5. write the expression $17^{\frac{3}{4}}$ using radicals.

6. write each expression in the form $a^b$, without using any radicals.
1. $\sqrt{5^9}$
2. $\frac{1}{\sqrt3{12}}$

7. write $32^{-\frac{2}{5}}$ without using exponents or radicals.

8.
match the equivalent expressions.
a: $8^{\frac{1}{3}}$ 1: $\frac{1}{8}$
b: $8^{-\frac{1}{3}}$ 2: $\frac{1}{4}$
c: $8^{-1}$ 3: $\frac{1}{2}$
d: $16^{\frac{1}{2}}$ 4: $1$
e: $16^{-\frac{1}{2}}$ 5: $2$
f: $16^0$ 6: $4$

Question

5. write the expression $17^{\frac{3}{4}}$ using radicals.

6. write each expression in the form $a^b$, without using any radicals.
1. $\sqrt{5^9}$
2. $\frac{1}{\sqrt3{12}}$

7. write $32^{-\frac{2}{5}}$ without using exponents or radicals.

8.
match the equivalent expressions.
a: $8^{\frac{1}{3}}$ 1: $\frac{1}{8}$
b: $8^{-\frac{1}{3}}$ 2: $\frac{1}{4}$
c: $8^{-1}$ 3: $\frac{1}{2}$
d: $16^{\frac{1}{2}}$ 4: $1$
e: $16^{-\frac{1}{2}}$ 5: $2$
f: $16^0$ 6: $4$

Accepted Answer

### Question 5
# Explanation:
## Step1: Recall the exponent - radical relationship
The formula for converting an expression with a fractional exponent to a radical is $a^{\frac{m}{n}}=\sqrt[n]{a^{m}}$ (where $n\gt0$, $a$ is a real number, and $m,n$ are integers). For the expression $17^{\frac{3}{4}}$, we have $a = 17$, $m=3$ and $n = 4$.
## Step2: Apply the formula
Using the formula $a^{\frac{m}{n}}=\sqrt[n]{a^{m}}$, we substitute $a = 17$, $m = 3$ and $n=4$ into it. So $17^{\frac{3}{4}}=\sqrt[4]{17^{3}}$

# Answer: $\sqrt[4]{17^{3}}$

### Question 6 (1)
# Explanation:
## Step1: Recall the radical - exponent relationship
The formula for converting a radical to an exponent is $\sqrt[n]{a^{m}}=a^{\frac{m}{n}}$. For the expression $\sqrt{5^{9}}$, the square root is the same as the $\frac{1}{2}$-th root, so $n = 2$ and $m=9$.
## Step2: Apply the formula
Using the formula $\sqrt[n]{a^{m}}=a^{\frac{m}{n}}$, we substitute $a = 5$, $m = 9$ and $n = 2$ into it. So $\sqrt{5^{9}}=5^{\frac{9}{2}}$

# Answer: $5^{\frac{9}{2}}$

### Question 6 (2)
# Explanation:
## Step1: Recall the radical - exponent and negative exponent rules
First, recall that $\frac{1}{\sqrt[n]{a}}=\frac{1}{a^{\frac{1}{n}}}=a^{-\frac{1}{n}}$. For the expression $\frac{1}{\sqrt[3]{12}}$, the cube root is the same as the $\frac{1}{3}$-th root, so $n = 3$ and $a = 12$.
## Step2: Apply the rules
Using the rule $\frac{1}{\sqrt[n]{a}}=a^{-\frac{1}{n}}$, we substitute $a = 12$ and $n = 3$ into it. So $\frac{1}{\sqrt[3]{12}}=12^{-\frac{1}{3}}$

# Answer: $12^{-\frac{1}{3}}$

### Question 7
# Explanation:
## Step1: Recall the exponent rules
First, recall the rules $a^{-n}=\frac{1}{a^{n}}$ and $a^{\frac{m}{n}}=\sqrt[n]{a^{m}}$ or $(\sqrt[n]{a})^{m}$. Also, note that $32 = 2^{5}$. For the expression $32^{-\frac{2}{5}}$, we can first use the negative exponent rule $a^{-n}=\frac{1}{a^{n}}$ to get $\frac{1}{32^{\frac{2}{5}}}$. Then, since $32=2^{5}$, we substitute $32$ with $2^{5}$ in the expression $\frac{1}{32^{\frac{2}{5}}}$, so we have $\frac{1}{(2^{5})^{\frac{2}{5}}}$
## Step2: Simplify the exponent
Using the power - of - a - power rule $(a^{m})^{n}=a^{mn}$, for $(2^{5})^{\frac{2}{5}}$, we have $5\times\frac{2}{5}=2$, so $(2^{5})^{\frac{2}{5}}=2^{2}=4$. Then $\frac{1}{(2^{5})^{\frac{2}{5}}}=\frac{1}{4}$

# Answer: $\frac{1}{4}$

### Question 8
# Explanation:
- For $A:8^{\frac{1}{3}}$: Recall that $8 = 2^{3}$, so $8^{\frac{1}{3}}=(2^{3})^{\frac{1}{3}}$. Using the power - of - a - power rule $(a^{m})^{n}=a^{mn}$, we get $2^{3\times\frac{1}{3}}=2^{1}=2$, so $A$ matches $5$.
- For $B:8^{-\frac{1}{3}}$: Using the negative exponent rule $a^{-n}=\frac{1}{a^{n}}$, we have $8^{-\frac{1}{3}}=\frac{1}{8^{\frac{1}{3}}}$. Since $8^{\frac{1}{3}} = 2$, then $\frac{1}{8^{\frac{1}{3}}}=\frac{1}{2}$, so $B$ matches $3$.
- For $C:8^{-1}$: Using the negative exponent rule $a^{-n}=\frac{1}{a^{n}}$, we have $8^{-1}=\frac{1}{8}$, so $C$ matches $1$.
- For $D:16^{\frac{1}{2}}$: The square root of $16$ is $4$, because $4\times4 = 16$, so $16^{\frac{1}{2}}=\sqrt{16}=4$, so $D$ matches $6$.
- For $E:16^{-\frac{1}{2}}$: Using the negative exponent rule $a^{-n}=\frac{1}{a^{n}}$, we have $16^{-\frac{1}{2}}=\frac{1}{16^{\frac{1}{2}}}$. Since $16^{\frac{1}{2}} = 4$, then $\frac{1}{16^{\frac{1}{2}}}=\frac{1}{4}$, so $E$ matches $2$.
- For $F:16^{0}$: Any non - zero number to the power of $0$ is $1$, so $16^{0}=1$, so $F$ matches $4$.

# Answer:
- $A$ matches $5$
- $B$ matches $3$
- $C$ matches $1$
- $D$ matches $6$
- $E$ matches $2$
- $F$ matches $4$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 5

Step1: Recall the exponent - radical relationship

Step2: Apply the formula

Step1: Recall the radical - exponent relationship

Step2: Apply the formula

Step1: Recall the radical - exponent and negative exponent rules

Step2: Apply the rules

Answer:

Question 6 (1)