Exponents - Riverside Math - Free Printable

Problem: Simplify the given expressions. Do not leave any negative exponents in your final answer.

We will solve each problem step by step, using the rules of exponents:

#### Rule Recap:
1. $ x^a \cdot x^b = x^{a+b} $
2. $ \frac{x^a}{x^b} = x^{a-b} $
3. $ (x^a)^b = x^{a \cdot b} $
4. $ x^{-a} = \frac{1}{x^a} $
5. $ (xy)^a = x^a y^a $
6. $ \left(\frac{x}{y}\right)^a = \frac{x^a}{y^a} $
7. $ x^0 = 1 $ (for $ x \neq 0 $)

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Problem 1: $ x^4 \cdot x^7 $

Using the rule $ x^a \cdot x^b = x^{a+b} $:
$$
x^4 \cdot x^7 = x^{4+7} = x^{11}
$$

Answer: $ \boxed{x^{11}} $

---

Problem 2: $ 2x^6 \cdot x^{11} $

First, combine the powers of $ x $ using $ x^a \cdot x^b = x^{a+b} $:
$$
x^6 \cdot x^{11} = x^{6+11} = x^{17}
$$
Now, include the coefficient $ 2 $:
$$
2x^6 \cdot x^{11} = 2x^{17}
$$

Answer: $ \boxed{2x^{17}} $

---

Problem 3: $ \frac{s^{14}}{s^8} $

Using the rule $ \frac{x^a}{x^b} = x^{a-b} $:
$$
\frac{s^{14}}{s^8} = s^{14-8} = s^6
$$

Answer: $ \boxed{s^6} $

---

Problem 4: $ (5p^4)^3 $

Use the rule $ (xy)^a = x^a y^a $ and $ (x^a)^b = x^{a \cdot b} $:
$$
(5p^4)^3 = 5^3 \cdot (p^4)^3 = 5^3 \cdot p^{4 \cdot 3} = 125 \cdot p^{12}
$$

Answer: $ \boxed{125p^{12}} $

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Problem 5: $ (3x^3y^5)^2 $

Use the rule $ (xy)^a = x^a y^a $ and $ (x^a)^b = x^{a \cdot b} $:
$$
(3x^3y^5)^2 = 3^2 \cdot (x^3)^2 \cdot (y^5)^2 = 9 \cdot x^{3 \cdot 2} \cdot y^{5 \cdot 2} = 9x^6y^{10}
$$

Answer: $ \boxed{9x^6y^{10}} $

---

Problem 6: $ (2m)^{-4} $

Use the rule $ x^{-a} = \frac{1}{x^a} $:
$$
(2m)^{-4} = \frac{1}{(2m)^4}
$$
Now, expand $ (2m)^4 $ using $ (xy)^a = x^a y^a $:
$$
(2m)^4 = 2^4 \cdot m^4 = 16m^4
$$
Thus:
$$
(2m)^{-4} = \frac{1}{16m^4}
$$

Answer: $ \boxed{\frac{1}{16m^4}} $

---

Problem 7: $ a^{-5} \cdot a^{-11} $

Use the rule $ x^a \cdot x^b = x^{a+b} $:
$$
a^{-5} \cdot a^{-11} = a^{-5 + (-11)} = a^{-16}
$$
Convert the negative exponent to a positive exponent:
$$
a^{-16} = \frac{1}{a^{16}}
$$

Answer: $ \boxed{\frac{1}{a^{16}}} $

---

Problem 8: $ \frac{25x^9}{5x^{12}} $

Simplify the coefficients and the powers of $ x $ separately:
$$
\frac{25x^9}{5x^{12}} = \frac{25}{5} \cdot \frac{x^9}{x^{12}} = 5 \cdot x^{9-12} = 5 \cdot x^{-3}
$$
Convert the negative exponent to a positive exponent:
$$
5 \cdot x^{-3} = \frac{5}{x^3}
$$

