Add exponents & roots bundle (quant/arithmetic)

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

@@ -0,0 +1,87 @@
+---
+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-property questions, in
+compound-growth word problems, and as fast-simplification steps. Almost every trap here comes
+from mixing up **adding**, **multiplying**, and **not touching** an exponent. Learn the handful of
+rules below cold and most "scary-looking" powers collapse to a small integer.
+
+## Core concepts
+
+**The exponent laws (same base).** These are the whole game. For any nonzero base \(a\):
+
+\[a^{m}\cdot a^{n} = a^{m+n} \qquad \frac{a^{m}}{a^{n}} = a^{m-n} \qquad \left(a^{m}\right)^{n} = a^{mn}\]
+
+Read them out loud: **multiply → add exponents**, **divide → subtract exponents**, **power of a
+power → multiply exponents**. The last one is the single most common slip: \(\left(a^{m}\right)^{n}\)
+is \(a^{mn}\), *not* \(a^{m+n}\).
+
+**Zero and negative exponents.** These are definitions, not something to "compute":
+
+\[a^{0} = 1 \ (a\neq 0) \qquad a^{-n} = \frac{1}{a^{n}}\]
+
+A negative exponent means **reciprocal**, never a negative number. So \(2^{-2} = \tfrac{1}{4}\),
+not \(-4\).
+
+**Roots are fractional exponents.** A root is just an exponent with a fraction:
+
+\[\sqrt[n]{a} = a^{1/n} \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^{m} = \sqrt[n]{a^{m}}\]
+
+So \(27^{2/3} = \left(\sqrt[3]{27}\right)^{2} = 3^{2} = 9\). Take the root **first** (numbers stay
+small), then apply the power.
+
+**Simplifying square roots.** Pull out perfect-square factors:
+
+\[\sqrt{50} = \sqrt{25\cdot 2} = 5\sqrt{2}\]
+
+You can only **add or subtract** roots that already have the same radical: \(5\sqrt{2}+3\sqrt{2} =
+8\sqrt{2}\). But \(\sqrt{a}+\sqrt{b} \neq \sqrt{a+b}\) — you cannot add under the root.
+Multiplication and division *do* combine: \(\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}\) and
+\(\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\tfrac{a}{b}}\).
+
+**Solving exponential equations.** Rewrite both sides with the **same base**, then set the
+exponents equal. If \(9^{x} = 3^{x+4}\), write \(9 = 3^{2}\) so the left side is \(3^{2x}\); then
+\(2x = x+4\), giving \(x = 4\).
+
+## Worked examples
+
+**Combining laws.** \(\dfrac{5^{6}}{5^{2}\cdot 5} = \dfrac{5^{6}}{5^{3}} = 5^{6-3} = 5^{3} = 125.\)
+Note the lone \(5\) is \(5^{1}\) — don't forget to count it.
+
+**Fractional exponent.** \(27^{2/3}\): cube-root first, \(\sqrt[3]{27} = 3\), then square,
+\(3^{2} = 9\). Doing it the other way (\(27^{2} = 729\), then \(\sqrt[3]{729} = 9\)) gives the same
+answer but with uglier numbers.
+
+**Factoring out a power.** \(2^{x+2} + 2^{x} = 2^{x}\left(2^{2}+1\right) = 5\cdot 2^{x}\). If that
+equals \(40\), then \(2^{x} = 8\), so \(x = 3\). When you see a **sum** of powers, factor out the
+smallest one — you cannot add the exponents of a sum.
+
+## Common traps
+
+- **Power of a power vs. product.** \(\left(a^{2}\right)^{3} = a^{6}\), but \(a^{2}\cdot a^{3} = a^{5}\). Multiply vs. add.
+- **Negative exponent ≠ negative number.** \(3^{-1} = \tfrac{1}{3}\), a positive fraction.
+- **\(a^{0} = 1\), not \(0\).** Any nonzero base to the zero power is 1.
+- **Adding under a root.** \(\sqrt{9+16} = \sqrt{25} = 5\), which is *not* \(\sqrt{9}+\sqrt{16} = 3+4 = 7\).
+- **Base times exponent.** \(2^{5}\) is \(32\), not \(2\times 5 = 10\).
+- **Power towers.** \(27^{2}\) means \(27\) squared \(\big(=3^{6}\big)\); it is not \(3^{(3^{2})} = 3^{9}\).
+
+## Key takeaways
+
+- Same base: multiply → add exponents, divide → subtract, power of a power → multiply.
+- \(a^{0}=1\); \(a^{-n}=\tfrac{1}{a^{n}}\) (reciprocal, never negative).
+- Roots are fractional exponents; take the root first to keep numbers small.
+- You can add/subtract roots only when the radical matches; you can never add under the root.
+- Solve exponential equations by rewriting to a common base and equating exponents.

