8th Grade Exponent Practice Worksheet - Simplify expressions with positive exponents.

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There are 20 problems on this worksheet. Each asks to simplify an expression with exponents, using only positive exponents in the answer.

I’ll go one by one, applying exponent rules:

- When dividing powers with same base: subtract exponents → a^m / a^n = a^(m−n)
- Negative exponent? Move it to the other side of the fraction → a^(-n) = 1/a^n
- Coefficients (numbers) are handled separately from variables.
- If a variable is in numerator and denominator, subtract exponents accordingly.

Let’s solve each:

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1) 2³ / 2⁷ = 2^(3−7) = 2^(-4) = 1/2⁴ → 1/16

2) 3 / 3⁵ = 3¹ / 3⁵ = 3^(1−5) = 3^(-4) = 1/3⁴ → 1/81

3) 4⁶ / 4¹ = 4^(6−1) = 4⁵ → 1024 (but since it says “only positive exponents”, we leave as 4⁵ unless asked to evaluate — but typically in these worksheets, they want simplified exponential form unless specified. Let me check instruction: “Your answer should contain only positive exponents.” So 4⁵ is fine. But sometimes they expect evaluated. Looking at problem 1, 2³/2⁷ → 1/16 is evaluated. Hmm. Actually, let's see: Problem 1: 2³/2⁷ = 1/2⁴ = 1/16 — they probably want numerical if simple. But for larger numbers like 4⁵, maybe leave as power? Wait, no — in middle school math, they often leave as power if not too big. But 4⁵ is 1024 — acceptable. However, looking at problem 4: 3¹⁰ / 3⁵ = 3⁵ — that’s 243. But again, perhaps they want exponential form. Let me re-read: “Simplify. Your answer should contain only positive exponents.” So as long as no negative exponents, it’s ok. So 4⁵ is acceptable. But to be safe, I’ll write both forms? No — better to follow typical worksheet style. In most such worksheets, they leave as power unless it’s very small. For example, 2⁴ = 16 is written, but 4⁵ might be left as 4⁵. Actually, let’s look at problem 5: 2⁴ / 2⁰ = 2⁴ / 1 = 16 — so they evaluate when easy. But 4⁵ is 1024 — still manageable. I think for consistency, I’ll evaluate when the result is a small integer, otherwise leave as power. But actually, in problem 1, 2^(-4) became 1/16 — evaluated. Similarly, problem 2: 1/81. So I think they want evaluated form if possible. But 4⁵ is 1024 — I’ll write 1024. Wait, but let’s check problem 6: 4a / ab³ — here variables are involved, so can’t evaluate. So for pure numbers, evaluate; for variables, leave in exponent form.

Actually, let me standardize: For expressions with only numbers, compute the value. For expressions with variables, simplify using exponent rules and leave in exponential form with positive exponents.

So:

1) 2³/2⁷ = 1/2⁴ = 1/16 → 1/16

2) 3/3⁵ = 1/3⁴ = 1/81 → 1/81

3) 4⁶/4¹ = 4⁵ = 1024 → 1024

4) 3¹⁰/3⁵ = 3⁵ = 243 → 243

5) 2⁴/2⁰ = 2⁴ / 1 = 16 → 16 (since any number to 0 power is 1)

6) 4a / ab³ = 4 / b³ → 4/b³ (a cancels)

7) 4x⁻¹ / x³y⁰ = 4 / (x * x³ * 1) = 4 / x⁴ → 4/x⁴ (y⁰=1, x⁻¹ in numerator becomes x in denominator)

8) 4x² / 2x⁴ = (4/2) * x^(2-4) = 2 * x^(-2) = 2/x² → 2/x²

9) 4m⁴ / m⁻²n² = 4 m^(4 - (-2)) / n² = 4 m⁶ / n² → 4m⁶/n²

10) m⁻²n⁻³ / 2m² = (1/2) m^(-2-2) n^(-3) = (1/2) m^(-4) n^(-3) = 1/(2 m⁴ n³) → 1/(2m⁴n³)

11) 2x⁹ / 3x⁸y² = (2/3) x^(9-8) / y² = (2/3) x / y² → 2x/(3y²)

12) 3x⁵y³ / 3x⁸y⁰ = (3/3) x^(5-8) y^(3-0) = 1 * x^(-3) * y³ = y³ / x³ → y³/x³

13) 3x⁵y⁻⁴ / 4x³y⁴ = (3/4) x^(5-3) y^(-4-4) = (3/4) x² y^(-8) = 3x²/(4y⁸) → 3x²/(4y⁸)

