More Properties of Exponents - Kuta Software - Free Printable

Let's solve each problem step by step using the properties of exponents. We'll simplify each expression so that the final answer contains only positive exponents.

---

1) $(x^{-2}x^{-3})^4$

Step 1: Combine the terms inside the parentheses using the rule:
$$
x^a \cdot x^b = x^{a+b}
$$
So:
$$
x^{-2} \cdot x^{-3} = x^{-2 + (-3)} = x^{-5}
$$

Step 2: Apply the exponent outside:
$$
(x^{-5})^4 = x^{-5 \cdot 4} = x^{-20}
$$

Step 3: Convert to positive exponent:
$$
x^{-20} = \frac{1}{x^{20}}
$$

✔ Answer: $\boxed{\frac{1}{x^{20}}}$

---

2) $(x^4)^{-1} \cdot 2x^4$

Step 1: Simplify $(x^4)^{-1} = x^{-4}$

Now we have:
$$
x^{-4} \cdot 2x^4
$$

Step 2: Multiply coefficients and variables:
$$
2 \cdot x^{-4} \cdot x^4 = 2 \cdot x^{-4+4} = 2 \cdot x^0 = 2 \cdot 1 = 2
$$

✔ Answer: $\boxed{2}$

---

3) $(m^3)^3 \cdot 2m^{-4}$

Step 1: Simplify $(m^3)^3 = m^{3\cdot3} = m^9$

Now:
$$
m^9 \cdot 2m^{-4} = 2 \cdot m^{9 + (-4)} = 2m^5
$$

✔ Answer: $\boxed{2m^5}$

---

4) $(2y)^3 \cdot 2y^{-2}$

Step 1: Expand $(2y)^3 = 2^3 \cdot y^3 = 8y^3$

Now:
$$
8y^3 \cdot 2y^{-2} = (8 \cdot 2) \cdot y^{3 + (-2)} = 16y^1 = 16y
$$

✔ Answer: $\boxed{16y}$

---

5) $\frac{2x^2y^4 \cdot 4x^2y^4 \cdot 3x}{3x^{-3}y^2}$

Step 1: Combine numerator:
- Coefficients: $2 \cdot 4 \cdot 3 = 24$
- $x$: $x^2 \cdot x^2 \cdot x = x^{2+2+1} = x^5$
- $y$: $y^4 \cdot y^4 = y^{8}$

Numerator: $24x^5y^8$

Denominator: $3x^{-3}y^2$

Now:
$$
\frac{24x^5y^8}{3x^{-3}y^2} = \frac{24}{3} \cdot x^{5 - (-3)} \cdot y^{8 - 2} = 8 \cdot x^{8} \cdot y^6
$$

✔ Answer: $\boxed{8x^8y^6}$

---

6) $\frac{2x^3 \cdot 3xy^2}{3x^2y^4}$

Step 1: Numerator: $2x^3 \cdot 3xy^2 = 6x^{3+1}y^2 = 6x^4y^2$

Denominator: $3x^2y^4$

Now:
$$
\frac{6x^4y^2}{3x^2y^4} = \frac{6}{3} \cdot x^{4-2} \cdot y^{2-4} = 2x^2y^{-2}
$$

Convert negative exponent:
$$
y^{-2} = \frac{1}{y^2}
$$

✔ Answer: $\boxed{\frac{2x^2}{y^2}}$

---

7) $\frac{x^5y^3 \cdot x^2}{4x^2}$

Step 1: Numerator: $x^5y^3 \cdot x^2 = x^{5+2}y^3 = x^7y^3$

Now:
$$
\frac{x^7y^3}{4x^2} = \frac{1}{4} \cdot x^{7-2} \cdot y^3 = \frac{1}{4}x^5y^3
$$

✔ Answer: $\boxed{\frac{1}{4}x^5y^3}$

---

8) $\frac{3x^2y^2}{2x^{-1} \cdot 4yx^2}$

Step 1: Simplify denominator:
$$
2x^{-1} \cdot 4yx^2 = (2 \cdot 4) \cdot x^{-1} \cdot x^2 \cdot y = 8 \cdot x^{1} \cdot y = 8xy
$$

Now:
$$
\frac{3x^2y^2}{8xy} = \frac{3}{8} \cdot x^{2-1} \cdot y^{2-1} = \frac{3}{8}xy
$$

✔ Answer: $\boxed{\frac{3}{8}xy}$

---

9) $\frac{x}{(2x^0)^2}$

Step 1: Recall $x^0 = 1$, so:
$$
(2x^0)^2 = (2 \cdot 1)^2 = 2^2 = 4
$$

So:
$$
\frac{x}{4}
$$

✔ Answer: $\boxed{\frac{x}{4}}$

---

10) $\frac{2m^{-4}}{(2m^{-4})^2}$

Step 1: Simplify denominator:
$$
(2m^{-4})^2 = 2^2 \cdot (m^{-4})^2 = 4m^{-8}
$$

Now:
$$
\frac{2m^{-4}}{4m^{-8}} = \frac{2}{4} \cdot m^{-4 - (-8)} = \frac{1}{2} \cdot m^{4}
$$

✔ Answer: $\boxed{\frac{1}{2}m^4}$

---

✔ Final Answers:

1) $\boxed{\frac{1}{x^{20}}} $
2) $\boxed{2} $
3) $\boxed{2m^5} $
4) $\boxed{16y} $
5) $\boxed{8x^8y^6} $
6) $\boxed{\frac{2x^2}{y^2}} $
7) $\boxed{\frac{1}{4}x^5y^3} $
8) $\boxed{\frac{3}{8}xy} $
9) $\boxed{\frac{x}{4}} $
10) $\boxed{\frac{1}{2}m^4} $

Let me know if you'd like these explained in video format or with visual steps!