Question

write each expression in exponential form.

1. $(sqrt4{m})^3$
2. $(sqrt7{6x})^4$
3. $sqrt5{v}$
4. $sqrt{6p}$
5. $(sqrt3{3a})^4$
6. $\frac{1}{(sqrt{3k})^3}$

simplify:

1. $9^{\frac{1}{2}}$
2. $343^{-\frac{4}{3}}$
3. $1000000^{\frac{1}{6}}$
4. $36^{\frac{1}{2}}$
5. $(x^6)^{\frac{1}{2}}$
6. $(9n^4)^{\frac{1}{2}}$
7. $(64n^{12})^{-\frac{1}{6}}$
8. $(81m^6)^{\frac{1}{2}}$

Explanation:

Let's solve these problems one by one. We'll start with writing expressions in exponential form and then simplifying the exponential expressions.

Writing Expressions in Exponential Form (Problems 19 - 24)

Recall the rule: $\sqrt[n]{a} = a^{\frac{1}{n}}$ and $(\sqrt[n]{a})^m = a^{\frac{m}{n}}$. Also, $\frac{1}{a^m} = a^{-m}$.

Problem 19: $(\sqrt[5]{m})^3$

Using the rule $(\sqrt[n]{a})^m = a^{\frac{m}{n}}$, here $n = 5$, $m = 3$, and $a = m$. So:
$(\sqrt[5]{m})^3 = m^{\frac{3}{5}}$

Problem 20: $(\sqrt[7]{6x})^4$

Using the same rule, $n = 7$, $m = 4$, $a = 6x$. So:
$(\sqrt[7]{6x})^4 = (6x)^{\frac{4}{7}}$

Problem 21: $\sqrt[8]{v}$

Using $\sqrt[n]{a} = a^{\frac{1}{n}}$, here $n = 8$, $a = v$. So:
$\sqrt[8]{v} = v^{\frac{1}{8}}$

Problem 22: $\sqrt{6p}$

Recall that $\sqrt{a} = a^{\frac{1}{2}}$, so here $n = 2$, $a = 6p$. So:
$\sqrt{6p} = (6p)^{\frac{1}{2}}$

Problem 23: $(\sqrt[3]{3a})^4$

Using $(\sqrt[n]{a})^m = a^{\frac{m}{n}}$, $n = 3$, $m = 4$, $a = 3a$. So:
$(\sqrt[3]{3a})^4 = (3a)^{\frac{4}{3}}$

Problem 24: $\frac{1}{(\sqrt{3k})^5}$

First, rewrite $(\sqrt{3k})^5$ as $((3k)^{\frac{1}{2}})^5$. Using the power of a power rule $(a^m)^n = a^{mn}$, we get $(3k)^{\frac{5}{2}}$. Then, $\frac{1}{(3k)^{\frac{5}{2}}} = (3k)^{-\frac{5}{2}}$

Simplifying Exponential Expressions (Problems 25 - 32)

Recall the rules: $a^{\frac{m}{n}} = \sqrt[n]{a^m}$, $a^{-\frac{m}{n}} = \frac{1}{a^{\frac{m}{n}}} = \frac{1}{\sqrt[n]{a^m}}$, and $(ab)^n = a^n b^n$.

Problem 25: $9^{\frac{1}{2}}$

$9^{\frac{1}{2}} = \sqrt{9} = 3$ (since $3^2 = 9$)

Problem 26: $343^{-\frac{1}{3}}$

First, rewrite the negative exponent: $343^{-\frac{1}{3}} = \frac{1}{343^{\frac{1}{3}}}$. Now, $343^{\frac{1}{3}} = \sqrt[3]{343} = 7$ (since $7^3 = 343$). So $\frac{1}{7}$

Problem 27: $1000000^{\frac{1}{6}}$

$1000000^{\frac{1}{6}} = \sqrt[6]{1000000}$. We know that $10^6 = 1000000$, so $\sqrt[6]{10^6} = 10$

Problem 28: $36^{\frac{1}{2}}$

$36^{\frac{1}{2}} = \sqrt{36} = 6$ (since $6^2 = 36$)

Problem 29: $(x^6)^{\frac{1}{2}}$

Using the power of a power rule: $(x^6)^{\frac{1}{2}} = x^{6 \times \frac{1}{2}} = x^3$

