Add exponents & roots bundle (quant/arithmetic)

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, laws-of-exponents]
+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 — in number properties, algebra, and word
+problems — and almost every mistake comes from misapplying one of a small set of rules. Learn the
+laws cold, know that a root is just a fractional exponent, and practice the two power moves the
+exam loves: **factoring out a common power** and **rewriting both sides with the same base**.
+
+## Core concepts
+
+**The laws of exponents (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: \((a^m)^n = a^{mn}\).
+- Power of a product → distribute: \((ab)^n = a^n b^n\).
+
+The single most common error is mixing these up — adding when you should multiply, or vice versa.
+Multiplying like bases adds exponents; raising a power to a power multiplies them.
+
+**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. \(2^{-3} = \tfrac18\), which is
+positive. For a fraction, \(\left(\tfrac{a}{b}\right)^{-n} = \left(\tfrac{b}{a}\right)^{n}\) — flip,
+then apply the positive power.
+
+**Fractional exponents are roots.**
+
+\[a^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^{m} = \sqrt[n]{a^m}\]
+
+So \(27^{2/3}\) means "cube root, then square": \((\sqrt[3]{27})^2 = 3^2 = 9\). Take the root of the
+**base**; the numerator of the fraction is the power.
+
+**Simplifying radicals.** Pull out perfect-square factors: \(\sqrt{ab} = \sqrt{a}\,\sqrt{b}\). Find
+the largest perfect square that divides the number and take its square root out front:
+
+\[\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25}\,\sqrt{2} = 5\sqrt{2}\]
+
+The number you pull out is the **square root** of the perfect square (5, not 25), and the factor
+you leave inside must have no perfect-square factors left.
+
+## Worked examples
+
+**Same base, solve for the exponent.** Solve \(4^{x+1} = 8^{x-1}\). Rewrite both sides with base 2:
+\(4 = 2^2\) and \(8 = 2^3\), so \(2^{2(x+1)} = 2^{3(x-1)}\). With equal bases the exponents must be
+equal: \(2(x+1) = 3(x-1)\Rightarrow 2x + 2 = 3x - 3 \Rightarrow x = 5\). Check: \(4^6 = 8^4 = 4096\).
+
+**Factor out the common power.** Simplify \(\dfrac{3^{12} - 3^{10}}{3^{10}}\). You cannot subtract
+\(3^{12} - 3^{10}\) directly, but you can factor the smaller power out of the numerator:
+\(3^{12} - 3^{10} = 3^{10}(3^{2} - 1) = 3^{10}\cdot 8\). Then
+\(\dfrac{3^{10}\cdot 8}{3^{10}} = 8\).
+
+## Common traps
+
+- **Adding vs. multiplying exponents.** \(2^3 \cdot 2^4 = 2^7\) (add), but \((2^3)^4 = 2^{12}\)
+  (multiply). Know which rule you're using.
+- **Negative exponent ≠ negative number.** \(3^{-2} = \tfrac19\), not \(-9\). It flips to a
+  reciprocal.
+- **Fractional exponent applied to the wrong part.** In \(27^{2/3}\), the 3 is a cube *root* of 27
+  and the 2 is a *power* — it is not \(27 \times \tfrac23 = 18\).
+- **Leaving the perfect square inside — or pulling the whole thing out.** \(\sqrt{50} = 5\sqrt2\):
+  you pull out \(\sqrt{25} = 5\), not 25, and 2 has no square factor left to remove.
+- **Combining a sum or difference of powers by "canceling."** In \(\tfrac{3^{12}-3^{10}}{3^{10}}\)
+  you must factor first; you cannot cancel one term of a difference.
+
+## Key takeaways
+
+- Same base: multiply → add exponents, divide → subtract, power of a power → multiply.
+- \(a^0 = 1\); a negative exponent is a reciprocal; a fractional exponent is a root.
+- Simplify radicals by pulling out the square root of the largest perfect-square factor.
+- To solve \(a^{f(x)} = b^{g(x)}\), rewrite both sides as the same base and set the exponents equal.
+- Sums and differences of powers get simplified by **factoring out the common power**, not canceling.

