write each expression in exponential form.
19) $(sqrt4{m})^3$
20) $(sqrt3{6x})^4$
21) $sqrt4{v}$
22) $sqrt{6p}$
23) $(sqrt3{3a})^4$
24) $\frac{1}{(sqrt{3k})^5}$
simplify.
25) $9^{\frac{1}{2}}$
26) $343^{-\frac{4}{3}}$
27) $1000000^{\frac{1}{6}}$
28) $36^{\frac{3}{2}}$
29) $(x^6)^{\frac{1}{2}}$
30) $(9n^4)^{\frac{1}{2}}$
31) $(64n^{12})^{-\frac{1}{6}}$
32) $(81m^6)^{\frac{1}{2}}$

Question

Accepted Answer

### Problem 19: $(\sqrt[4]{m})^3$
# Explanation:
## Step1: Recall the radical to exponent rule
The $n$-th root of a number $a$ can be written as $a^{\frac{1}{n}}$. So, $\sqrt[4]{m}=m^{\frac{1}{4}}$.
## Step2: Apply the power of a power rule
When we raise a power to a power, we multiply the exponents. So, $(m^{\frac{1}{4}})^3 = m^{\frac{1}{4}\times3}$.
## Step3: Simplify the exponent
$\frac{1}{4}\times3=\frac{3}{4}$, so the expression becomes $m^{\frac{3}{4}}$.
# Answer:
$m^{\frac{3}{4}}$

### Problem 20: $(\sqrt[3]{6x})^4$
# Explanation:
## Step1: Convert the cube root to exponent form
$\sqrt[3]{6x}=(6x)^{\frac{1}{3}}$ (using the rule $\sqrt[n]{a}=a^{\frac{1}{n}}$).
## Step2: Apply the power of a power rule
$((6x)^{\frac{1}{3}})^4=(6x)^{\frac{1}{3}\times4}$ (since $(a^m)^n = a^{m\times n}$).
## Step3: Simplify the exponent
$\frac{1}{3}\times4 = \frac{4}{3}$, so the expression is $(6x)^{\frac{4}{3}}$.
# Answer:
$(6x)^{\frac{4}{3}}$

### Problem 21: $\sqrt[4]{\sqrt{v}}$
# Explanation:
## Step1: Convert the inner square root to exponent form
$\sqrt{v}=v^{\frac{1}{2}}$ (using $\sqrt[n]{a}=a^{\frac{1}{n}}$ with $n = 2$).
## Step2: Convert the outer fourth root to exponent form
Now we have $\sqrt[4]{v^{\frac{1}{2}}}$, which can be written as $(v^{\frac{1}{2}})^{\frac{1}{4}}$ (using $\sqrt[n]{a}=a^{\frac{1}{n}}$ with $n = 4$).
## Step3: Apply the power of a power rule
Multiply the exponents: $\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}$, so the expression is $v^{\frac{1}{8}}$.
# Answer:
$v^{\frac{1}{8}}$

### Problem 22: $\sqrt{6p}$
# Explanation:
## Step1: Recall the square root to exponent rule
The square root of a number $a$ is $a^{\frac{1}{2}}$. So, $\sqrt{6p}=(6p)^{\frac{1}{2}}$ (since $\sqrt[n]{a}=a^{\frac{1}{n}}$ with $n = 2$).
# Answer:
$(6p)^{\frac{1}{2}}$

### Problem 23: $(\sqrt[3]{3a})^4$
# Explanation:
## Step1: Convert the cube root to exponent form
$\sqrt[3]{3a}=(3a)^{\frac{1}{3}}$ (using $\sqrt[n]{a}=a^{\frac{1}{n}}$ with $n = 3$).
## Step2: Apply the power of a power rule
$((3a)^{\frac{1}{3}})^4=(3a)^{\frac{1}{3}\times4}$ (since $(a^m)^n=a^{m\times n}$).
## Step3: Simplify the exponent
$\frac{1}{3}\times4=\frac{4}{3}$, so the expression is $(3a)^{\frac{4}{3}}$.
# Answer:
$(3a)^{\frac{4}{3}}$

### Problem 24: $\frac{1}{(\sqrt{3k})^5}$
# Explanation:
## Step1: Convert the square root to exponent form
$\sqrt{3k}=(3k)^{\frac{1}{2}}$ (using $\sqrt[n]{a}=a^{\frac{1}{n}}$ with $n = 2$).
## Step2: Apply the power of a power rule to the denominator
$((3k)^{\frac{1}{2}})^5=(3k)^{\frac{1}{2}\times5}=(3k)^{\frac{5}{2}}$ (since $(a^m)^n=a^{m\times n}$).
## Step3: Recall the negative exponent rule
$\frac{1}{a^m}=a^{-m}$, so $\frac{1}{(3k)^{\frac{5}{2}}}=(3k)^{-\frac{5}{2}}$.
# Answer:
$(3k)^{-\frac{5}{2}}$

### Problem 25: $9^{\frac{1}{2}}$
# Explanation:
## Step1: Recall the exponent rule for square roots
$a^{\frac{1}{2}}=\sqrt{a}$, so $9^{\frac{1}{2}}=\sqrt{9}$.
## Step2: Evaluate the square root
$\sqrt{9} = 3$ (since $3\times3 = 9$).
# Answer:
$3$

