#superpowers practice in working with powers & exponents worksheet 4.2 a this worksheet focuses on a product raised to a power, exponents are negative. questions 1) which of the following statements are true about $(2 \times 3)^{-5}$? a) it is a fraction b) $ = -5(2 \times 3)$ c) $ = \frac{1}{2^5 3^5}$ d) $ = \frac{1}{5(2 \times 3)}$ e) it has a negative answer 2) simplify. use the ‘product raised to a power’ rule if needed. a) $((ab)^2)^5$ b) $(a^2 b^2)^5$ c) $(a^3 b^{-2})^2$ d) $((ab)^5)^{-2}$ e) $(a^5 b^5)^2$ f) $(a^3 b^2)^{-2}$ 3) which of the powers below simplify to give an answer of 1? a) $((xy)^2)^0$ b) $(x^2 y^2)^0$ c) $(x^0 y^{-2})^2$ d) $((xy)^0)^{-2}$ e) $(x^0 y^0)^2$ f) $(x^0 y^2)^{-2}$ 4) two grade 9 learners were given this question to simplify: $(ab^2)^{-5} \cdot (ab^2)^{-5}$ they used different approaches. a) copy and complete each approach learner 1 learner 2 $(ab^2)^{-5} \cdot (ab^2)^{-5}$ $(ab^2)^{-5} \cdot (ab^2)^{-5}$ $ = a^{\square} b^{2 \times \square} \cdot a^{\square} b^{2 \times \square}$ $ = (ab^2)^{\square + \square}$ $ = a^{(-5 + \square)} b^{(-10 + (-10))}$ $ = a^{-10} b^{2 \times (-10)}$ $ = a^{\square} b^{\square}$ $ = a^{\square} b^{\square}$ $ = \frac{1}{a^{10} b^{20}}$ $ = \frac{1}{a^{10} b^{20}}$ b) describe the main differences between the approaches of the two learners. 5) simplify, leave answers with negative exponents. try to use the ‘product raised to a power’ rule as often as you can. a) $(2 \times 3)^{-4}$ b) $(2^2 3^3)^{-3}$ c) $(2^2 x^{-3})^3$ d) $(x^0 y^{-2})^2 \cdot (x^0 y^2)^{-2}$ e) $(3ab^5)^{-2} b$ f) $(a^3 \cdot b^4)^2 \cdot (a^3 \cdot b^4)^{-5}$ g) $((cd)^{-2})^3 \cdot ((cd)^{-2})^3$ h) $((cd)^{-2})^3 + ((cd)^{-2})^3$

Question

Accepted Answer

### Question 1: Analyze $\boldsymbol{(2 \times 3)^{-5}}$
# Explanation:
## Step 1: Recall negative exponent rule
The rule for negative exponents is $a^{-n}=\frac{1}{a^{n}}$ (where $a
eq0$ and $n$ is a positive integer). Also, $(ab)^n = a^n b^n$ (product - to - power rule). For $(2\times3)^{-5}$, first, we know that $2\times3 = 6$, so $(2\times3)^{-5}=6^{-5}$. Using the negative exponent rule, $6^{-5}=\frac{1}{6^{5}}$. But $6 = 2\times3$, so $6^{5}=(2\times3)^{5}=2^{5}\times3^{5}$ (by the product - to - power rule $(ab)^n=a^{n}b^{n}$). So $(2\times3)^{-5}=\frac{1}{2^{5}\times3^{5}}$.
- Option a: Since $(2\times3)^{-5}=\frac{1}{2^{5}\times3^{5}}$, it is a fraction. So option a is true.
- Option b: The expression $- 5(2\times3)$ is equal to $-30$, while $(2\times3)^{-5}=\frac{1}{2^{5}\times3^{5}}
eq - 30$. So option b is false.
- Option c: As we derived above, $(2\times3)^{-5}=\frac{1}{2^{5}\times3^{5}}$. So option c is true.
- Option d: $\frac{1}{5(2\times3)}=\frac{1}{30}$, and $(2\times3)^{-5}=\frac{1}{2^{5}\times3^{5}}=\frac{1}{7776}
eq\frac{1}{30}$. So option d is false.
- Option e: $(2\times3)^{-5}=\frac{1}{2^{5}\times3^{5}}$, which is a positive number (since it is a reciprocal of a positive number). So option e is false.

