Exponents and Radicals

Master exponent rules and radical simplification for SAT

🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Start Interactive Lesson →

Exponents and Radicals (SAT)

Exponent Rules

Product Rule

am⋅an=am+na^m \cdot a^n = a^{m+n}

Quotient Rule

aman=am−n\frac{a^m}{a^n} = a^{m-n}

Power Rule

(am)n=amn(a^m)^n = a^{mn}

Negative Exponents

a−n=1ana^{-n} = \frac{1}{a^n}

Zero Exponent

a0=1a^0 = 1 (if a≠0a \neq 0)

Radicals

Simplifying

50=25⋅2=52\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}

Converting Between Forms

amn=am/n\sqrt[n]{a^m} = a^{m/n}

Examples:

  • x=x1/2\sqrt{x} = x^{1/2}
  • x23=x2/3\sqrt[3]{x^2} = x^{2/3}

SAT Tricks

  • Fractional exponents: x3/2=(x)3=x3x^{3/2} = (\sqrt{x})^3 = \sqrt{x^3}
  • Rationalizing: 12=22\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}
  • Watch for: (2x)3=8x3(2x)^3 = 8x^3, not 2x32x^3!

📚 Practice Problems

1Problem 1easy

❓ Question:

Simplify: x5⋅x3x^5 \cdot x^3

💡 Show Solution

Solution:

Use product rule (add exponents): x5⋅x3=x5+3=x8x^5 \cdot x^3 = x^{5+3} = x^8

Answer: x8x^8

2Problem 2medium

❓ Question:

Simplify: 72\sqrt{72}

💡 Show Solution

Solution:

Factor to find perfect squares: 72=36⋅2\sqrt{72} = \sqrt{36 \cdot 2}

=36⋅2= \sqrt{36} \cdot \sqrt{2}

=62= 6\sqrt{2}

Answer: 626\sqrt{2}

3Problem 3hard

❓ Question:

Simplify: (x3)4x7\frac{(x^3)^4}{x^7}

💡 Show Solution

Solution:

Step 1: Power rule in numerator (x3)4=x12(x^3)^4 = x^{12}

Step 2: Quotient rule x12x7=x12−7=x5\frac{x^{12}}{x^7} = x^{12-7} = x^5

Answer: x5x^5

SAT Tip: Apply power rule before quotient rule!

4Problem 4easy

❓ Question:

Simplify: x5x2\frac{x^5}{x^2}

💡 Show Solution

Rule: xaxb=xa−b\frac{x^a}{x^b} = x^{a-b}

x5x2=x5−2=x3\frac{x^5}{x^2} = x^{5-2} = x^3

Answer: x3x^3

5Problem 5easy

❓ Question:

Simplify: x5x2\frac{x^5}{x^2}

💡 Show Solution

Rule: xaxb=xa−b\frac{x^a}{x^b} = x^{a-b}

x5x2=x5−2=x3\frac{x^5}{x^2} = x^{5-2} = x^3

Answer: x3x^3

6Problem 6medium

❓ Question:

Simplify: (2x3)4(2x^3)^4

💡 Show Solution

Rule: (ab)n=an⋅bn(ab)^n = a^n \cdot b^n and (xa)b=xab(x^a)^b = x^{ab}

(2x3)4=24⋅(x3)4=16⋅x12=16x12(2x^3)^4 = 2^4 \cdot (x^3)^4 = 16 \cdot x^{12} = 16x^{12}

Answer: 16x1216x^{12}

Common mistake: Forgetting to raise the coefficient (2) to the power as well.

7Problem 7medium

❓ Question:

Simplify: (2x3)4(2x^3)^4

💡 Show Solution

Rule: (ab)n=an⋅bn(ab)^n = a^n \cdot b^n and (xa)b=xab(x^a)^b = x^{ab}

(2x3)4=24⋅(x3)4=16⋅x12=16x12(2x^3)^4 = 2^4 \cdot (x^3)^4 = 16 \cdot x^{12} = 16x^{12}

Answer: 16x1216x^{12}

Common mistake: Forgetting to raise the coefficient (2) to the power as well.

8Problem 8medium

❓ Question:

Rewrite x53\sqrt[3]{x^5} using rational exponents.

💡 Show Solution

Rule: xmn=xm/n\sqrt[n]{x^m} = x^{m/n}

x53=x5/3\sqrt[3]{x^5} = x^{5/3}

Check: x5/3=x1+2/3=x⋅x2/3=xx23x^{5/3} = x^{1 + 2/3} = x \cdot x^{2/3} = x\sqrt[3]{x^2}

Answer: x5/3x^{5/3}

SAT Tip: The SAT frequently asks you to convert between radical and exponential notation.

