Practice worksheet for simplifying exponential expressions using the product and quotient rules.

To solve the problems involving the laws of exponents, we will use the following key rules:

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

Let's solve each problem step by step.

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Problem 1: $ a \cdot a^3 $

- Use the Product Rule: $ a^m \cdot a^n = a^{m+n} $
- Here, $ a = a^1 $ and $ a^3 $:
$$
a \cdot a^3 = a^{1+3} = a^4
$$

Answer: $ \boxed{a^4} $

---

Problem 2: $ \frac{b^7}{b^4} $

- Use the Quotient Rule: $ \frac{a^m}{a^n} = a^{m-n} $
- Here, $ b^7 $ and $ b^4 $:
$$
\frac{b^7}{b^4} = b^{7-4} = b^3
$$

Answer: $ \boxed{b^3} $

---

Problem 3: $ a^3b^2(a^2b) $

- Use the Product Rule for both $ a $ and $ b $ terms.
- First, group the $ a $ terms and the $ b $ terms:
$$
a^3b^2(a^2b) = (a^3 \cdot a^2)(b^2 \cdot b)
$$
- Apply the Product Rule:
$$
a^3 \cdot a^2 = a^{3+2} = a^5
$$
$$
b^2 \cdot b = b^{2+1} = b^3
$$
- Combine the results:
$$
a^5b^3
$$

Answer: $ \boxed{a^5b^3} $

---

Problem 4: $ 7q^8 \cdot 7q^3 $

- Use the Product Rule for both the coefficients and the $ q $ terms.
- First, multiply the coefficients:
$$
7 \cdot 7 = 49
$$
- Then, apply the Product Rule for $ q $:
$$
q^8 \cdot q^3 = q^{8+3} = q^{11}
$$
- Combine the results:
$$
49q^{11}
$$

Answer: $ \boxed{49q^{11}} $

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Problem 5: $ \frac{5x^4}{6x^2} $

- Use the Quotient Rule for the $ x $ terms and simplify the coefficients.
- Simplify the coefficients:
$$
\frac{5}{6}
$$
- Apply the Quotient Rule for $ x $:
$$
\frac{x^4}{x^2} = x^{4-2} = x^2
$$
- Combine the results:
$$
\frac{5}{6}x^2
$$

Answer: $ \boxed{\frac{5}{6}x^2} $

---

Problem 6: $ \frac{g^5h^3}{g^3h^5} $

- Use the Quotient Rule for both $ g $ and $ h $ terms.
- For $ g $:
$$
\frac{g^5}{g^3} = g^{5-3} = g^2
$$
- For $ h $:
$$
\frac{h^3}{h^5} = h^{3-5} = h^{-2}
$$
- Combine the results:
$$
g^2h^{-2}
$$
- Using the Negative Exponent Rule, rewrite $ h^{-2} $ as $ \frac{1}{h^2} $:
$$
g^2h^{-2} = \frac{g^2}{h^2}
$$

Answer: $ \boxed{\frac{g^2}{h^2}} $

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Problem 7: $ 5^3a^2b \cdot 5^2b^7 $

- Use the Product Rule for both the base $ 5 $ and the variables $ a $ and $ b $.
- For the base $ 5 $:
$$
5^3 \cdot 5^2 = 5^{3+2} = 5^5
$$
- For $ a $:
$$
a^2 \quad \text{(no other $ a $ term to combine with)}
$$
- For $ b $:
$$
b \cdot b^7 = b^{1+7} = b^8
$$
- Combine the results:
$$
5^5a^2b^8
$$

Answer: $ \boxed{5^5a^2b^8} $

---

Problem 8: $ \frac{9hab}{3b^2} $

- Simplify the coefficients and use the Quotient Rule for the variables.
- Simplify the coefficients:
$$
\frac{9}{3} = 3
$$
- For $ h $:
$$
h \quad \text{(no other $ h $ term to combine with)}
$$
- For $ a $:
$$
a \quad \text{(no other $ a $ term to combine with)}
$$
- For $ b $:
$$
\frac{b}{b^2} = b^{1-2} = b^{-1}
$$
- Combine the results:
$$
3ha \cdot b^{-1}
$$
- Using the Negative Exponent Rule, rewrite $ b^{-1} $ as $ \frac{1}{b} $:
$$
3ha \cdot b^{-1} = \frac{3ha}{b}
$$

Answer: $ \boxed{\frac{3ha}{b}} $

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Problem 9: $ a^3b^2c \cdot a^2b^3c^4 $