# Answer:
The true statements are a) It is a fraction and c) $=\frac{1}{2^{5}3^{5}}$

### Question 2a: Simplify $\boldsymbol{((ab)^{2})^{5}}$
# Explanation:
## Step 1: Recall power - of - a - power rule
The power - of - a - power rule states that $(a^{m})^{n}=a^{m\times n}$. For $((ab)^{2})^{5}$, let $a = ab$, $m = 2$ and $n = 5$.
Using the power - of - a - power rule $(a^{m})^{n}=a^{m\times n}$, we have $((ab)^{2})^{5}=(ab)^{2\times5}$
## Step 2: Simplify the exponent
$2\times5 = 10$, so $(ab)^{2\times5}=(ab)^{10}$
## Step 3: Apply product - to - power rule
Using the product - to - power rule $(ab)^{n}=a^{n}b^{n}$, $(ab)^{10}=a^{10}b^{10}$

# Answer:
$a^{10}b^{10}$

### Question 2b: Simplify $\boldsymbol{(a^{2}b^{2})^{5}}$
# Explanation:
## Step 1: Apply product - to - power rule
The product - to - power rule is $(ab)^{n}=a^{n}b^{n}$. For $(a^{2}b^{2})^{5}$, we can apply this rule. Let $a = a^{2}$, $b = b^{2}$ and $n = 5$.
So $(a^{2}b^{2})^{5}=(a^{2})^{5}(b^{2})^{5}$
## Step 2: Apply power - of - a - power rule
Using the power - of - a - power rule $(a^{m})^{n}=a^{m\times n}$, $(a^{2})^{5}=a^{2\times5}=a^{10}$ and $(b^{2})^{5}=b^{2\times5}=b^{10}$
## Step 3: Combine the results
$(a^{2})^{5}(b^{2})^{5}=a^{10}b^{10}$

# Answer:
$a^{10}b^{10}$

### Question 2c: Simplify $\boldsymbol{(a^{3}b^{-2})^{2}}$
# Explanation:
## Step 1: Apply product - to - power rule
Using the product - to - power rule $(ab)^{n}=a^{n}b^{n}$, we have $(a^{3}b^{-2})^{2}=(a^{3})^{2}(b^{-2})^{2}$
## Step 2: Apply power - of - a - power rule
Using the power - of - a - power rule $(a^{m})^{n}=a^{m\times n}$, $(a^{3})^{2}=a^{3\times2}=a^{6}$ and $(b^{-2})^{2}=b^{-2\times2}=b^{-4}$
## Step 3: Combine the results
$(a^{3})^{2}(b^{-2})^{2}=a^{6}b^{-4}$

# Answer:
$a^{6}b^{-4}$

### Question 2d: Simplify $\boldsymbol{((ab)^{5})^{-2}}$
# Explanation:
## Step 1: Apply power - of - a - power rule
Using the power - of - a - power rule $(a^{m})^{n}=a^{m\times n}$, for $((ab)^{5})^{-2}$, let $a = ab$, $m = 5$ and $n=-2$.
So $((ab)^{5})^{-2}=(ab)^{5\times(-2)}$
## Step 2: Simplify the exponent
$5\times(-2)=-10$, so $(ab)^{5\times(-2)}=(ab)^{-10}$
## Step 3: Apply product - to - power rule
Using the product - to - power rule $(ab)^{n}=a^{n}b^{n}$, $(ab)^{-10}=a^{-10}b^{-10}$

# Answer:
$a^{-10}b^{-10}$

### Question 2e: Simplify $\boldsymbol{(a^{5}b^{5})^{2}}$
# Explanation:
## Step 1: Apply product - to - power rule
Using the product - to - power rule $(ab)^{n}=a^{n}b^{n}$, for $(a^{5}b^{5})^{2}$, we get $(a^{5})^{2}(b^{5})^{2}$
## Step 2: Apply power - of - a - power rule
Using the power - of - a - power rule $(a^{m})^{n}=a^{m\times n}$, $(a^{5})^{2}=a^{5\times2}=a^{10}$ and $(b^{5})^{2}=b^{5\times2}=b^{10}$
## Step 3: Combine the results
$(a^{5})^{2}(b^{5})^{2}=a^{10}b^{10}$

