find the value of each power.
1. $5^3 =$
2. $7^{-2} =$
3. $51^1 =$
4. $3^{-4} =$
5. $1^{12} =$
6. $64^0 =$
7. $4^{-3} =$
8. $4^3 =$
9. $10^5 =$
find the missing exponent.
10. $n^3 = n^{square} \cdot n^{-3}$
11. $\frac{a^{square}}{a^2} = a^4$
12. $(r^4)^{square} = r^{12}$
simplify each expression.
13. $(9 - 3)^2 - (5 \cdot 4)^0 =$
14. $(2 + 3)^5 \div (5^2)^2 =$
15. $4^2 \div (6 - 2)^4 =$
16. $(1 + 7)^2^2 \cdot (12^2)^0 =$
use the description below to complete exercises 17–20.

Question

Accepted Answer

### Problem 2: $7^{-2}$
# Explanation:
## Step 1: Recall the negative exponent rule
The negative exponent rule states that $a^{-n}=\frac{1}{a^{n}}$ for any non - zero number $a$ and positive integer $n$. For $a = 7$ and $n=2$, we have $7^{-2}=\frac{1}{7^{2}}$
## Step 2: Calculate $7^{2}$
We know that $7^{2}=7\times7 = 49$. So $\frac{1}{7^{2}}=\frac{1}{49}$
# Answer: $\frac{1}{49}$

### Problem 4: $3^{-4}$
# Explanation:
## Step 1: Apply the negative exponent rule
Using the rule $a^{-n}=\frac{1}{a^{n}}$ with $a = 3$ and $n = 4$, we get $3^{-4}=\frac{1}{3^{4}}$
## Step 2: Compute $3^{4}$
$3^{4}=3\times3\times3\times3=81$. So $\frac{1}{3^{4}}=\frac{1}{81}$
# Answer: $\frac{1}{81}$

### Problem 7: $4^{-3}$
# Explanation:
## Step 1: Use the negative exponent rule
By the rule $a^{-n}=\frac{1}{a^{n}}$, for $a = 4$ and $n=3$, we have $4^{-3}=\frac{1}{4^{3}}$
## Step 2: Calculate $4^{3}$
$4^{3}=4\times4\times4 = 64$. So $\frac{1}{4^{3}}=\frac{1}{64}$
# Answer: $\frac{1}{64}$

### Problem 10: Find the missing exponent in $n^{3}=n^{\square}\cdot n^{-3}$
# Explanation:
## Step 1: Recall the product of powers rule
The product of powers rule states that $a^{m}\cdot a^{n}=a^{m + n}$ for any non - zero number $a$ and integers $m$ and $n$. Let the missing exponent be $x$. Then $n^{3}=n^{x}\cdot n^{-3}=n^{x+( - 3)}=n^{x - 3}$
## Step 2: Solve for $x$
Since the bases are the same ($n$), we can set the exponents equal to each other. So $3=x - 3$. Adding 3 to both sides of the equation, we get $x=3 + 3=6$
# Answer: $6$

### Problem 11: Find the missing exponent in $\frac{a^{\square}}{a^{2}}=a^{4}$
# Explanation:
## Step 1: Recall the quotient of powers rule
The quotient of powers rule is $\frac{a^{m}}{a^{n}}=a^{m - n}$ for $a
eq0$ and integers $m$ and $n$. Let the missing exponent be $x$. Then $\frac{a^{x}}{a^{2}}=a^{x-2}$
## Step 2: Solve for $x$
Since $\frac{a^{x}}{a^{2}}=a^{4}$, we have $a^{x - 2}=a^{4}$. Setting the exponents equal (because the bases are the same), we get $x-2 = 4$. Adding 2 to both sides, $x=4 + 2=6$
# Answer: $6$

### Problem 12: Find the missing exponent in $(r^{4})^{\square}=r^{12}$
# Explanation:
## Step 1: Recall the power of a power rule
The power of a power rule is $(a^{m})^{n}=a^{m\times n}$ for any non - zero number $a$ and integers $m$ and $n$. Let the missing exponent be $x$. Then $(r^{4})^{x}=r^{4x}$
## Step 2: Solve for $x$
Since $(r^{4})^{x}=r^{12}$, we have $4x = 12$. Dividing both sides by 4, $x=\frac{12}{4}=3$
# Answer: $3$

### Problem 13: $(9 - 3)^{2}-(5\cdot4)^{0}$
# Explanation:
## Step 1: Simplify the expressions inside the parentheses
First, calculate $9 - 3=6$ and $5\cdot4 = 20$
## Step 2: Evaluate the powers
For $(9 - 3)^{2}$, we have $6^{2}=6\times6 = 36$. For $(5\cdot4)^{0}$, any non - zero number to the power of 0 is 1, so $20^{0}=1$
## Step 3: Subtract the two results
$36-1 = 35$ (Note: There was a mistake in the original hand - written answer. The correct calculation is as above)
# Answer: $35$

### Problem 14: $(2 + 3)^{5}\div(5^{2})^{2}$
# Explanation:
## Step 1: Simplify the expressions inside the parentheses and the powers
First, $2 + 3=5$, so $(2 + 3)^{5}=5^{5}$. Then $(5^{2})^{2}$ using the power of a power rule $(a^{m})^{n}=a^{mn}$, we have $5^{2\times2}=5^{4}$
## Step 2: Use the quotient of powers rule
$5^{5}\div5^{4}$ using $\frac{a^{m}}{a^{n}}=a^{m - n}$ (where $a = 5$, $m = 5$, $n = 4$) gives $5^{5-4}=5^{1}=5$
# Answer: $5$

### Problem 15: $4^{2}\div(6 - 2)^{4}$
# Explanation:
## Step 1: Simplify the expressions inside the parentheses and the powers
First, $6 - 2=4$, $4^{2}=16$ and $(6 - 2)^{4}=4^{4}=4\times4\times4\times4 = 256$
## Step 2: Perform the division
$16\div256=\frac{16}{256}=\frac{1}{16}$ (Note: There was a mistake in the original hand - written answer. The correct calculation is as above)
# Answer: $\frac{1}{16}$

### Problem 16: $[(1 + 7)^{2}]^{2}\cdot(12^{2})^{0}$
# Explanation:
## Step 1: Simplify the expressions inside the parentheses and the powers
First, $1+7 = 8$, so $(1 + 7)^{2}=8^{2}=64$. Then $[(1 + 7)^{2}]^{2}=64^{2}$ or using the power of a power rule $(8^{2})^{2}=8^{4}=4096$. Also, $(12^{2})^{0}=1$ (any non - zero number to the power of 0 is 1)
## Step 2: Multiply the two results
$4096\times1 = 4096$
# Answer: $4096$

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QUESTION IMAGE

Question

Explanation:

Problem 2: \(7^{-2}\)

Step 1: Recall the negative exponent rule

Step 2: Calculate \(7^{2}\)

Step 1: Apply the negative exponent rule

Step 2: Compute \(3^{4}\)

Step 1: Use the negative exponent rule

Step 2: Calculate \(4^{3}\)

Answer:

Problem 4: \(3^{-4}\)