8th grade math (H) - Exponent Practice Sept 21 - Free Printable

Problem: Simplify the given expressions so that your answer contains only positive exponents.

We will solve each problem step by step using the rules of exponents. The key rules we will use are:

1. Quotient Rule: \( \frac{a^m}{a^n} = a^{m-n} \)
2. Negative Exponent Rule: \( a^{-n} = \frac{1}{a^n} \)
3. Product Rule: \( a^m \cdot a^n = a^{m+n} \)
4. Power of a Power Rule: \( (a^m)^n = a^{m \cdot n} \)

---

#### Problem 1: \( \frac{2^3}{2^2} \)

Using the Quotient Rule:
\[
\frac{2^3}{2^2} = 2^{3-2} = 2^1 = 2
\]

Answer: \( 2 \)

---

#### Problem 2: \( \frac{3}{4^4} \)

This expression is already in its simplest form with positive exponents. No further simplification is needed.

Answer: \( \frac{3}{4^4} \)

---

#### Problem 3: \( \frac{4^0}{4^4} \)

Using the Zero Exponent Rule (\( 4^0 = 1 \)) and the Quotient Rule:
\[
\frac{4^0}{4^4} = \frac{1}{4^4} = 4^{-4}
\]
Since we need positive exponents, we rewrite it as:
\[
4^{-4} = \frac{1}{4^4}
\]

Answer: \( \frac{1}{4^4} \)

---

#### Problem 4: \( \frac{3^0}{3^4} \)

Using the Zero Exponent Rule (\( 3^0 = 1 \)) and the Quotient Rule:
\[
\frac{3^0}{3^4} = \frac{1}{3^4} = 3^{-4}
\]
Since we need positive exponents, we rewrite it as:
\[
3^{-4} = \frac{1}{3^4}
\]

Answer: \( \frac{1}{3^4} \)

---

#### Problem 5: \( \frac{2^4}{2^0} \)

Using the Zero Exponent Rule (\( 2^0 = 1 \)) and the Quotient Rule:
\[
\frac{2^4}{2^0} = \frac{2^4}{1} = 2^4
\]

Answer: \( 2^4 \)

---

#### Problem 6: \( \frac{4a}{ab^3} \)

Simplify by canceling common factors:
\[
\frac{4a}{ab^3} = \frac{4}{b^3}
\]

Answer: \( \frac{4}{b^3} \)

---

#### Problem 7: \( \frac{4x^{-1}}{x^3 y^0} \)

Using the Zero Exponent Rule (\( y^0 = 1 \)) and the Negative Exponent Rule:
\[
\frac{4x^{-1}}{x^3 y^0} = \frac{4x^{-1}}{x^3 \cdot 1} = \frac{4x^{-1}}{x^3} = 4x^{-1-3} = 4x^{-4}
\]
Rewrite with positive exponents:
\[
4x^{-4} = \frac{4}{x^4}
\]

Answer: \( \frac{4}{x^4} \)

---

#### Problem 8: \( \frac{4x^2}{2x^4} \)

Simplify by canceling common factors and using the Quotient Rule:
\[
\frac{4x^2}{2x^4} = \frac{4}{2} \cdot \frac{x^2}{x^4} = 2 \cdot x^{2-4} = 2x^{-2}
\]
Rewrite with positive exponents:
\[
2x^{-2} = \frac{2}{x^2}
\]

Answer: \( \frac{2}{x^2} \)

---

#### Problem 9: \( \frac{4m^4}{m^{-3} n^2} \)

Using the Negative Exponent Rule and the Quotient Rule:
\[
\frac{4m^4}{m^{-3} n^2} = 4 \cdot \frac{m^4}{m^{-3}} \cdot \frac{1}{n^2} = 4 \cdot m^{4 - (-3)} \cdot \frac{1}{n^2} = 4 \cdot m^{4+3} \cdot \frac{1}{n^2} = 4m^7 \cdot \frac{1}{n^2} = \frac{4m^7}{n^2}
\]

Answer: \( \frac{4m^7}{n^2} \)