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, laws-of-exponents, same-base]
+choices:
+  A: "729"
+  B: "243"
+  C: "6561"
+  D: "9"
+  E: "90"
+answer: A
+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 powers of the **same base**, you **add** the exponents:
+
+\[3^4 \cdot 3^2 = 3^{4+2} = 3^6\]
+
+Now evaluate \(3^6\): \(3^4 = 81\) and \(3^2 = 9\), so \(3^6 = 81 \times 9 = 729\).
+
+**The traps:**
+
+- **C (6561)** = \(3^8\), from *multiplying* the exponents (\(4 \times 2\)). You multiply exponents
+  only when raising a power to a power, \((3^4)^2\) — not when multiplying two powers.
+- **D (9)** = \(3^2\), from *subtracting* the exponents; that's the rule for division, not
+  multiplication.
+- **E (90)** = \(81 + 9\), from adding the two values instead of combining the powers.
+- **B (243)** = \(3^5\), a miscount of \(4 + 2\).
+
+## Hints
+
+- Same base, and the powers are multiplied — which exponent rule applies?
+- \(a^m \cdot a^n = a^{m+n}\). Add the exponents, then evaluate.

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, power-of-a-power]
+choices:
+  A: "128"
+  B: "24"
+  C: "4096"
+  D: "81"
+  E: "512"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(\left(2^3\right)^4\)?
+
+## Explanation
+
+To raise a power to a power, **multiply** the exponents:
+
+\[\left(2^3\right)^4 = 2^{3 \times 4} = 2^{12}\]
+
+And \(2^{12} = 4096\) (since \(2^{10} = 1024\), and \(2^{12} = 1024 \times 4\)).
+
+**The traps:**
+
+- **A (128)** = \(2^{3+4} = 2^7\), from *adding* the exponents. Adding is for multiplying like
+  bases (\(2^3 \cdot 2^4\)); a power of a power multiplies them.
+- **B (24)** = \(2 \times 3 \times 4\), from multiplying all three numbers together.
+- **D (81)** = \(3^4\), from raising the *exponent* to the 4th power and dropping the base.
+- **E (512)** = \(2^9\), a miscount of \(3 \times 4\).
+
+## Hints
+
+- A power raised to another power — do you add or multiply the exponents?
+- \((a^m)^n = a^{mn}\). Multiply, then evaluate the single power.

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, negative-exponents, fractions]
+choices:
+  A: "4/9"
+  B: "-9/4"
+  C: "3/2"
+  D: "-4/9"
+  E: "9/4"
+answer: E
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(\left(\dfrac{2}{3}\right)^{-2}\)?
+
+## Explanation
+
+A **negative exponent** means take the reciprocal, then apply the positive power. For a fraction,
+flip it:
+
+\[\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^{2} = \frac{3^2}{2^2} = \frac{9}{4}\]
+
+**The traps:**
+
+- **A (4/9)** = \(\left(\tfrac23\right)^2\), from squaring without first taking the reciprocal —
+  ignoring the negative sign on the exponent.
+- **C (3/2)** = the reciprocal without squaring — you flipped but forgot the exponent is 2.
+- **B (−9/4) and D (−4/9)** treat the negative exponent as making the *value* negative. A negative
+  exponent flips the base to a reciprocal; it never changes the sign of the result.
+
+## Hints
+
+- A negative exponent does not make the answer negative — it means "reciprocal."
+- Flip the fraction first, then apply the positive power of 2.

