Question

simplify. your answer should contain only positive exponents.

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

$= x^{-2\cdot4} \cdot x^{-3\cdot4}$
$= x^{-8} \cdot x^{-12}$
$= x^{-20}$

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

$= x^{-12} \cdot 2x^{4}$
$= 2x^{-8}$
$= \frac{2}{x^{8}}$

1. $(n^{3})^{3} \cdot 2n^{-1}$

$= n^{3\cdot3} \cdot 2n^{-1}$
$= n^{9} \cdot 2n^{-1}$
$= 2n^{8}$

1. $(2v)^{2} \cdot 2v^{2}$

$= 4v^{2} \cdot 2v^{2}$
$= 8v^{4}$

1. $\frac{2x^{2}y^{4} \cdot 4x^{2}y^{4} \cdot 3x}{3x^{-3}y^{2}} = \frac{24x^{5}y^{8}}{3x^{-3}y^{2}} = \frac{24x^{5}y^{8} \cdot x^{3}}{3y^{2}}$

$= 8x^{8}y^{6}$

1. $\frac{2y^{3} \cdot 3xy^{3}}{3x^{2}y^{4}} = \frac{6xy^{6}}{3x^{2}y^{4}} = 2x^{-1}y^{2} = \frac{2y^{2}}{x}$
2. $\frac{x^{3}y^{3} \cdot x^{3}}{4x^{2}} = \frac{x^{6}y^{3}}{4x^{2}} = \frac{x^{4}y^{3}}{4}$
3. $\frac{3x^{2}y^{2}}{2x^{-1} \cdot 4yx^{2}} = \frac{3x^{2}y^{2}}{8xy} = \frac{3xy}{8}$
4. $\frac{x}{(2x^{0})^{2}}$
5. $\frac{2m^{-4}}{(2m^{-4})^{3}}$

Explanation:

Let's solve each problem step by step using the properties of exponents (product rule: \(a^m \cdot a^n = a^{m + n}\), power rule: \((a^m)^n = a^{m \cdot n}\), negative exponent rule: \(a^{-n}=\frac{1}{a^n}\) and \(\frac{1}{a^{-n}} = a^n\)):

Problem 1: \((x^{-2}x^{-3})^4\)

Step 1: Use the product rule inside the parentheses

When multiplying exponents with the same base, add the exponents: \(x^{-2} \cdot x^{-3}=x^{-2 + (-3)} = x^{-5}\)

Step 2: Use the power rule

Raise the result to the 4th power: \((x^{-5})^4 = x^{-5 \cdot 4}=x^{-20}\)

Step 3: Convert to positive exponent

Using the negative exponent rule: \(x^{-20}=\frac{1}{x^{20}}\)

Answer:

\(\frac{1}{x^{20}}\)

Problem 2: \((x^4)^{-3} \cdot 2x^4\)

Step 1: Use the power rule

\((x^4)^{-3}=x^{4 \cdot (-3)} = x^{-12}\)

Step 2: Multiply with \(2x^4\)

Using the product rule: \(x^{-12} \cdot 2x^4 = 2x^{-12 + 4}=2x^{-8}\)

Step 3: Convert to positive exponent

\(2x^{-8}=\frac{2}{x^8}\)