# Answer:
$a^{10}b^{10}$

### Question 2f: Simplify $\boldsymbol{(a^{3}b^{2})^{-2}}$
# Explanation:
## Step 1: Apply product - to - power rule
Using the product - to - power rule $(ab)^{n}=a^{n}b^{n}$, we have $(a^{3}b^{2})^{-2}=(a^{3})^{-2}(b^{2})^{-2}$
## Step 2: Apply power - of - a - power rule
Using the power - of - a - power rule $(a^{m})^{n}=a^{m\times n}$, $(a^{3})^{-2}=a^{3\times(-2)}=a^{-6}$ and $(b^{2})^{-2}=b^{2\times(-2)}=b^{-4}$
## Step 3: Combine the results
$(a^{3})^{-2}(b^{2})^{-2}=a^{-6}b^{-4}$

# Answer:
$a^{-6}b^{-4}$

### Question 3a: Simplify $\boldsymbol{((xy)^{2})^{0}}$
# Explanation:
## Step 1: Recall the zero - exponent rule
The zero - exponent rule states that $a^{0}=1$ for $a
eq0$. For $((xy)^{2})^{0}$, first, use the power - of - a - power rule $(a^{m})^{n}=a^{m\times n}$. Let $a = xy$, $m = 2$ and $n = 0$. Then $((xy)^{2})^{0}=(xy)^{2\times0}=(xy)^{0}$
## Step 2: Apply zero - exponent rule
Using the zero - exponent rule $a^{0}=1$ (where $a = xy$ and $xy
eq0$), $(xy)^{0}=1$

# Answer:
$1$

### Question 3b: Simplify $\boldsymbol{(x^{2}y^{2})^{0}}$
# Explanation:
## Step 1: Apply zero - exponent rule
By the zero - exponent rule $a^{0}=1$ for $a
eq0$. Here $a=x^{2}y^{2}$, and as long as $x$ and $y$ are not both zero, $x^{2}y^{2}
eq0$. So $(x^{2}y^{2})^{0}=1$

# Answer:
$1$

### Question 3c: Simplify $\boldsymbol{(x^{0}y^{-2})^{2}}$
# Explanation:
## Step 1: Apply zero - exponent rule
By the zero - exponent rule, $x^{0}=1$ (for $x
eq0$). So the expression becomes $(1\times y^{-2})^{2}=(y^{-2})^{2}$
## Step 2: Apply power - of - a - power rule
Using the power - of - a - power rule $(a^{m})^{n}=a^{m\times n}$, $(y^{-2})^{2}=y^{-2\times2}=y^{-4}$
## Step 3: Recall negative exponent rule (optional for simplification check)
$y^{-4}=\frac{1}{y^{4}}$, but we can leave it as $y^{-4}$. However, let's check the value: $(x^{0}y^{-2})^{2}=(1\times y^{-2})^{2}=y^{-4}
eq1$

Wait, we made a mistake. Wait, $(x^{0}y^{-2})^{2}=(1\times y^{-2})^{2}=y^{-4}$, but we need to check if it equals $1$. Since $y^{-4}=\frac{1}{y^{4}}$, unless $y = \pm1$, it is not equal to $1$. Wait, no, let's re - do:

Wait, $(x^{0}y^{-2})^{2}=(x^{0})^{2}(y^{-2})^{2}$ (product - to - power rule). $(x^{0})^{2}=x^{0\times2}=x^{0}=1$ (zero - exponent rule), $(y^{-2})^{2}=y^{-4}$. So $(x^{0}y^{-2})^{2}=1\times y^{-4}=y^{-4}
eq1$ in general. But wait, maybe we misread. Wait, the question is which powers simplify to $1$.

Wait, let's start over.

## Step 1: Expand the expression
$(x^{0}y^{-2})^{2}=x^{0\times2}y^{-2\times2}$ (by product - to - power rule $(ab)^{n}=a^{n}b^{n}$ and power - of - a - power rule $(a^{m})^{n}=a^{m\times n}$)
## Step 2: Simplify exponents
$x^{0\times2}=x^{0}=1$ (zero - exponent rule), $y^{-2\times2}=y^{-4}$. So $(x^{0}y^{-2})^{2}=1\times y^{-4}=y^{-4}
eq1$ (unless $y=\pm1$, but in general, it is not equal to $1$)