9Problem 9medium

❓ Question:

Rewrite x53\sqrt[3]{x^5} using rational exponents.

💡 Show Solution

Rule: xmn=xm/n\sqrt[n]{x^m} = x^{m/n}

x53=x5/3\sqrt[3]{x^5} = x^{5/3}

Check: x5/3=x1+2/3=x⋅x2/3=xx23x^{5/3} = x^{1 + 2/3} = x \cdot x^{2/3} = x\sqrt[3]{x^2}

Answer: x5/3x^{5/3}

SAT Tip: The SAT frequently asks you to convert between radical and exponential notation.

10Problem 10hard

❓ Question:

If 27x=9x+127^x = 9^{x+1}, what is the value of xx?

💡 Show Solution

Strategy: Express both sides as powers of 3.

27=3327 = 3^3, so 27x=(33)x=33x27^x = (3^3)^x = 3^{3x} 9=329 = 3^2, so 9x+1=(32)x+1=32(x+1)=32x+29^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x+2}

Set the exponents equal (same base): 3x=2x+23x = 2x + 2 x=2x = 2

Check: 272=72927^2 = 729 and 93=7299^3 = 729 ✓

Answer: x=2x = 2

Strategy: When you have exponential equations, try to make the bases the same, then set exponents equal.

11Problem 11hard

❓ Question:

If 27x=9x+127^x = 9^{x+1}, what is the value of xx?

💡 Show Solution

Strategy: Express both sides as powers of 3.

27=3327 = 3^3, so 27x=(33)x=33x27^x = (3^3)^x = 3^{3x} 9=329 = 3^2, so 9x+1=(32)x+1=32(x+1)=32x+29^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x+2}

Set the exponents equal (same base): 3x=2x+23x = 2x + 2 x=2x = 2

Check: 272=72927^2 = 729 and 93=7299^3 = 729 ✓

Answer: x=2x = 2

Strategy: When you have exponential equations, try to make the bases the same, then set exponents equal.

12Problem 12expert

❓ Question:

Simplify: (3x2y−1)39x−1y4\frac{(3x^2y^{-1})^3}{9x^{-1}y^4}

💡 Show Solution

Step 1: Simplify the numerator: (3x2y−1)3=33⋅x2⋅3⋅y−1⋅3=27x6y−3(3x^2y^{-1})^3 = 3^3 \cdot x^{2 \cdot 3} \cdot y^{-1 \cdot 3} = 27x^6y^{-3}

Step 2: Write the full fraction: 27x6y−39x−1y4\frac{27x^6y^{-3}}{9x^{-1}y^4}

Step 3: Simplify coefficients: 279=3\frac{27}{9} = 3

Step 4: Apply quotient rule for each variable: x6−(−1)=x7x^{6-(-1)} = x^7 y−3−4=y−7=1y7y^{-3-4} = y^{-7} = \frac{1}{y^7}

Step 5: Combine: 3⋅x7⋅y−7=3x7y73 \cdot x^7 \cdot y^{-7} = \frac{3x^7}{y^7}

Answer: 3x7y7\frac{3x^7}{y^7}

13Problem 13expert

❓ Question:

Simplify: (3x2y−1)39x−1y4\frac{(3x^2y^{-1})^3}{9x^{-1}y^4}

💡 Show Solution

Step 1: Simplify the numerator: (3x2y−1)3=33⋅x2⋅3⋅y−1⋅3=27x6y−3(3x^2y^{-1})^3 = 3^3 \cdot x^{2 \cdot 3} \cdot y^{-1 \cdot 3} = 27x^6y^{-3}

Step 2: Write the full fraction: 27x6y−39x−1y4\frac{27x^6y^{-3}}{9x^{-1}y^4}

Step 3: Simplify coefficients: 279=3\frac{27}{9} = 3

Step 4: Apply quotient rule for each variable: x6−(−1)=x7x^{6-(-1)} = x^7 y−3−4=y−7=1y7y^{-3-4} = y^{-7} = \frac{1}{y^7}

Step 5: Combine: 3⋅x7⋅y−7=3x7y73 \cdot x^7 \cdot y^{-7} = \frac{3x^7}{y^7}

Answer: 3x7y7\frac{3x^7}{y^7}

← Previous Topic
Quadratic Equations
Next Topic →
Polynomial and Rational Expressions
🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore more SAT Prep topics