Answer: $ \boxed{\frac{5}{x^3}} $

---

Problem 9: $ (156u^4w^4)^0 $

Any nonzero number raised to the power of 0 is 1:
$$
(156u^4w^4)^0 = 1
$$

Answer: $ \boxed{1} $

---

Problem 10: $ (5x^5y^4)^2 $

Use the rule $ (xy)^a = x^a y^a $ and $ (x^a)^b = x^{a \cdot b} $:
$$
(5x^5y^4)^2 = 5^2 \cdot (x^5)^2 \cdot (y^4)^2 = 25 \cdot x^{5 \cdot 2} \cdot y^{4 \cdot 2} = 25x^{10}y^8
$$

Answer: $ \boxed{25x^{10}y^8} $

---

Problem 11: $ (-q^3b^5)^7 $

Use the rule $ (xy)^a = x^a y^a $ and $ (x^a)^b = x^{a \cdot b} $:
$$
(-q^3b^5)^7 = (-1)^7 \cdot (q^3)^7 \cdot (b^5)^7 = -1 \cdot q^{3 \cdot 7} \cdot b^{5 \cdot 7} = -q^{21}b^{35}
$$

Answer: $ \boxed{-q^{21}b^{35}} $

---

Problem 12: $ (2x^5y)(3x^4y^7) $

Combine the coefficients and the powers of $ x $ and $ y $ separately:
$$
(2x^5y)(3x^4y^7) = (2 \cdot 3) \cdot (x^5 \cdot x^4) \cdot (y \cdot y^7) = 6 \cdot x^{5+4} \cdot y^{1+7} = 6x^9y^8
$$

Answer: $ \boxed{6x^9y^8} $

---

Problem 13: $ (2y^4)^2(5y^6)^2 $

Simplify each term separately using $ (xy)^a = x^a y^a $ and $ (x^a)^b = x^{a \cdot b} $:
$$
(2y^4)^2 = 2^2 \cdot (y^4)^2 = 4 \cdot y^{4 \cdot 2} = 4y^8
$$
$$
(5y^6)^2 = 5^2 \cdot (y^6)^2 = 25 \cdot y^{6 \cdot 2} = 25y^{12}
$$
Now multiply the results:
$$
(2y^4)^2(5y^6)^2 = (4y^8)(25y^{12}) = (4 \cdot 25) \cdot (y^8 \cdot y^{12}) = 100 \cdot y^{8+12} = 100y^{20}
$$

Answer: $ \boxed{100y^{20}} $

---

Problem 14: $ \left(\frac{10x^2}{x^{18}}\right)^{-5} $

First simplify the fraction inside the parentheses:
$$
\frac{10x^2}{x^{18}} = 10 \cdot \frac{x^2}{x^{18}} = 10 \cdot x^{2-18} = 10 \cdot x^{-16}
$$
So:
$$
\left(\frac{10x^2}{x^{18}}\right)^{-5} = (10x^{-16})^{-5}
$$
Use the rule $ (xy)^a = x^a y^a $ and $ (x^a)^b = x^{a \cdot b} $:
$$
(10x^{-16})^{-5} = 10^{-5} \cdot (x^{-16})^{-5} = 10^{-5} \cdot x^{-16 \cdot -5} = 10^{-5} \cdot x^{80}
$$
Convert $ 10^{-5} $ to a positive exponent:
$$
10^{-5} = \frac{1}{10^5}
$$
Thus:
$$
(10x^{-16})^{-5} = \frac{x^{80}}{10^5}
$$

Answer: $ \boxed{\frac{x^{80}}{10^5}} $

---

Problem 15: $ \left(\frac{50x^4}{2x^3}\right)^2 $

Simplify the fraction inside the parentheses:
$$
\frac{50x^4}{2x^3} = \frac{50}{2} \cdot \frac{x^4}{x^3} = 25 \cdot x^{4-3} = 25x
$$
Now square the result:
$$
\left(\frac{50x^4}{2x^3}\right)^2 = (25x)^2 = 25^2 \cdot x^2 = 625x^2
$$