---

#### Problem 10: \( \frac{m^{-2} n^{-3}}{2m^2} \)

Using the Negative Exponent Rule and the Quotient Rule:
\[
\frac{m^{-2} n^{-3}}{2m^2} = \frac{1}{2} \cdot \frac{m^{-2}}{m^2} \cdot n^{-3} = \frac{1}{2} \cdot m^{-2-2} \cdot n^{-3} = \frac{1}{2} \cdot m^{-4} \cdot n^{-3}
\]
Rewrite with positive exponents:
\[
\frac{1}{2} \cdot m^{-4} \cdot n^{-3} = \frac{1}{2m^4 n^3}
\]

Answer: \( \frac{1}{2m^4 n^3} \)

---

#### Problem 11: \( \frac{2x^6}{3x^4 y^2} \)

Simplify by canceling common factors and using the Quotient Rule:
\[
\frac{2x^6}{3x^4 y^2} = \frac{2}{3} \cdot \frac{x^6}{x^4} \cdot \frac{1}{y^2} = \frac{2}{3} \cdot x^{6-4} \cdot \frac{1}{y^2} = \frac{2}{3} \cdot x^2 \cdot \frac{1}{y^2} = \frac{2x^2}{3y^2}
\]

Answer: \( \frac{2x^2}{3y^2} \)

---

#### Problem 12: \( \frac{3x^2 y^3}{3x^2 y^2} \)

Simplify by canceling common factors:
\[
\frac{3x^2 y^3}{3x^2 y^2} = \frac{3}{3} \cdot \frac{x^2}{x^2} \cdot \frac{y^3}{y^2} = 1 \cdot 1 \cdot y^{3-2} = y^1 = y
\]

Answer: \( y \)

---

#### Problem 13: \( \frac{3x^4 y^4}{4x^4 y^4} \)

Simplify by canceling common factors:
\[
\frac{3x^4 y^4}{4x^4 y^4} = \frac{3}{4} \cdot \frac{x^4}{x^4} \cdot \frac{y^4}{y^4} = \frac{3}{4} \cdot 1 \cdot 1 = \frac{3}{4}
\]

Answer: \( \frac{3}{4} \)

---

#### Problem 14: \( \frac{4x^{-4}}{4y^{-1}} \)

Using the Negative Exponent Rule:
\[
\frac{4x^{-4}}{4y^{-1}} = \frac{4}{4} \cdot \frac{x^{-4}}{y^{-1}} = 1 \cdot x^{-4} \cdot y^1 = \frac{y}{x^4}
\]

Answer: \( \frac{y}{x^4} \)

---

#### Problem 15: \( \frac{4u^4 v^{-1}}{4v} \)

Simplify by canceling common factors and using the Negative Exponent Rule:
\[
\frac{4u^4 v^{-1}}{4v} = \frac{4}{4} \cdot \frac{u^4}{1} \cdot \frac{v^{-1}}{v} = 1 \cdot u^4 \cdot v^{-1-1} = u^4 \cdot v^{-2}
\]
Rewrite with positive exponents:
\[
u^4 \cdot v^{-2} = \frac{u^4}{v^2}
\]

Answer: \( \frac{u^4}{v^2} \)

---

#### Problem 16: \( \frac{2x^4}{2x^4 y^{-4}} \)

Using the Negative Exponent Rule and the Quotient Rule:
\[
\frac{2x^4}{2x^4 y^{-4}} = \frac{2}{2} \cdot \frac{x^4}{x^4} \cdot \frac{1}{y^{-4}} = 1 \cdot 1 \cdot y^4 = y^4
\]

Answer: \( y^4 \)

---

#### Problem 17: \( \frac{4a^{-1} v^2}{4a^{-1} v^{-1} \cdot u^{-4}} \)

Simplify step by step:
\[
\frac{4a^{-1} v^2}{4a^{-1} v^{-1} \cdot u^{-4}} = \frac{4}{4} \cdot \frac{a^{-1}}{a^{-1}} \cdot \frac{v^2}{v^{-1}} \cdot \frac{1}{u^{-4}} = 1 \cdot 1 \cdot v^{2 - (-1)} \cdot u^4 = v^{2+1} \cdot u^4 = v^3 u^4
\]

