x-1} = 2^{3(x-1)}\]
+
+With a common base, the exponents must be equal:
+
+\[2(x+1) = 3(x-1)\]
+\[2x + 2 = 3x - 3\]
+\[x = 5\]
+
+The answer is **5**.
+
+**Check:** \(4^{6} = 4096\) and \(8^{4} = 4096\). ✓
+
+**The trap:** trying to compare 4 and 8 directly, or guessing a small value, misses the key move —
+express both sides as powers of the same prime, then set the exponents equal.
+
+## Hints
+
+- Both 4 and 8 are powers of 2. Rewrite each side with base 2.
+- Once the bases match, the exponents must be equal — solve the resulting linear equation.