Answer: $ \boxed{625x^2} $

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Problem 16: $ (8k^2j^{12})(4k^{-7}j^3) $

Combine the coefficients and the powers of $ k $ and $ j $ separately:
$$
(8k^2j^{12})(4k^{-7}j^3) = (8 \cdot 4) \cdot (k^2 \cdot k^{-7}) \cdot (j^{12} \cdot j^3) = 32 \cdot k^{2+(-7)} \cdot j^{12+3}
$$
Simplify the exponents:
$$
32 \cdot k^{2-7} \cdot j^{15} = 32 \cdot k^{-5} \cdot j^{15}
$$
Convert the negative exponent to a positive exponent:
$$
32 \cdot k^{-5} \cdot j^{15} = \frac{32j^{15}}{k^5}
$$

Answer: $ \boxed{\frac{32j^{15}}{k^5}} $

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Problem 17: $ \frac{a^5b^2}{4b} \cdot \frac{16a^2b^4}{a^2b^7} $

Simplify each fraction separately:
$$
\frac{a^5b^2}{4b} = \frac{a^5}{4} \cdot \frac{b^2}{b} = \frac{a^5}{4} \cdot b^{2-1} = \frac{a^5b}{4}
$$
$$
\frac{16a^2b^4}{a^2b^7} = \frac{16}{1} \cdot \frac{a^2}{a^2} \cdot \frac{b^4}{b^7} = 16 \cdot a^{2-2} \cdot b^{4-7} = 16 \cdot a^0 \cdot b^{-3} = 16 \cdot 1 \cdot b^{-3} = \frac{16}{b^3}
$$
Now multiply the simplified fractions:
$$
\frac{a^5b}{4} \cdot \frac{16}{b^3} = \frac{a^5b \cdot 16}{4 \cdot b^3} = \frac{16a^5b}{4b^3}
$$
Simplify the coefficients and the powers of $ b $:
$$
\frac{16a^5b}{4b^3} = \frac{16}{4} \cdot \frac{a^5b}{b^3} = 4 \cdot a^5 \cdot b^{1-3} = 4a^5b^{-2}
$$
Convert the negative exponent to a positive exponent:
$$
4a^5b^{-2} = \frac{4a^5}{b^2}
$$

Answer: $ \boxed{\frac{4a^5}{b^2}} $

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Problem 18: $ \frac{9x^7y^3}{xy} \cdot \frac{2x^4y^8}{2x^2y^4} $

Simplify each fraction separately:
$$
\frac{9x^7y^3}{xy} = \frac{9}{1} \cdot \frac{x^7}{x} \cdot \frac{y^3}{y} = 9 \cdot x^{7-1} \cdot y^{3-1} = 9x^6y^2
$$
$$
\frac{2x^4y^8}{2x^2y^4} = \frac{2}{2} \cdot \frac{x^4}{x^2} \cdot \frac{y^8}{y^4} = 1 \cdot x^{4-2} \cdot y^{8-4} = x^2y^4
$$
Now multiply the simplified fractions:
$$
9x^6y^2 \cdot x^2y^4 = 9 \cdot (x^6 \cdot x^2) \cdot (y^2 \cdot y^4) = 9 \cdot x^{6+2} \cdot y^{2+4} = 9x^8y^6
$$

Answer: $ \boxed{9x^8y^6} $

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Problem 19: $ (3a^2b^6)^2(3a^{10}b^4)^3 $