- Use the Product Rule for $ a $, $ b $, and $ c $.
- For $ a $:
$$
a^3 \cdot a^2 = a^{3+2} = a^5
$$
- For $ b $:
$$
b^2 \cdot b^3 = b^{2+3} = b^5
$$
- For $ c $:
$$
c \cdot c^4 = c^{1+4} = c^5
$$
- Combine the results:
$$
a^5b^5c^5
$$

Answer: $ \boxed{a^5b^5c^5} $

---

Problem 10: $ \frac{a^8b^5c^3}{a^4b^2c^5} $

- Use the Quotient Rule for $ a $, $ b $, and $ c $.
- For $ a $:
$$
\frac{a^8}{a^4} = a^{8-4} = a^4
$$
- For $ b $:
$$
\frac{b^5}{b^2} = b^{5-2} = b^3
$$
- For $ c $:
$$
\frac{c^3}{c^5} = c^{3-5} = c^{-2}
$$
- Combine the results:
$$
a^4b^3c^{-2}
$$
- Using the Negative Exponent Rule, rewrite $ c^{-2} $ as $ \frac{1}{c^2} $:
$$
a^4b^3c^{-2} = \frac{a^4b^3}{c^2}
$$

Answer: $ \boxed{\frac{a^4b^3}{c^2}} $

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Problem 11: $ \frac{2x^4y^3}{12xy^5} $

- Simplify the coefficients and use the Quotient Rule for the variables.
- Simplify the coefficients:
$$
\frac{2}{12} = \frac{1}{6}
$$
- For $ x $:
$$
\frac{x^4}{x} = x^{4-1} = x^3
$$
- For $ y $:
$$
\frac{y^3}{y^5} = y^{3-5} = y^{-2}
$$
- Combine the results:
$$
\frac{1}{6}x^3y^{-2}
$$
- Using the Negative Exponent Rule, rewrite $ y^{-2} $ as $ \frac{1}{y^2} $:
$$
\frac{1}{6}x^3y^{-2} = \frac{x^3}{6y^2}
$$

Answer: $ \boxed{\frac{x^3}{6y^2}} $

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Problem 12: $ \left( \frac{pq^2}{r^3} \right)^2 \cdot \left( \frac{p^2q}{r^2} \right)^3 $

- Use the Power of a Quotient Rule and the Power of a Product Rule.
- First, apply the Power of a Quotient Rule to each term:
$$
\left( \frac{pq^2}{r^3} \right)^2 = \frac{(pq^2)^2}{(r^3)^2}
$$
$$
\left( \frac{p^2q}{r^2} \right)^3 = \frac{(p^2q)^3}{(r^2)^3}
$$
- Simplify each part using the Power of a Product Rule:
$$
(pq^2)^2 = p^2(q^2)^2 = p^2q^4
$$
$$
(r^3)^2 = r^{3 \cdot 2} = r^6
$$
So:
$$
\left( \frac{pq^2}{r^3} \right)^2 = \frac{p^2q^4}{r^6}
$$
Similarly:
$$
(p^2q)^3 = (p^2)^3 \cdot q^3 = p^6q^3
$$
$$
(r^2)^3 = r^{2 \cdot 3} = r^6
$$
So:
$$
\left( \frac{p^2q}{r^2} \right)^3 = \frac{p^6q^3}{r^6}
$$
- Now multiply the two results:
$$
\frac{p^2q^4}{r^6} \cdot \frac{p^6q^3}{r^6}
$$
- Use the Product Rule for the numerator and the Quotient Rule for the denominator:
$$
\frac{p^2q^4 \cdot p^6q^3}{r^6 \cdot r^6} = \frac{p^{2+6}q^{4+3}}{r^{6+6}} = \frac{p^8q^7}{r^{12}}
$$

Answer: $ \boxed{\frac{p^8q^7}{r^{12}}} $

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Final Answers:

1. $ \boxed{a^4} $
2. $ \boxed{b^3} $
3. $ \boxed{a^5b^3} $
4. $ \boxed{49q^{11}} $
5. $ \boxed{\frac{5}{6}x^2} $
6. $ \boxed{\frac{g^2}{h^2}} $
7. $ \boxed{5^5a^2b^8} $
8. $ \boxed{\frac{3ha}{b}} $
9. $ \boxed{a^5b^5c^5} $
10. $ \boxed{\frac{a^4b^3}{c^2}} $
11. $ \boxed{\frac{x^3}{6y^2}} $
12. $ \boxed{\frac{p^8q^7}{r^{12}}} $