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

@@ -0,0 +1,49 @@
+---
+id: quant-arithmetic-exponents-roots-0017
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: easy
+tags: [exponents, exponent-laws]
+choices:
+  A: "15"
+  B: "25"
+  C: "125"
+  D: "625"
+  E: "3125"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(\dfrac{5^{6}}{5^{2}\cdot 5}\)?
+
+## Explanation
+
+Combine the powers in the denominator first, then subtract exponents.
+
+The denominator is \(5^{2}\cdot 5 = 5^{2}\cdot 5^{1} = 5^{2+1} = 5^{3}\). (The lone \(5\) is
+\(5^{1}\) — it's easy to skip.)
+
+Now divide, subtracting exponents:
+
+\[\frac{5^{6}}{5^{3}} = 5^{6-3} = 5^{3} = 125\]
+
+So the answer is **125**.
+
+**The traps:**
+- **E (3125 = \(5^{5}\))** comes from *adding* the lone \(5\) instead of subtracting it: \(5^{6-2+1}\).
+- **D (625 = \(5^{4}\))** comes from ignoring the lone \(5\) entirely: \(5^{6-2}\).
+- **B (25)** comes from *dividing* the exponents (\(6\div 3\)) instead of subtracting them.
+- **A (15)** comes from reading \(5^{3}\) as \(5\times 3\).
+
+## Hints
+
+- Rewrite the lone \(5\) as \(5^{1}\) and combine the whole denominator into a single power.
+- Dividing powers of the same base means **subtracting** exponents, not dividing them.

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

@@ -0,0 +1,54 @@
+---
+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: "-3"
+  B: "1/4"
+  C: "5/4"
+  D: "3/2"
+  E: "1"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(3^{0} + 2^{-2}\)?
+
+## Explanation
+
+Handle each term with its definition — these are not things to "multiply out."
+
+Any nonzero base raised to the zero power is \(1\):
+
+\[3^{0} = 1\]
+
+A negative exponent means the **reciprocal**, not a negative number:
+
+\[2^{-2} = \frac{1}{2^{2}} = \frac{1}{4}\]
+
+Add them:
+
+\[1 + \frac{1}{4} = \frac{5}{4}\]
+
+So the answer is **5/4**.
+
+**The traps:**
+- **B (1/4)** treats \(3^{0}\) as \(0\) instead of \(1\).
+- **A (−3)** treats the negative exponent as a negative value: \(2^{-2} = -4\), then \(1 + (-4)\).
+- **D (3/2)** takes \(2^{-2} = \tfrac{1}{2}\), forgetting to square the base.
+- **E (1)** correctly gets \(3^{0} = 1\) but drops the \(2^{-2}\) term as if it were \(0\).
+
+## Hints
+
+- \(a^{0} = 1\) for any nonzero \(a\); a zero exponent never gives \(0\).
+- A negative exponent flips the base into a fraction — it never makes the value negative.

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

@@ -0,0 +1,49 @@
+---
+id: quant-arithmetic-exponents-roots-0019
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: medium
+tags: [roots, radicals, simplifying]
+choices:
+  A: "8√2"
+  B: "15√2"
+  C: "14√2"
+  D: "2√17"
+  E: "34√2"
+answer: A
+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 can only add square roots once they share the same radical, so simplify each one first by
+pulling out perfect-square factors.
+
+\[\sqrt{50} = \sqrt{25\cdot 2} = 5\sqrt{2} \qquad \sqrt{18} = \sqrt{9\cdot 2} = 3\sqrt{2}\]
+
+Now both terms are multiples of \(\sqrt{2}\), so add the coefficients:
+
+\[5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\]
+
+So the answer is **8√2**.
+
+**The traps:**
+- **D (2√17)** comes from adding *under* the root: \(\sqrt{50+18} = \sqrt{68} = 2\sqrt{17}\). But \(\sqrt{a}+\sqrt{b}\neq\sqrt{a+b}\).
+- **B (15√2)** comes from *multiplying* the coefficients (\(5\times 3\)) instead of adding them.
+- **C (14√2)** comes from simplifying \(\sqrt{18}\) wrongly as \(9\sqrt{2}\) (dividing the radicand by 2), giving \(5\sqrt{2}+9\sqrt{2}\).
+- **E (34√2)** comes from mis-extracting both roots by dividing the radicand by 2: \(25\sqrt{2}+9\sqrt{2}\).
+
+## Hints
+
+- Factor each radicand into a perfect square times what's left, e.g. \(50 = 25\times 2\).
+- Roots add like "like terms": only combine them when the part under the radical matches.