14) 4x⁻ / 4y⁻¹ = (4/4) x^(-4) y^(1) = 1 * x^(-4) y = y / x⁴ → y/x⁴

15) 4a⁸y⁻¹ / 4v = (4/4) a⁸ y^(-1) / v = a⁸ / (v y) → a⁸/(vy)

16) 2y⁴ / [2x⁻² * 2x⁻¹ y⁻⁴] — wait, let's parse: denominator is 2x⁻² * 2x⁻¹ y⁻⁴ = 4 x^(-3) y^(-4)

So overall: 2y⁴ / [4 x^{-3} y^{-4}] = (2/4) y^{4 - (-4)} x^{3} = (1/2) y^8 x^3 → x³y⁸/2

17) 4a⁻¹v² / [4w⁻³v⁻¹ * w⁻¹v⁴] — first, denominator: 4 w^{-3} v^{-1} * w^{-1} v^4 = 4 w^{-4} v^{3}

Numerator: 4 a^{-1} v²

So: [4 a^{-1} v²] / [4 w^{-4} v^{3}] = a^{-1} v^{2-3} w^{4} = a^{-1} v^{-1} w^4 = w⁴ / (a v) → w⁴/(av)

18) c⁻²y⁻⁴ * 3x²y² / 4yx² — first, numerator: 3 c^{-2} x² y^{-4} y² = 3 c^{-2} x² y^{-2}

Denominator: 4 y x²

So: [3 c^{-2} x² y^{-2}] / [4 x² y] = (3/4) c^{-2} x^{2-2} y^{-2-1} = (3/4) c^{-2} y^{-3} = 3/(4 c² y³) → 3/(4c²y³)

19) 4a⁴b² / [2a⁷b² * 2b⁻⁴] — denominator: 2*2 a⁷ b² b^{-4} = 4 a⁷ b^{-2}

Numerator: 4 a⁴ b²

So: [4 a⁴ b²] / [4 a⁷ b^{-2}] = a^{4-7} b^{2 - (-2)} = a^{-3} b^{4} = b⁴ / a³ → b⁴/a³

20) 4x⁻¹y⁰ / [2x³y⁻⁴ * x⁻²y⁻²] — first, y⁰ = 1

Numerator: 4 x^{-1} * 1 = 4 x^{-1}

Denominator: 2 x³ y^{-4} * x^{-2} y^{-2} = 2 x^{3-2} y^{-4-2} = 2 x¹ y^{-6}

So: [4 x^{-1}] / [2 x y^{-6}] = (4/2) x^{-1-1} y^{6} = 2 x^{-2} y^6 = 2 y^6 / x² → 2y⁶/x²

Now, let me double-check a few tricky ones:

Problem 16: 2y⁴ / (2x⁻² * 2x⁻¹ y⁻⁴) = 2y⁴ / (4 x^{-3} y^{-4}) = (2/4) y^{4+4} x^{3} = (1/2) y^8 x^3 → yes, x³y⁸/2

Problem 17: Denominator: 4w⁻³v⁻¹ * w⁻¹v⁴ = 4 w^{-4} v^{3}, numerator 4a⁻¹v² → so 4a⁻¹v² / 4w⁻⁴v³ = a⁻¹ v^{-1} w⁴ → w⁴/(av) — correct.

Problem 18: Numerator: c⁻²y⁻⁴ * 3x²y² = 3c⁻²x²y⁻²; denominator 4yx² → so 3c⁻²x²y⁻² / 4x²y = 3c⁻² y^{-3} / 4 = 3/(4c²y³) — correct.

Problem 20: Numerator 4x⁻¹; denominator 2x³y⁻⁴ * x⁻²y⁻² = 2x^{1}y^{-6}; so 4x⁻¹ / (2x y^{-6}) = 2 x^{-2} y^6 = 2y⁶/x² — correct.

All seem good.

Final Answers:

1) 1/16
2) 1/81
3) 1024
4) 243
5) 16
6) 4/b³
7) 4/x⁴
8) 2/x²
9) 4m⁶/n²
10) 1/(2m⁴n³)
11) 2x/(3y²)
12) y³/x³
13) 3x²/(4y⁸)
14) y/x⁴
15) a⁸/(vy)
16) x³y⁸/2
17) w⁴/(av)
18) 3/(4c²y³)
19) b⁴/a³
20) 2y⁶/x²