ALG-2 Solutions: Exponent Laws and Radicals

Problem 1 — Identifying Exponent Laws (C1)

Variant 0 — Which law justifies \( x^5 \cdot x^3 = x^8 \)? Answer: Product rule.

The product rule states \( a^m \cdot a^n = a^{m+n} \). When two powers with the same base are multiplied, the exponents add: \( 5 + 3 = 8 \).

• Quotient rule: applies to division (\( a^m / a^n = a^{m-n} \)), not multiplication.
• Power of a power: applies to \( (a^m)^n \) — an exponent applied to an already-exponentiated expression.
• Power of a product: applies when a product \( (ab)^n \) is raised to a power — two different bases under one exponent.

Variant 1 — Which law justifies \( a^8 / a^3 = a^5 \)? Answer: Quotient rule.

The quotient rule states \( a^m / a^n = a^{m-n} \). Dividing same-base powers subtracts the exponents: \( 8 - 3 = 5 \).

Variant 2 — Which law justifies \( (b^3)^4 = b^12 \)? Answer: Power of a power.

The power-of-a-power rule states \( (a^m)^n = a^{mn} \). When an exponent is raised to another exponent, multiply: \( 3 \times 4 = 12 \).

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Problem 2 — Evaluating Rational Exponents (C2)

Variant 0 — \( 27^0.3333333333333333 \): Answer: 3.

\( 27^{1/3} = \sqrt[3]{27} = 3 \) because \( 3^3 = 27 \).

• 9: confuses the exponent 1/3 with dividing by 3 (gives 9, not 3).
• 1/27: would be \( 27^-1 \) — the reciprocal, not the cube root.
• 81: is \( 3^4 \) — unrelated to this calculation.

Variant 1 — \( 16^0.75 \): Answer: 8.

Strategy: take the 4th root first (keeps numbers small), then cube. \( 16^{3/4} = (\sqrt[4]{16})^3 = 2^3 = 8 \).

• 12: multiplies base by exponent (16 × 3/4 = 12) — incorrect use of the exponent.
• 4: gives only \( 16^0.25 \) — the 4th root without the cube.
• 2: gives only \( \\sqrt[4]16 \) — the 4th root, stops before cubing.

Variant 2 — \( 1/x^3 = x^? \): Answer: \( x^{-3} \).

The negative exponent rule: \( a^{-n} = 1/a^n \), so \( 1/x^3 = x^{-3} \).

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Problem 3 — Freshman's Dream (C1, C3)

Correct answer: C — \( \sqrt{4 \cdot 9} = \sqrt{4} \cdot \sqrt{9} = 6 \).

• A: \( \sqrt{9+16} = \sqrt{25} = 5 \neq 7 \). The product rule does not apply to sums.
• B: \( (x+y)^2 = x^2 + 2xy + y^2 \neq x^2 + y^2 \). The middle term \( 2xy \) is always missing from the "Freshman's Dream" version.
• C: Correct. \( \sqrt{4 \cdot 9} = \sqrt{36} = 6 \), and the product rule confirms \( \sqrt{4} \cdot \sqrt{9} = 2 \cdot 3 = 6 \). Radicals split over products.
• D: \( \sqrt{x^2+y^2} \neq x+y \) in general. Counterexample: \( x=3, y=4 \) gives \( \sqrt{9+16} = 5 \neq 7 \).

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Problem 4 — Rationalizing with a Conjugate (C4)

Correct answer: B — multiply by the conjugate \( (\sqrt{3}-1)/(\sqrt{3}-1) \) to get \( 3(\sqrt{3}-1) = 3\sqrt{3}-3 \).

Full calculation:

\[ \frac{6}{\sqrt{3}+1} \cdot \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{6(\sqrt{3}-1)}{(\sqrt{3})^2 - 1^2} = \frac{6(\sqrt{3}-1)}{3-1} = \frac{6(\sqrt{3}-1)}{2} = 3(\sqrt{3}-1) \]

• A: Multiplying by the same expression (not the conjugate) gives \( (\sqrt{3}+1)^2 \) in the denominator — still has a radical.
• C: Multiplying only by \( \sqrt{3}/\sqrt{3} \) doesn't remove the \( +1 \) term from the denominator.
• D: Simply rewriting the denominator sign doesn't preserve the value of the fraction.

Problem 1 — Simplify Using Exponent Laws (Generator)

Generator problems vary, but all follow this template: simplify \( v^m \cdot v^n / v^p \).

Method:

1. Product rule first: \( v^m \cdot v^n = v^{m+n} \)
2. Quotient rule: \( v^{m+n} / v^p = v^{m+n-p} \)

The generator is designed so \( m + n - p \geq 1 \), so no negative exponents appear in the answer. If your answer has a negative exponent, recheck the subtraction.