Problem 30: $(9n^4)^{\frac{1}{2}}$

Using the product rule for exponents: $(9n^4)^{\frac{1}{2}} = 9^{\frac{1}{2}} \times (n^4)^{\frac{1}{2}}$. We know $9^{\frac{1}{2}} = 3$ and $(n^4)^{\frac{1}{2}} = n^{4 \times \frac{1}{2}} = n^2$. So $3n^2$

Problem 31: $(64n^{12})^{-\frac{1}{6}}$

First, rewrite the negative exponent: $(64n^{12})^{-\frac{1}{6}} = \frac{1}{(64n^{12})^{\frac{1}{6}}}$. Now, apply the product rule: $\frac{1}{64^{\frac{1}{6}} \times (n^{12})^{\frac{1}{6}}}$. Calculate $64^{\frac{1}{6}}$: since $2^6 = 64$, $64^{\frac{1}{6}} = 2$. Calculate $(n^{12})^{\frac{1}{6}} = n^{12 \times \frac{1}{6}} = n^2$. So $\frac{1}{2n^2}$

Problem 32: $(81m^6)^{\frac{1}{2}}$

Using the product rule: $(81m^6)^{\frac{1}{2}} = 81^{\frac{1}{2}} \times (m^6)^{\frac{1}{2}}$. $81^{\frac{1}{2}} = 9$ (since $9^2 = 81$) and $(m^6)^{\frac{1}{2}} = m^{6 \times \frac{1}{2}} = m^3$. So $9m^3$

Final Answers

Writing in Exponential Form:

1. $\boldsymbol{m^{\frac{3}{5}}}$
2. $\boldsymbol{(6x)^{\frac{4}{7}}}$
3. $\boldsymbol{v^{\frac{1}{8}}}$
4. $\boldsymbol{(6p)^{\frac{1}{2}}}$
5. $\boldsymbol{(3a)^{\frac{4}{3}}}$
6. $\boldsymbol{(3k)^{-\frac{5}{2}}}$

Simplifying:

1. $\boldsymbol{3}$
2. $\boldsymbol{\frac{1}{7}}$
3. $\boldsymbol{10}$
4. $\boldsymbol{6}$
5. $\boldsymbol{x^3}$
6. $\boldsymbol{3n^2}$
7. $\boldsymbol{\frac{1}{2n^2}}$
8. $\boldsymbol{9m^3}$

Answer:

Let's solve these problems one by one. We'll start with writing expressions in exponential form and then simplifying the exponential expressions.

Writing Expressions in Exponential Form (Problems 19 - 24)

Recall the rule: $\sqrt[n]{a} = a^{\frac{1}{n}}$ and $(\sqrt[n]{a})^m = a^{\frac{m}{n}}$. Also, $\frac{1}{a^m} = a^{-m}$.

Problem 19: $(\sqrt[5]{m})^3$

Using the rule $(\sqrt[n]{a})^m = a^{\frac{m}{n}}$, here $n = 5$, $m = 3$, and $a = m$. So:
$(\sqrt[5]{m})^3 = m^{\frac{3}{5}}$

Problem 20: $(\sqrt[7]{6x})^4$

Using the same rule, $n = 7$, $m = 4$, $a = 6x$. So:
$(\sqrt[7]{6x})^4 = (6x)^{\frac{4}{7}}$

Problem 21: $\sqrt[8]{v}$

Using $\sqrt[n]{a} = a^{\frac{1}{n}}$, here $n = 8$, $a = v$. So:
$\sqrt[8]{v} = v^{\frac{1}{8}}$

Problem 22: $\sqrt{6p}$

Recall that $\sqrt{a} = a^{\frac{1}{2}}$, so here $n = 2$, $a = 6p$. So:
$\sqrt{6p} = (6p)^{\frac{1}{2}}$

Problem 23: $(\sqrt[3]{3a})^4$

Using $(\sqrt[n]{a})^m = a^{\frac{m}{n}}$, $n = 3$, $m = 4$, $a = 3a$. So:
$(\sqrt[3]{3a})^4 = (3a)^{\frac{4}{3}}$