Answer: \( v^3 u^4 \)

---

#### Problem 18: \( \frac{x^{-2} y^2 - 3x^3 y^2}{4yx^2} \)

Split the fraction into two terms:
\[
\frac{x^{-2} y^2 - 3x^3 y^2}{4yx^2} = \frac{x^{-2} y^2}{4yx^2} - \frac{3x^3 y^2}{4yx^2}
\]

Simplify each term separately:
1. For \( \frac{x^{-2} y^2}{4yx^2} \):
\[
\frac{x^{-2} y^2}{4yx^2} = \frac{1}{4} \cdot \frac{x^{-2}}{x^2} \cdot \frac{y^2}{y} = \frac{1}{4} \cdot x^{-2-2} \cdot y^{2-1} = \frac{1}{4} \cdot x^{-4} \cdot y^1 = \frac{y}{4x^4}
\]

2. For \( \frac{3x^3 y^2}{4yx^2} \):
\[
\frac{3x^3 y^2}{4yx^2} = \frac{3}{4} \cdot \frac{x^3}{x^2} \cdot \frac{y^2}{y} = \frac{3}{4} \cdot x^{3-2} \cdot y^{2-1} = \frac{3}{4} \cdot x^1 \cdot y^1 = \frac{3xy}{4}
\]

Combine the results:
\[
\frac{x^{-2} y^2 - 3x^3 y^2}{4yx^2} = \frac{y}{4x^4} - \frac{3xy}{4}
\]

Answer: \( \frac{y}{4x^4} - \frac{3xy}{4} \)

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#### Problem 19: \( \frac{4a^2 b^2}{2a^2 b^2 \cdot 2b^4} \)

Simplify by canceling common factors:
\[
\frac{4a^2 b^2}{2a^2 b^2 \cdot 2b^4} = \frac{4}{2 \cdot 2} \cdot \frac{a^2}{a^2} \cdot \frac{b^2}{b^2} \cdot \frac{1}{b^4} = \frac{4}{4} \cdot 1 \cdot 1 \cdot b^{-4} = 1 \cdot b^{-4} = b^{-4}
\]
Rewrite with positive exponents:
\[
b^{-4} = \frac{1}{b^4}
\]

Answer: \( \frac{1}{b^4} \)

---

#### Problem 20: \( \frac{4x^4 y^6}{2x^4 y^2 \cdot x^2 y^2} \)

Simplify the denominator first:
\[
2x^4 y^2 \cdot x^2 y^2 = 2 \cdot x^{4+2} \cdot y^{2+2} = 2x^6 y^4
\]

Now simplify the fraction:
\[
\frac{4x^4 y^6}{2x^6 y^4} = \frac{4}{2} \cdot \frac{x^4}{x^6} \cdot \frac{y^6}{y^4} = 2 \cdot x^{4-6} \cdot y^{6-4} = 2 \cdot x^{-2} \cdot y^2
\]
Rewrite with positive exponents:
\[
2 \cdot x^{-2} \cdot y^2 = \frac{2y^2}{x^2}
\]

Answer: \( \frac{2y^2}{x^2} \)

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Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & 2 \\
2. & \frac{3}{4^4} \\
3. & \frac{1}{4^4} \\
4. & \frac{1}{3^4} \\
5. & 2^4 \\
6. & \frac{4}{b^3} \\
7. & \frac{4}{x^4} \\
8. & \frac{2}{x^2} \\
9. & \frac{4m^7}{n^2} \\
10. & \frac{1}{2m^4 n^3} \\
11. & \frac{2x^2}{3y^2} \\
12. & y \\
13. & \frac{3}{4} \\
14. & \frac{y}{x^4} \\
15. & \frac{u^4}{v^2} \\
16. & y^4 \\
17. & v^3 u^4 \\
18. & \frac{y}{4x^4} - \frac{3xy}{4} \\
19. & \frac{1}{b^4} \\
20. & \frac{2y^2}{x^2}
\end{array}
}
\]