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Problem 2 — Evaluate a Rational Exponent (Generator)

Generator problems use this method: \( a^{m/n} = (\sqrt[n]{a})^m \). Always take the root first to keep numbers manageable.

All generated cases have integer answers. If your computation produces a non-integer, recheck which value is the root index (denominator of the exponent) and which is the power (numerator).

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Problem 3 — Simplify a Radical Expression (C3)

Variant 0: \( \sqrt{50} \)

\( 50 = 25 \cdot 2 \), so \( \sqrt{50} = 5\sqrt{2} \).

Variant 1: \( \sqrt{72} \)

\( 72 = 36 \cdot 2 \), so \( \sqrt{72} = 6\sqrt{2} \).

Alternatively: \( 72 = 4 \cdot 18 = 4 \cdot 9 \cdot 2 \), giving \( \sqrt{4} \cdot \sqrt{9} \cdot \sqrt{2} = 2 \cdot 3 \cdot \sqrt{2} = 6\sqrt{2} \). Same answer.

Variant 2: \( \sqrt{48} \)

\( 48 = 16 \cdot 3 \), so \( \sqrt{48} = 4\sqrt{3} \).

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Problem 4 — Rationalize the Denominator (C4)

Variant 0: \( 4/(\sqrt{7}+1) \)

\[ \frac{4}{\sqrt{7}+1} \cdot \frac{\sqrt{7}-1}{\sqrt{7}-1} = \frac{4(\sqrt{7}-1)}{7-1} = \frac{4(\sqrt{7}-1)}{6} = \frac{2(\sqrt{7}-1)}{3} \]

Variant 1: \( 6/(\sqrt{5}+\sqrt{2}) \)

\[ \frac{6}{\sqrt{5}+\sqrt{2}} \cdot \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} = \frac{6(\sqrt{5}-\sqrt{2})}{5-2} = 2(\sqrt{5}-\sqrt{2}) \]

Variant 2: \( 1/(\sqrt{3}-1) \)

\[ \frac{1}{\sqrt{3}-1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{\sqrt{3}+1}{3-1} = \frac{\sqrt{3}+1}{2} \]

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Problem 5 — Factoring with Fractional Exponents (C2, C3)

Simplify \( x^{-1/2} + x^{1/2} \) by factoring out \( x^{-1/2} \).

\[ x^{-1/2} + x^{1/2} = x^{-1/2}(1 + x^{1/2 - (-1/2)}) = x^{-1/2}(1 + x) = \frac{1+x}{\sqrt{x}} \]

Checking the exponent: factoring \( x^-0.5 \) from \( x^0.5 \) leaves \( x^1 = x^1 = x \). ✓

Why does this matter in calculus? When differentiating a product using the product rule, you frequently get two terms with fractional exponents like \( x^-0.5 + x^0.5 \). Factoring and simplifying them is required before your answer is in acceptable form.

Feynman Test — Model Explanation

A strong explanation connects the quotient rule to the zero-exponent rule:

"The quotient rule says \( a^m / a^n = a^{m-n} \). If I let \( m = 0 \), I get \( a^0 / a^n = a^{-n} \). But \( a^0 = 1 \) from the zero-exponent rule, so \( 1/a^n = a^{-n} \). This forces the definition — there's no other value \( a^{-n} \) could have if the quotient rule is going to be consistent."

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Apply — Simplify \( (2x^-1y^2)^3 / (4x^2 y^-1) \)

Correct answer: \( 2y^7/x^5 \).

Full solution:

Step 1 (power of a product): \( (2x^{-1}y^2)^3 = 2^3 (x^{-1})^3 (y^2)^3 = 8 x^{-3} y^6 \)

Step 2 (divide): \( \dfrac{8 x^{-3} y^6}{4 x^2 y^{-1}} = 2 \cdot x^{-3-2} \cdot y^{6-(-1)} = 2 x^{-5} y^7 \)

Step 3 (positive exponents): \( 2 x^{-5} y^7 = \dfrac{2y^7}{x^5} \)

Common errors in this problem:

• Misapplying the \( y \) quotient step: \( y^6 / y^{-1} = y^{6-(-1)} = y^7 \), not \( y^5 \). Dividing by \( y^{-1} \) is the same as multiplying by \( y^1 \).
• Forgetting to apply the power of a product to the coefficient: \( 2^3 = 8 \), not \( 2 \cdot 3 = 6 \).

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Analyze — Error in \( \sqrt{x^2+4} \)

The error: The student wrote \( \sqrt{x^2+4} = \sqrt{x^2} + \sqrt{4} \). This applies the product rule for radicals \( \sqrt{ab} = \sqrt{a} \cdot \sqrt{b} \) to a sum — which is the Freshman's Dream error.