### Question 3d: Simplify $\boldsymbol{((xy)^{0})^{-2}}$
# Explanation:
## Step 1: Apply zero - exponent rule
By the zero - exponent rule, $(xy)^{0}=1$ (for $xy
eq0$). So the expression becomes $(1)^{-2}$
## Step 2: Apply negative - exponent rule
Using the negative - exponent rule $a^{-n}=\frac{1}{a^{n}}$, for $a = 1$ and $n = 2$, $1^{-2}=\frac{1}{1^{2}} = 1$

# Answer:
$1$

### Question 3e: Simplify $\boldsymbol{(x^{0}y^{0})^{2}}$
# Explanation:
## Step 1: Apply zero - exponent rule
By the zero - exponent rule, $x^{0}=1$ (for $x
eq0$) and $y^{0}=1$ (for $y
eq0$). So $x^{0}y^{0}=1\times1 = 1$
## Step 2: Apply power - of - a - power rule
$(x^{0}y^{0})^{2}=1^{2}=1$

# Answer:
$1$

### Question 3f: Simplify $\boldsymbol{(x^{0}y^{2})^{-2}}$
# Explanation:
## Step 1: Apply zero - exponent rule
By the zero - exponent rule, $x^{0}=1$ (for $x
eq0$). So the expression becomes $(1\times y^{2})^{-2}=(y^{2})^{-2}$
## Step 2: Apply power - of - a - power rule
Using the power - of - a - power rule $(a^{m})^{n}=a^{m\times n}$, $(y^{2})^{-2}=y^{2\times(-2)}=y^{-4}$
## Step 3: Check if equal to 1
$y^{-4}=\frac{1}{y^{4}}
eq1$ (unless $y = \pm1$, in general, it is not equal to $1$)

So the powers that simplify to $1$ are:
- a) $((xy)^{2})^{0}$
- b) $(x^{2}y^{2})^{0}$
- d) $((xy)^{0})^{-2}$
- e) $(x^{0}y^{0})^{2}$

### Question 4a: Complete Learner 1's approach
For Learner 1:
$(ab^{2})^{-5}\cdot(ab^{2})^{-5}$
$=a^{-5}b^{2\times(-5)}\cdot a^{-5}b^{2\times(-5)}$ (using the power - of - a - product rule $(ab^{n})^{m}=a^{m}b^{n\times m}$)
$=a^{(-5 + (-5))}b^{(-10+(-10))}$ (using the product rule for exponents $a^{m}\cdot a^{n}=a^{m + n}$)
$=a^{-10}b^{-20}$
$=\frac{1}{a^{10}b^{20}}$

Filling in the blanks:
- First blank in $a^{\square}b^{2\times\square}$: $-5$
- Second blank in $a^{\square}b^{2\times\square}$: $-5$
- Third blank in $a^{(-5+\square)}b^{(-10 + (-10))}$: $-5$
- Fourth blank in $a^{\square}b^{\square}$: $-10$
- Fifth blank in $a^{\square}b^{\square}$: $-20$

For Learner 2:
$(ab^{2})^{-5}\cdot(ab^{2})^{-5}$
$=(ab^{2})^{-5+(-5)}$ (using the product rule for exponents $a^{m}\cdot a^{n}=a^{m + n}$, here $a = ab^{2}$, $m=-5$, $n = - 5$)
$=a^{-10}b^{2\times(-10)}$
$=a^{-10}b^{-20}$
$=\frac{1}{a^{10}b^{20}}$

Filling in the blanks:
- First blank in $(ab^{2})^{\square+\square}$: $-5$
- Second blank in $(ab^{2})^{\square+\square}$: $-5$
- Third blank in $a^{\square}b^{\square}$: $-10$
- Fourth blank in $a^{\square}b^{\square}$: $-20$

### Question 4b: Describe the difference between the two

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QUESTION IMAGE

Question

Explanation:

Question 1: Analyze \(\boldsymbol{(2 \times 3)^{-5}}\)

Step 1: Recall negative exponent rule

Step 1: Recall power - of - a - power rule

Step 2: Simplify the exponent

Step 3: Apply product - to - power rule

Step 1: Apply product - to - power rule

Step 2: Apply power - of - a - power rule

Step 3: Combine the results

Answer:

Question 2a: Simplify \(\boldsymbol{((ab)^{2})^{5}}\)