### Problem 26: $343^{-\frac{4}{3}}$
# Explanation:
## Step1: Recall the negative exponent rule
$a^{-m}=\frac{1}{a^m}$, so $343^{-\frac{4}{3}}=\frac{1}{343^{\frac{4}{3}}}$.
## Step2: Recall the exponent rule for cube roots
$a^{\frac{1}{3}}=\sqrt[3]{a}$, and $343 = 7^3$, so $343^{\frac{1}{3}}=\sqrt[3]{343}=\sqrt[3]{7^3}=7$.
## Step3: Apply the power of a power rule
$343^{\frac{4}{3}}=(343^{\frac{1}{3}})^4$. Since $343^{\frac{1}{3}} = 7$, then $(343^{\frac{1}{3}})^4=7^4$.
## Step4: Evaluate $7^4$
$7^4=7\times7\times7\times7 = 2401$.
## Step5: So, $\frac{1}{343^{\frac{4}{3}}}=\frac{1}{2401}$
# Answer:
$\frac{1}{2401}$

### Problem 27: $1000000^{\frac{1}{6}}$
# Explanation:
## Step1: Recall the exponent rule for $n$-th roots
$a^{\frac{1}{n}}=\sqrt[n]{a}$, so $1000000^{\frac{1}{6}}=\sqrt[6]{1000000}$.
## Step2: Notice that $1000000 = 10^6$
So, $\sqrt[6]{10^6}=10$ (since the $n$-th root of $a^n$ is $a$ for real numbers $a$ and positive integers $n$).
# Answer:
$10$

### Problem 28: $36^{\frac{3}{2}}$
# Explanation:
## Step1: Recall the exponent rule for square roots
$a^{\frac{1}{2}}=\sqrt{a}$, so $36^{\frac{3}{2}}=(36^{\frac{1}{2}})^3$ (using $(a^m)^n=a^{m\times n}$ with $m=\frac{1}{2}$ and $n = 3$).
## Step2: Evaluate $36^{\frac{1}{2}}$
$36^{\frac{1}{2}}=\sqrt{36}=6$ (since $6\times6 = 36$).
## Step3: Evaluate $6^3$
$6^3=6\times6\times6 = 216$.
# Answer:
$216$

### Problem 29: $(x^6)^{\frac{1}{2}}$
# Explanation:
## Step1: Apply the power of a power rule
$(a^m)^n=a^{m\times n}$, so $(x^6)^{\frac{1}{2}}=x^{6\times\frac{1}{2}}$.
## Step2: Simplify the exponent
$6\times\frac{1}{2}=3$, so the expression is $x^3$.
# Answer:
$x^3$

### Problem 30: $(9n^4)^{\frac{1}{2}}$
# Explanation:
## Step1: Apply the power of a product rule
$(ab)^m=a^m\times b^m$, so $(9n^4)^{\frac{1}{2}}=9^{\frac{1}{2}}\times(n^4)^{\frac{1}{2}}$.
## Step2: Evaluate $9^{\frac{1}{2}}$
$9^{\frac{1}{2}}=\sqrt{9}=3$ (since $3\times3 = 9$).
## Step3: Apply the power of a power rule to $n^4$
$(n^4)^{\frac{1}{2}}=n^{4\times\frac{1}{2}}=n^2$ (using $(a^m)^n=a^{m\times n}$).
## Step4: Multiply the results
$3\times n^2 = 3n^2$.
# Answer:
$3n^2$

### Problem 31: $(64n^{12})^{-\frac{1}{6}}$
# Explanation:
## Step1: Apply the power of a product rule
$(ab)^m=a^m\times b^m$, so $(64n^{12})^{-\frac{1}{6}}=64^{-\frac{1}{6}}\times(n^{12})^{-\frac{1}{6}}$.
## Step2: Recall the negative exponent rule and evaluate $64^{-\frac{1}{6}}$
$a^{-m}=\frac{1}{a^m}$, so $64^{-\frac{1}{6}}=\frac{1}{64^{\frac{1}{6}}}$. Also, $64 = 2^6$, so $64^{\frac{1}{6}}=\sqrt[6]{64}=\sqrt[6]{2^6}=2$, thus $\frac{1}{64^{\frac{1}{6}}}=\frac{1}{2}$.
## Step3: Apply the power of a power rule to $n^{12}$
$(n^{12})^{-\frac{1}{6}}=n^{12\times(-\frac{1}{6})}=n^{-2}$ (using $(a^m)^n=a^{m\times n}$).
## Step4: Recall the negative exponent rule for $n^{-2}$
$n^{-2}=\frac{1}{n^2}$, so now we have $\frac{1}{2}\times\frac{1}{n^2}=\frac{1}{2n^2}$.
# Answer:
$\frac{1}{2n^2}$

### Problem 32: $(81m^6)^{\frac{1}{2}}$
# Explanation:
## Step1: Apply the power of a product rule
$(ab)^m=a^m\times b^m$, so $(81m^6)^{\frac{1}{2}}=81^{\frac{1}{2}}\times(m^6)^{\frac{1}{2}}$.
## Step2: Evaluate $81^{\frac{1}{2}}$
$81^{\frac{1}{2}}=\sqrt{81}=9$ (since $9\times9 = 81$).
## Step3: Apply the power of a power rule to $m^6$
$(m^6)^{\frac{1}{2}}=m^{6\times\frac{1}{2}}=m^3$ (using $(a^m)^n=a^{m\times n}$).
## Step4: Multiply the results
$9\times m^3 = 9m^3$.
# Answer:
$9m^3$

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QUESTION IMAGE

Question

Explanation:

Problem 19: \((\sqrt[4]{m})^3\)

Step1: Recall the radical to exponent rule

Step2: Apply the power of a power rule

Step3: Simplify the exponent

Step1: Convert the cube root to exponent form

Step2: Apply the power of a power rule

Step3: Simplify the exponent

Step1: Convert the inner square root to exponent form

Step2: Convert the outer fourth root to exponent form

Step3: Apply the power of a power rule

Answer:

Problem 20: \((\sqrt[3]{6x})^4\)