Correct value at \( x = 3 \): \( \sqrt{3^2 + 4} = \sqrt{9+4} = \sqrt{13} \approx 3.606 \), not 5.

The expression \( \sqrt{x^2+4} \) cannot be simplified because \( x^2 \) and \( 4 \) are added, not multiplied. No factorization removes the radical.

Path A — The Analyst 🔬

Simplify \( \dfrac{(4x^{3/2} y^{-1})^2}{2x^2 \cdot \sqrt{y^3}} \).

Step 1 — Power of a product: \( (4x^{3/2}y^{-1})^2 = 16 x^3 y^{-2} \)

Step 2 — Convert radical: \( \sqrt{y^3} = y^{3/2} \)

Step 3 — Denominator: \( 2x^2 \cdot y^{3/2} \)

Step 4 — Quotient rule: \( \dfrac{16x^3 y^{-2}}{2x^2 y^{3/2}} = 8 x^{3-2} y^{-2-3/2} = 8x y^{-7/2} \)

Check exponent on y: \( -2 - 3/2 = -4/2 - 3/2 = -7/2 \) ✓

Step 5 — Positive exponents: \( 8x y^{-7/2} = \dfrac{8x}{y^{7/2}} \)

Step 6 — Rationalize denominator: \( y^{7/2} = y^3 \sqrt{y} \), so \( \dfrac{8x}{y^3\sqrt{y}} \cdot \dfrac{\sqrt{y}}{\sqrt{y}} = \dfrac{8x\sqrt{y}}{y^4} \) \( \checkmark \)

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Path B — The Architect 🏗️

Simplify \( \dfrac{\sqrt{x+h} - \sqrt{x}}{h} \) and evaluate the limit as \( h \to 0 \).

Step 1: Conjugate of \( (\sqrt{x+h} - \sqrt{x}) \) is \( (\sqrt{x+h} + \sqrt{x}) \).

Step 2: Multiply numerator and denominator by the conjugate. Numerator: difference of squares gives \( (x+h) - x = h \). Denominator: \( h(\sqrt{x+h} + \sqrt{x}) \).

Step 3: Cancel \( h \): \( \dfrac{1}{\sqrt{x+h} + \sqrt{x}} \).

Step 4: Substitute \( h = 0 \): \( \dfrac{1}{2\sqrt{x}} \). This is the derivative of \( \sqrt{x} \). \( \checkmark \)

Challenge 1 — The Freshman's Dream

(a) \( (a+b)^2 = a^2 + 2ab + b^2 \).

(b) Equality \( (a+b)^2 = a^2 + b^2 \) holds only when \( 2ab = 0 \), i.e., \( a = 0 \) or \( b = 0 \).

(c) The power-of-a-product rule \( (ab)^n = a^n b^n \) works because each factor is independent — multiplying \( a \) by itself \( n \) times and \( b \) by itself \( n \) times gives the same result as multiplying \( ab \) by itself \( n \) times. When you have a sum \( (a+b)^n \), the factors are no longer independent: each time you multiply \( (a+b) \) by itself, you get cross-products (\( ab \) terms) that can't be eliminated or ignored. The distributive property forces them into the expansion.

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Challenge 2 — Fractional Exponent Simplification

\( \dfrac{x^{-1/2} + x^{1/2}}{x^{3/2}} \)

Step 1 — Factor numerator: \( x^{-1/2}(1+x) \)

Step 2 — Divide: \( \dfrac{x^{-1/2}(1+x)}{x^{3/2}} = (1+x) \cdot x^{-1/2 - 3/2} = (1+x) x^{-2} = \dfrac{1+x}{x^2} \)

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Challenge 3 — Cube-Root Difference Quotient

Using \( A^3 - B^3 = (A-B)(A^2+AB+B^2) \) with \( A = (x+h)^{1/3} \), \( B = x^{1/3} \):

Multiply by \( \dfrac{A^2+AB+B^2}{A^2+AB+B^2} \). The numerator becomes \( A^3 - B^3 = (x+h) - x = h \). Cancel \( h \):

\[ \frac{1}{(x+h)^{2/3} + (x+h)^{1/3} x^{1/3} + x^{2/3}} \]

As \( h \to 0 \): denominator \( \to 3x^{2/3} \), so the limit is \( \dfrac{1}{3x^{2/3}} = \dfrac{1}{3} x^{-2/3} \). This is the derivative of \( x^{1/3} \) — consistent with the power rule \( \frac{d}{dx}[x^n] = nx^{n-1} \) at \( n = 1/3 \). \( \checkmark \)