17) $(2b^{4})^{-1}$
18) $(x^{3}y^{-1})^{2}$
19) $(2x^{4}y^{-3})^{-1}$
20) $(3m)^{-2}$
21) $\frac{r^{2}}{2r^{3}}$
22) $\frac{x^{-1}}{4x^{4}}$
23) $\frac{3n^{4}}{3n^{3}}$
24) $\frac{m^{4}}{2m^{4}}$
25) $\frac{3m^{-4}}{m^{3}}$
26) $\frac{2x^{4}y^{-4}z^{-3}}{3x^{2}y^{-3}z^{4}}$
27) $\frac{4x^{0}y^{-2}z^{3}}{4x}$
28) $\frac{2h^{3}j^{-3}k^{4}}{3jk}$
29) $\frac{4m^{4}n^{3}p^{3}}{3m^{2}n^{2}p^{4}}$
30) $\frac{3x^{3}y^{-1}z^{-1}}{x^{-4}y^{0}z^{0}}$

Question

Accepted Answer

Let's solve these problems one by one using the properties of exponents. The key properties we'll use are:
- $(ab)^n = a^n b^n$
- $(a^m)^n = a^{mn}$
- $a^m \cdot a^n = a^{m + n}$
- $a^m \div a^n = a^{m - n}$
- $a^{-n} = \frac{1}{a^n}$ (and $\frac{1}{a^{-n}} = a^n$)
- $a^0 = 1$ (for $a
eq 0$)

### Problem 17: $(2b^4)^{-1}$
## Step 1: Apply the power of a product rule
Using $(ab)^n = a^n b^n$, we get:
$(2b^4)^{-1} = 2^{-1} \cdot (b^4)^{-1}$
## Step 2: Simplify the exponents
Using $(a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$:
$2^{-1} = \frac{1}{2}$ and $(b^4)^{-1} = b^{-4} = \frac{1}{b^4}$
Multiply them together:
$2^{-1} \cdot (b^4)^{-1} = \frac{1}{2} \cdot \frac{1}{b^4} = \frac{1}{2b^4}$
# Answer: $\frac{1}{2b^4}$

### Problem 18: $(x^3 y^{-1})^2$
## Step 1: Apply the power of a product rule
$(x^3 y^{-1})^2 = (x^3)^2 \cdot (y^{-1})^2$
## Step 2: Simplify the exponents
Using $(a^m)^n = a^{mn}$:
$(x^3)^2 = x^{6}$ and $(y^{-1})^2 = y^{-2} = \frac{1}{y^2}$
Multiply them together:
$(x^3)^2 \cdot (y^{-1})^2 = x^6 \cdot \frac{1}{y^2} = \frac{x^6}{y^2}$
# Answer: $\frac{x^6}{y^2}$

### Problem 19: $(2x^4 y^{-3})^{-1}$
## Step 1: Apply the power of a product rule
$(2x^4 y^{-3})^{-1} = 2^{-1} \cdot (x^4)^{-1} \cdot (y^{-3})^{-1}$
## Step 2: Simplify the exponents
Using $(a^m)^n = a^{mn}$ and $a^{-n} = \frac{1}{a^n}$:
$2^{-1} = \frac{1}{2}$, $(x^4)^{-1} = x^{-4} = \frac{1}{x^4}$, $(y^{-3})^{-1} = y^{3}$
Multiply them together:
$2^{-1} \cdot (x^4)^{-1} \cdot (y^{-3})^{-1} = \frac{1}{2} \cdot \frac{1}{x^4} \cdot y^3 = \frac{y^3}{2x^4}$
# Answer: $\frac{y^3}{2x^4}$

### Problem 20: $(3m)^{-2}$
## Step 1: Apply the power of a product rule
$(3m)^{-2} = 3^{-2} \cdot m^{-2}$
## Step 2: Simplify the exponents
Using $a^{-n} = \frac{1}{a^n}$:
$3^{-2} = \frac{1}{3^2} = \frac{1}{9}$ and $m^{-2} = \frac{1}{m^2}$
Multiply them together:
$3^{-2} \cdot m^{-2} = \frac{1}{9} \cdot \frac{1}{m^2} = \frac{1}{9m^2}$
# Answer: $\frac{1}{9m^2}$