Simplify each term separately using $ (xy)^a = x^a y^a $ and $ (x^a)^b = x^{a \cdot b} $:
$$
(3a^2b^6)^2 = 3^2 \cdot (a^2)^2 \cdot (b^6)^2 = 9 \cdot a^{2 \cdot 2} \cdot b^{6 \cdot 2} = 9a^4b^{12}
$$
$$
(3a^{10}b^4)^3 = 3^3 \cdot (a^{10})^3 \cdot (b^4)^3 = 27 \cdot a^{10 \cdot 3} \cdot b^{4 \cdot 3} = 27a^{30}b^{12}
$$
Now multiply the results:
$$
(3a^2b^6)^2(3a^{10}b^4)^3 = (9a^4b^{12})(27a^{30}b^{12}) = (9 \cdot 27) \cdot (a^4 \cdot a^{30}) \cdot (b^{12} \cdot b^{12})
$$
Simplify the coefficients and the powers of $ a $ and $ b $:
$$
(9 \cdot 27) \cdot (a^4 \cdot a^{30}) \cdot (b^{12} \cdot b^{12}) = 243 \cdot a^{4+30} \cdot b^{12+12} = 243a^{34}b^{24}
$$

Answer: $ \boxed{243a^{34}b^{24}} $

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Problem 20: $ \left(\frac{5a^{10}b^8c^{12}}{15a^2b^4c^{11}}\right)^3 $

Simplify the fraction inside the parentheses:
$$
\frac{5a^{10}b^8c^{12}}{15a^2b^4c^{11}} = \frac{5}{15} \cdot \frac{a^{10}}{a^2} \cdot \frac{b^8}{b^4} \cdot \frac{c^{12}}{c^{11}}
$$
Simplify each part:
$$
\frac{5}{15} = \frac{1}{3}, \quad \frac{a^{10}}{a^2} = a^{10-2} = a^8, \quad \frac{b^8}{b^4} = b^{8-4} = b^4, \quad \frac{c^{12}}{c^{11}} = c^{12-11} = c^1
$$
So:
$$
\frac{5a^{10}b^8c^{12}}{15a^2b^4c^{11}} = \frac{1}{3} \cdot a^8 \cdot b^4 \cdot c = \frac{a^8b^4c}{3}
$$
Now cube the result:
$$
\left(\frac{5a^{10}b^8c^{12}}{15a^2b^4c^{11}}\right)^3 = \left(\frac{a^8b^4c}{3}\right)^3
$$
Use the rule $ \left(\frac{x}{y}\right)^a = \frac{x^a}{y^a} $ and $ (xy)^a = x^a y^a $:
$$
\left(\frac{a^8b^4c}{3}\right)^3 = \frac{(a^8b^4c)^3}{3^3} = \frac{(a^8)^3 \cdot (b^4)^3 \cdot c^3}{27} = \frac{a^{8 \cdot 3} \cdot b^{4 \cdot 3} \cdot c^3}{27} = \frac{a^{24}b^{12}c^3}{27}
$$

Answer: $ \boxed{\frac{a^{24}b^{12}c^3}{27}} $

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Final Answers:

1. $ \boxed{x^{11}} $
2. $ \boxed{2x^{17}} $
3. $ \boxed{s^6} $
4. $ \boxed{125p^{12}} $
5. $ \boxed{9x^6y^{10}} $
6. $ \boxed{\frac{1}{16m^4}} $
7. $ \boxed{\frac{1}{a^{16}}} $
8. $ \boxed{\frac{5}{x^3}} $
9. $ \boxed{1} $
10. $ \boxed{25x^{10}y^8} $
11. $ \boxed{-q^{21}b^{35}} $
12. $ \boxed{6x^9y^8} $
13. $ \boxed{100y^{20}} $
14. $ \boxed{\frac{x^{80}}{10^5}} $
15. $ \boxed{625x^2} $
16. $ \boxed{\frac{32j^{15}}{k^5}} $
17. $ \boxed{\frac{4a^5}{b^2}} $
18. $ \boxed{9x^8y^6} $
19. $ \boxed{243a^{34}b^{24}} $
20. $ \boxed{\frac{a^{24}b^{12}c^3}{27}} $