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

@@ -0,0 +1,49 @@
+---
+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: "6"
+  C: "9"
+  D: "18"
+  E: "27"
+answer: C
+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 packages a root and a power together: the denominator is the root, the
+numerator is the power. Take the **root first** to keep the numbers small:
+
+\[27^{2/3} = \left(\sqrt[3]{27}\right)^{2}\]
+
+The cube root of \(27\) is \(3\), and then:
+
+\[3^{2} = 9\]
+
+So the answer is **9**. (You can also do it the other way — \(27^{2} = 729\), then \(\sqrt[3]{729} = 9\) — but the numbers are uglier.)
+
+**The traps:**
+- **A (3)** stops at the cube root and forgets to apply the power of \(2\).
+- **B (6)** takes the cube root (\(3\)) but then *multiplies* by \(2\) instead of squaring.
+- **D (18)** multiplies the base by the exponent: \(27 \times \tfrac{2}{3}\).
+- **E (27)** ignores the exponent altogether and leaves the base.
+
+## Hints
+
+- Read \(a^{m/n}\) as "take the \(n\)th root, then raise to the \(m\)th power."
+- The denominator of the fraction is the **root**; taking it first keeps the arithmetic easy.

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

@@ -0,0 +1,48 @@
+---
+id: quant-arithmetic-exponents-roots-0021
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: medium
+tags: [exponents, exponent-laws]
+choices:
+  A: "35"
+  B: "96"
+  C: "128"
+  D: "256"
+  E: "512"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(2^{x} = 32\), what is the value of \(2^{x+3}\)?
+
+## Explanation
+
+You don't even need to find \(x\) — split the power using the product rule.
+
+\[2^{x+3} = 2^{x}\cdot 2^{3}\]
+
+You're told \(2^{x} = 32\), and \(2^{3} = 8\), so:
+
+\[2^{x+3} = 32 \times 8 = 256\]
+
+So the answer is **256**. (As a check: \(32 = 2^{5}\), so \(x = 5\) and \(2^{x+3} = 2^{8} = 256\).)
+
+**The traps:**
+- **A (35)** *adds* the exponent as if it were a plain number: \(32 + 3\).
+- **B (96)** multiplies by \(3\) instead of by \(2^{3} = 8\): \(32 \times 3\).
+- **C (128)** multiplies by only \(2^{2} = 4\) (an off-by-one on the exponent): \(32 \times 4\).
+- **E (512)** multiplies by \(2^{4} = 16\), doubling one time too many.
+
+## Hints
+
+- \(2^{x+3}\) is not \(2^{x} + 3\); a sum in the exponent means a **product** of powers.
+- Rewrite \(2^{x+3}\) as \(2^{x}\cdot 2^{3}\) and substitute the value you're given.

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

@@ -0,0 +1,50 @@
+---
+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: "2"
+  B: "4"
+  C: "5"
+  D: "6"
+  E: "8"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(3^{\,x+1} = 27^{2}\), what is the value of \(x\)?
+
+## Explanation
+
+The trick is to write both sides with the **same base** — here, base \(3\).
+
+Since \(27 = 3^{3}\), the right side is a power of a power, so **multiply** the exponents:
+
+\[27^{2} = \left(3^{3}\right)^{2} = 3^{3\times 2} = 3^{6}\]
+
+Now both sides are powers of \(3\), so the exponents must be equal:
+
+\[x + 1 = 6 \quad\Longrightarrow\quad x = 5\]
+
+So the answer is **5**.
+
+**The traps:**
+- **D (6)** solves \(x + 1 = 6\) correctly but reports the exponent, forgetting to subtract \(1\).
+- **B (4)** *adds* the exponents on the right (\(3^{3+2} = 3^{5}\)) instead of multiplying them, giving \(x+1 = 5\).
+- **E (8)** reads \(27^{2}\) as a power tower \(3^{(3^{2})} = 3^{9}\), giving \(x+1 = 9\).
+- **A (2)** ignores the square entirely, treating the right side as \(27 = 3^{3}\), giving \(x+1 = 3\).
+
+## Hints
+
+- Rewrite \(27\) as a power of \(3\) so both sides share a base, then equate the exponents.
+- \(\left(3^{3}\right)^{2}\) means multiply the exponents — it is not \(3^{3+2}\) and not \(3^{3^{2}}\).