Problem 24: $\frac{1}{(\sqrt{3k})^5}$

First, rewrite $(\sqrt{3k})^5$ as $((3k)^{\frac{1}{2}})^5$. Using the power of a power rule $(a^m)^n = a^{mn}$, we get $(3k)^{\frac{5}{2}}$. Then, $\frac{1}{(3k)^{\frac{5}{2}}} = (3k)^{-\frac{5}{2}}$

Simplifying Exponential Expressions (Problems 25 - 32)

Recall the rules: $a^{\frac{m}{n}} = \sqrt[n]{a^m}$, $a^{-\frac{m}{n}} = \frac{1}{a^{\frac{m}{n}}} = \frac{1}{\sqrt[n]{a^m}}$, and $(ab)^n = a^n b^n$.

Problem 25: $9^{\frac{1}{2}}$

$9^{\frac{1}{2}} = \sqrt{9} = 3$ (since $3^2 = 9$)

Problem 26: $343^{-\frac{1}{3}}$

First, rewrite the negative exponent: $343^{-\frac{1}{3}} = \frac{1}{343^{\frac{1}{3}}}$. Now, $343^{\frac{1}{3}} = \sqrt[3]{343} = 7$ (since $7^3 = 343$). So $\frac{1}{7}$

Problem 27: $1000000^{\frac{1}{6}}$

$1000000^{\frac{1}{6}} = \sqrt[6]{1000000}$. We know that $10^6 = 1000000$, so $\sqrt[6]{10^6} = 10$

Problem 28: $36^{\frac{1}{2}}$

$36^{\frac{1}{2}} = \sqrt{36} = 6$ (since $6^2 = 36$)

Problem 29: $(x^6)^{\frac{1}{2}}$

Using the power of a power rule: $(x^6)^{\frac{1}{2}} = x^{6 \times \frac{1}{2}} = x^3$

Problem 30: $(9n^4)^{\frac{1}{2}}$

Using the product rule for exponents: $(9n^4)^{\frac{1}{2}} = 9^{\frac{1}{2}} \times (n^4)^{\frac{1}{2}}$. We know $9^{\frac{1}{2}} = 3$ and $(n^4)^{\frac{1}{2}} = n^{4 \times \frac{1}{2}} = n^2$. So $3n^2$

Problem 31: $(64n^{12})^{-\frac{1}{6}}$

First, rewrite the negative exponent: $(64n^{12})^{-\frac{1}{6}} = \frac{1}{(64n^{12})^{\frac{1}{6}}}$. Now, apply the product rule: $\frac{1}{64^{\frac{1}{6}} \times (n^{12})^{\frac{1}{6}}}$. Calculate $64^{\frac{1}{6}}$: since $2^6 = 64$, $64^{\frac{1}{6}} = 2$. Calculate $(n^{12})^{\frac{1}{6}} = n^{12 \times \frac{1}{6}} = n^2$. So $\frac{1}{2n^2}$

Problem 32: $(81m^6)^{\frac{1}{2}}$

Using the product rule: $(81m^6)^{\frac{1}{2}} = 81^{\frac{1}{2}} \times (m^6)^{\frac{1}{2}}$. $81^{\frac{1}{2}} = 9$ (since $9^2 = 81$) and $(m^6)^{\frac{1}{2}} = m^{6 \times \frac{1}{2}} = m^3$. So $9m^3$

Final Answers

Writing in Exponential Form:

1. $\boldsymbol{m^{\frac{3}{5}}}$
2. $\boldsymbol{(6x)^{\frac{4}{7}}}$
3. $\boldsymbol{v^{\frac{1}{8}}}$
4. $\boldsymbol{(6p)^{\frac{1}{2}}}$
5. $\boldsymbol{(3a)^{\frac{4}{3}}}$
6. $\boldsymbol{(3k)^{-\frac{5}{2}}}$

Simplifying:

1. $\boldsymbol{3}$
2. $\boldsymbol{\frac{1}{7}}$
3. $\boldsymbol{10}$
4. $\boldsymbol{6}$
5. $\boldsymbol{x^3}$
6. $\boldsymbol{3n^2}$
7. $\boldsymbol{\frac{1}{2n^2}}$
8. $\boldsymbol{9m^3}$