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, fractional-exponents, roots]
+choices:
+  A: "3"
+  B: "9"
+  C: "18"
+  D: "729"
+  E: "6"
+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 is a root and a power: the **denominator** is the root, the **numerator** is
+the power. So \(27^{2/3}\) means "take the cube root of 27, then square it":
+
+\[27^{2/3} = \left(\sqrt[3]{27}\right)^{2} = 3^2 = 9\]
+
+(Doing it in the other order gives the same result: \(\sqrt[3]{27^2} = \sqrt[3]{729} = 9\).)
+
+**The traps:**
+
+- **A (3)** = \(\sqrt[3]{27}\) — you took the cube root but forgot to square it.
+- **C (18)** = \(27 \times \tfrac23\), from *multiplying* the base by the fraction instead of
+  treating \(\tfrac23\) as a root-and-power.
+- **D (729)** = \(27^2\), from squaring but never taking the cube root.
+- **E (6)** = \(3 \times 2\), from taking the cube root (3) and multiplying by the numerator (2)
+  instead of raising to that power.
+
+## Hints
+
+- In \(a^{m/n}\), the denominator \(n\) is a root and the numerator \(m\) is a power.
+- Take the cube root of 27 first (it's a perfect cube), then apply the power of 2.

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: [roots, radicals, simplifying-radicals]
+choices:
+  A: "25√2"
+  B: "2√5"
+  C: "5√10"
+  D: "5√2"
+  E: "10√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 \(\sqrt{50}\)?
+
+## Explanation
+
+To simplify a radical, pull out the largest **perfect-square** factor. The largest perfect square
+dividing 50 is 25:
+
+\[\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25}\,\sqrt{2} = 5\sqrt{2}\]
+
+You take the *square root* of the perfect square (\(\sqrt{25} = 5\)) out front, and 2 has no
+perfect-square factor left, so \(5\sqrt2\) is fully simplified. (Check: \((5\sqrt2)^2 = 25 \cdot 2 = 50\).)
+
+**The traps:**
+
+- **A (25√2)** pulls the 25 out front without taking its square root — you must move out
+  \(\sqrt{25} = 5\), not 25.
+- **C (5√10)** factors 50 as \(5 \times 10\) and pulls out the 5, but 5 isn't a perfect square, so
+  nothing legitimately comes out and \(\sqrt{10}\) is not simplified.
+- **B (2√5)** = \(\sqrt{20}\), and **E (10√2)** = \(\sqrt{200}\); neither equals \(\sqrt{50}\).
+
+## Hints
+
+- Find the largest perfect square that divides 50.
+- Pull out its *square root*, not the perfect square itself.

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, common-base, equations]
+choices:
+  A: "1"
+  B: "2"
+  C: "5"
+  D: "-5"
+  E: "3"
+answer: C
+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 aren't equal, but both are powers of 2. Rewrite everything with base 2, using
+\(4 = 2^2\) and \(8 = 2^3\):
+
+\[4^{\,x+1} = \left(2^2\right)^{x+1} = 2^{\,2(x+1)}, \qquad 8^{\,x-1} = \left(2^3\right)^{x-1} = 2^{\,3(x-1)}\]
+
+With the same base, the exponents must be equal:
+
+\[2(x+1) = 3(x-1) \;\Rightarrow\; 2x + 2 = 3x - 3 \;\Rightarrow\; x = 5\]
+
+Check: \(4^{6} = 4096\) and \(8^{4} = 4096\). ✓
+
+**The traps:**
+
+- **E (3)** comes from writing \(8 = 2^4\) instead of \(2^3\): then \(2(x+1) = 4(x-1)\) gives
+  \(x = 3\). Remember \(8 = 2^3\).
+- **B (2)** comes from not distributing the multiplier over the constant — writing
+  \(2x + 1 = 3x - 1\) instead of \(2(x+1) = 3(x-1)\).
+- **A (1)** and **D (−5)** are sign slips when moving terms across the equation.
+
+## Hints
+
+- Rewrite both sides as powers of the same base (2), then set the exponents equal.
+- Be careful to distribute: \((2^2)^{x+1} = 2^{2(x+1)}\), and \(2(x+1)\) means \(2x+2\).