### Problem 21: $\frac{r^2}{2r^3}$
## Step 1: Apply the quotient rule for exponents
Using $a^m \div a^n = a^{m - n}$:
$\frac{r^2}{r^3} = r^{2 - 3} = r^{-1}$
## Step 2: Simplify the expression
$\frac{r^2}{2r^3} = \frac{1}{2} \cdot r^{-1} = \frac{1}{2r}$ (since $r^{-1} = \frac{1}{r}$)
# Answer: $\frac{1}{2r}$

### Problem 22: $\frac{x^{-1}}{4x^4}$
## Step 1: Apply the quotient rule for exponents
Using $a^m \div a^n = a^{m - n}$:
$\frac{x^{-1}}{x^4} = x^{-1 - 4} = x^{-5}$
## Step 2: Simplify the expression
$\frac{x^{-1}}{4x^4} = \frac{1}{4} \cdot x^{-5} = \frac{1}{4x^5}$ (since $x^{-5} = \frac{1}{x^5}$)
# Answer: $\frac{1}{4x^5}$

### Problem 23: $\frac{3n^4}{3n^3}$
## Step 1: Simplify the coefficients and apply the quotient rule for exponents
The coefficient $\frac{3}{3} = 1$. For the variable part, using $a^m \div a^n = a^{m - n}$:
$\frac{n^4}{n^3} = n^{4 - 3} = n^1 = n$
## Step 2: Combine the results
$\frac{3n^4}{3n^3} = 1 \cdot n = n$
# Answer: $n$

### Problem 24: $\frac{m^4}{2m^4}$
## Step 1: Simplify the coefficients and apply the quotient rule for exponents
The coefficient $\frac{1}{2}$ (since $\frac{m^4}{m^4} = 1$ using $a^m \div a^n = a^{m - n}$ with $m = n = 4$, so $m^{4 - 4} = m^0 = 1$)
## Step 2: Combine the results
$\frac{m^4}{2m^4} = \frac{1}{2} \cdot 1 = \frac{1}{2}$
# Answer: $\frac{1}{2}$

### Problem 25: $\frac{3m^{-4}}{m^3}$
## Step 1: Apply the quotient rule for exponents
Using $a^m \div a^n = a^{m - n}$:
$\frac{m^{-4}}{m^3} = m^{-4 - 3} = m^{-7}$
## Step 2: Simplify the expression
$\frac{3m^{-4}}{m^3} = 3 \cdot m^{-7} = \frac{3}{m^7}$ (since $m^{-7} = \frac{1}{m^7}$)
# Answer: $\frac{3}{m^7}$

### Problem 26: $\frac{2x^4 y^{-4} z^{-3}}{3x^2 y^{-3} z^4}$
## Step 1: Separate the coefficients and variables
We can write this as $\frac{2}{3} \cdot \frac{x^4}{x^2} \cdot \frac{y^{-4}}{y^{-3}} \cdot \frac{z^{-3}}{z^4}$
## Step 2: Apply the quotient rule for each variable
- For $x$: $\frac{x^4}{x^2} = x^{4 - 2} = x^2$
- For $y$: $\frac{y^{-4}}{y^{-3}} = y^{-4 - (-3)} = y^{-1}$
- For $z$: $\frac{z^{-3}}{z^4} = z^{-3 - 4} = z^{-7}$
## Step 3: Combine the results
$\frac{2}{3} \cdot x^2 \cdot y^{-1} \cdot z^{-7} = \frac{2x^2}{3y z^7}$ (since $y^{-1} = \frac{1}{y}$ and $z^{-7} = \frac{1}{z^7}$)
# Answer: $\frac{2x^2}{3y z^7}$

