simplify. your answer should contain only positive exponents.
1) $(x^{-2}x^{-3})^{4}$
$= x^{-2\cdot4} \cdot x^{-3\cdot4}$
$= x^{-8} \cdot x^{-12}$
$= x^{-20}$
2) $(x^{4})^{-3} \cdot 2x^{4}$
$= x^{-12} \cdot 2x^{4}$
$= 2x^{-8}$
$= \frac{2}{x^{8}}$
3) $(n^{3})^{3} \cdot 2n^{-1}$
$= n^{3\cdot3} \cdot 2n^{-1}$
$= n^{9} \cdot 2n^{-1}$
$= 2n^{8}$
4) $(2v)^{2} \cdot 2v^{2}$
$= 4v^{2} \cdot 2v^{2}$
$= 8v^{4}$
5) $\frac{2x^{2}y^{4} \cdot 4x^{2}y^{4} \cdot 3x}{3x^{-3}y^{2}} = \frac{24x^{5}y^{8}}{3x^{-3}y^{2}} = \frac{24x^{5}y^{8} \cdot x^{3}}{3y^{2}}$
$= 8x^{8}y^{6}$
6) $\frac{2y^{3} \cdot 3xy^{3}}{3x^{2}y^{4}} = \frac{6xy^{6}}{3x^{2}y^{4}} = 2x^{-1}y^{2} = \frac{2y^{2}}{x}$
7) $\frac{x^{3}y^{3} \cdot x^{3}}{4x^{2}} = \frac{x^{6}y^{3}}{4x^{2}} = \frac{x^{4}y^{3}}{4}$
8) $\frac{3x^{2}y^{2}}{2x^{-1} \cdot 4yx^{2}} = \frac{3x^{2}y^{2}}{8xy} = \frac{3xy}{8}$
9) $\frac{x}{(2x^{0})^{2}}$
10) $\frac{2m^{-4}}{(2m^{-4})^{3}}$

Question

Accepted Answer

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
$2x^{-8}=\frac{2}{x^8}$
# Answer: $\frac{2}{x^8}$

### Problem 3: $(n^3)^3 \cdot 2n^{-1}$
## Step 1: Use the power rule
$(n^3)^3 = n^{3 \cdot 3}=n^9$
## Step 2: Multiply with $2n^{-1}$
Using the product rule: $n^9 \cdot 2n^{-1}=2n^{9 + (-1)} = 2n^8$
# Answer: $2n^8$

### Problem 4: $(2v)^2 \cdot 2v^2$
## Step 1: Expand $(2v)^2$
Using the power of a product rule: $(2v)^2 = 2^2 \cdot v^2 = 4v^2$
## Step 2: Multiply with $2v^2$
Using the product rule: $4v^2 \cdot 2v^2=(4 \cdot 2)v^{2 + 2}=8v^4$
# Answer: $8v^4$

### Problem 5: $\frac{2x^2y^4 \cdot 4x^2y^4 \cdot 3x}{3x^{-3}y^2}$
## Step 1: Multiply the numerators
Multiply the coefficients and use the product rule for exponents:
- Coefficients: $2 \cdot 4 \cdot 3 = 24$
- For $x$: $x^2 \cdot x^2 \cdot x = x^{2 + 2 + 1}=x^5$
- For $y$: $y^4 \cdot y^4 = y^{4 + 4}=y^8$
So the numerator is $24x^5y^8$
## Step 2: Simplify the fraction
Divide the coefficients and use the quotient rule for exponents ($a^m \div a^n = a^{m - n}$):
- Coefficients: $\frac{24}{3}=8$
- For $x$: $x^5 \div x^{-3}=x^{5 - (-3)} = x^8$
- For $y$: $y^8 \div y^2 = y^{8 - 2}=y^6$
Putting it together: $8x^8y^6$
# Answer: $8x^8y^6$

### Problem 6: $\frac{2y^3 \cdot 3xy^3}{3x^2y^4}$
## Step 1: Multiply the numerators
- Coefficients: $2 \cdot 3 = 6$
- For $x$: $x$ (since there's one $x$ in the second term)
- For $y$: $y^3 \cdot y^3 = y^{3 + 3}=y^6$
So the numerator is $6xy^6$
## Step 2: Simplify the fraction
Divide the coefficients and use the quotient rule for exponents:
- Coefficients: $\frac{6}{3}=2$
- For $x$: $x \div x^2 = x^{1 - 2}=x^{-1}=\frac{1}{x}$
- For $y$: $y^6 \div y^4 = y^{6 - 4}=y^2$
Multiply the results: $2 \cdot \frac{1}{x} \cdot y^2=\frac{2y^2}{x}$
# Answer: $\frac{2y^2}{x}$

### Problem 7: $\frac{x^3y^3 \cdot x^3}{4x^2}$
## Step 1: Multiply the numerators (for $x$)
Using the product rule: $x^3 \cdot x^3 = x^{3 + 3}=x^6$ (the $y^3$ remains as is for now)
So the numerator is $x^6y^3$
## Step 2: Simplify the fraction (for $x$)
Using the quotient rule: $x^6 \div x^2 = x^{6 - 2}=x^4$
So we have $\frac{x^4y^3}{4}$
# Answer: $\frac{x^4y^3}{4}$

### Problem 8: $\frac{3x^2y^2}{2x^{-1} \cdot 4yx^2}$
## Step 1: Multiply the denominators
- Coefficients: $2 \cdot 4 = 8$
- For $x$: $x^{-1} \cdot x^2 = x^{-1 + 2}=x^1 = x$
- For $y$: $y$ (since there's one $y$ in the second term)
So the denominator is $8xy$
## Step 2: Simplify the fraction
- Coefficients: $\frac{3}{8}$
- For $x$: $x^2 \div x = x^{2 - 1}=x$
- For $y$: $y^2 \div y = y^{2 - 1}=y$
Multiply the results: $\frac{3}{8} \cdot x \cdot y=\frac{3xy}{8}$
# Answer: $\frac{3xy}{8}$

### Problem 9: $\frac{x}{(2x^0)^2}$
## Step 1: Simplify $x^0$
Any non - zero number to the power of 0 is 1, so $x^0 = 1$ (assuming $x
eq0$)
## Step 2: Simplify the denominator
$(2x^0)^2=(2 \cdot 1)^2 = 2^2=4$
## Step 3: Simplify the fraction
$\frac{x}{4}$
# Answer: $\frac{x}{4}$

### Problem 10: $\frac{2m^{-4}}{(2m^{-4})^3}$
## Step 1: Simplify the denominator using the power rule
$(2m^{-4})^3=2^3 \cdot (m^{-4})^3 = 8m^{-12}$
## Step 2: Rewrite the fraction
$\frac{2m^{-4}}{8m^{-12}}$
## Step 3: Simplify the fraction
- Coefficients: $\frac{2}{8}=\frac{1}{4}$
- For $m$: $m^{-4} \div m^{-12}=m^{-4-(-12)} = m^8$
Putting it together: $\frac{1}{4}m^8=\frac{m^8}{4}$
# Answer: $\frac{m^8}{4}$

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Question

Explanation:

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

Step 1: Use the product rule inside the parentheses

Step 2: Use the power rule

Step 3: Convert to positive exponent

Answer:

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

Step 1: Use the power rule

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

Step 3: Convert to positive exponent