### Problem 27: $\frac{4x^0 y^{-2} z^3}{4x}$
## Step 1: Simplify the coefficients and the $x$ term
The coefficient $\frac{4}{4} = 1$. For the $x$ term, $x^0 = 1$ (since $a^0 = 1$ for $a
eq 0$), so $\frac{x^0}{x} = \frac{1}{x} = x^{-1}$
## Step 2: Simplify the $y$ and $z$ terms
The $y$ term is $y^{-2}$ and the $z$ term is $z^3$
## Step 3: Combine the results
$\frac{4x^0 y^{-2} z^3}{4x} = 1 \cdot x^{-1} \cdot y^{-2} \cdot z^3 = \frac{z^3}{x y^2}$ (since $x^{-1} = \frac{1}{x}$ and $y^{-2} = \frac{1}{y^2}$)
# Answer: $\frac{z^3}{x y^2}$

### Problem 28: $\frac{2h^3 j^{-3} k^4}{3j k}$
## Step 1: Separate the coefficients and variables
We can write this as $\frac{2}{3} \cdot h^3 \cdot \frac{j^{-3}}{j} \cdot \frac{k^4}{k}$
## Step 2: Apply the quotient rule for $j$ and $k$
- For $j$: $\frac{j^{-3}}{j} = j^{-3 - 1} = j^{-4}$
- For $k$: $\frac{k^4}{k} = k^{4 - 1} = k^3$
## Step 3: Combine the results
$\frac{2}{3} \cdot h^3 \cdot j^{-4} \cdot k^3 = \frac{2h^3 k^3}{3j^4}$ (since $j^{-4} = \frac{1}{j^4}$)
# Answer: $\frac{2h^3 k^3}{3j^4}$

### Problem 29: $\frac{4m^4 n^3 p^3}{3m^2 n^2 p^4}$
## Step 1: Separate the coefficients and variables
We can write this as $\frac{4}{3} \cdot \frac{m^4}{m^2} \cdot \frac{n^3}{n^2} \cdot \frac{p^3}{p^4}$
## Step 2: Apply the quotient rule for each variable
- For $m$: $\frac{m^4}{m^2} = m^{4 - 2} = m^2$
- For $n$: $\frac{n^3}{n^2} = n^{3 - 2} = n$
- For $p$: $\frac{p^3}{p^4} = p^{3 - 4} = p^{-1}$
## Step 3: Combine the results
$\frac{4}{3} \cdot m^2 \cdot n \cdot p^{-1} = \frac{4m^2 n}{3p}$ (since $p^{-1} = \frac{1}{p}$)
# Answer: $\frac{4m^2 n}{3p}$

### Problem 30: $\frac{3x^3 y^{-1} z^{-1}}{x^{-4} y^0 z^0}$
## Step 1: Simplify the coefficients and constants
The coefficient is $3$. For $y^0$ and $z^0$, we know $a^0 = 1$ (for $a
eq 0$), so $y^0 = 1$ and $z^0 = 1$. So the denominator simplifies to $x^{-4} \cdot 1 \cdot 1 = x^{-4}$
## Step 2: Apply the quotient rule for $x$, $y$, and $z$
- For $x$: $\frac{x^3}{x^{-4}} = x^{3 - (-4)} = x^7$ (since $\frac{1}{x^{-4}} = x^4$, so $x^3 \cdot x^4 = x^7$)
- For $y$: $\frac{y^{-1}}{y^0} = y^{-1 - 0} = y^{-1}$ (but $y^0 = 1$, so it's just $y^{-1}$)
- For $z$: $\frac{z^{-1}}{z^0} = z^{-1 - 0} = z^{-1}$ (but $z^0 = 1$, so it's just $z^{-1}$)
## Step 3: Combine the results
$3 \cdot x^7 \cdot y^{-1} \cdot z^{-1} = \frac{3x^7}{y z}$ (since $y^{-1} = \frac{1}{y}$ and $z^{-1} = \frac{1}{z}$)
# Answer: $\frac{3x^7}{y z}$

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QUESTION IMAGE

Question

Explanation:

Problem 17: \((2b^4)^{-1}\)

Step 1: Apply the power of a product rule

Step 2: Simplify the exponents

Answer:

Problem 18: \((x^3 y^{-1})^2\)

Step 1: Apply the power of a product rule

Step 2: Simplify the exponents