Rational Exponents Worksheets - Math Monks - Free Printable

Explanation:
Let's solve each problem step by step. We need to simplify each expression and make sure the final answer has only positive exponents.

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Problem 1:
$(x^{\frac{1}{2}} y^{\frac{1}{2}})^{-1}$

- Use the rule: $(ab)^n = a^n b^n$
So, $(x^{\frac{1}{2}} y^{\frac{1}{2}})^{-1} = x^{-\frac{1}{2}} y^{-\frac{1}{2}}$

- Now, convert negative exponents to positive:
$x^{-\frac{1}{2}} y^{-\frac{1}{2}} = \frac{1}{x^{\frac{1}{2}} y^{\frac{1}{2}}}$

- This can also be written as $\frac{1}{\sqrt{x} \sqrt{y}}$ or $\frac{1}{\sqrt{xy}}$, but since we want only positive exponents, we leave it as $\frac{1}{x^{\frac{1}{2}} y^{\frac{1}{2}}}$

Final answer: $\frac{1}{x^{\frac{1}{2}} y^{\frac{1}{2}}}$

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Problem 2:
$(a \cdot ab^2)^0$

- Any non-zero number to the power of 0 is 1.
- Even if the expression inside is complicated, as long as it's not zero, the whole thing is 1.

So, $(a \cdot ab^2)^0 = 1$

Final answer: $1$

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Problem 3:
$(81m^6)^{\frac{1}{2}}$

- Use the rule: $(ab)^n = a^n b^n$
So, $81^{\frac{1}{2}} \cdot (m^6)^{\frac{1}{2}}$

- $81^{\frac{1}{2}} = \sqrt{81} = 9$

- $(m^6)^{\frac{1}{2}} = m^{6 \cdot \frac{1}{2}} = m^3$

Final answer: $9m^3$

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Problem 4:
$(64n^{12})^{\frac{1}{6}}$

- $64^{\frac{1}{6}} \cdot (n^{12})^{\frac{1}{6}}$

- $64 = 2^6$, so $64^{\frac{1}{6}} = (2^6)^{\frac{1}{6}} = 2^{6 \cdot \frac{1}{6}} = 2^1 = 2$

- $(n^{12})^{\frac{1}{6}} = n^{12 \cdot \frac{1}{6}} = n^2$

Final answer: $2n^2$

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Problem 5:
$\frac{(2x)^{\frac{7}{4}}}{(4x)^{\frac{4}{3}}}$

- First, write $4$ as $2^2$:
So, $(4x)^{\frac{4}{3}} = (2^2 x)^{\frac{4}{3}} = 2^{2 \cdot \frac{4}{3}} x^{\frac{4}{3}} = 2^{\frac{8}{3}} x^{\frac{4}{3}}$

- $(2x)^{\frac{7}{4}} = 2^{\frac{7}{4}} x^{\frac{7}{4}}$

- Now divide:
$\frac{2^{\frac{7}{4}} x^{\frac{7}{4}}}{2^{\frac{8}{3}} x^{\frac{4}{3}}}$

- Use the rule: $\frac{a^m}{a^n} = a^{m-n}$

- For base 2: $2^{\frac{7}{4} - \frac{8}{3}} = 2^{\frac{21 - 32}{12}} = 2^{-\frac{11}{12}}$

- For base x: $x^{\frac{7}{4} - \frac{4}{3}} = x^{\frac{21 - 16}{12}} = x^{\frac{5}{12}}$

- So we have: $2^{-\frac{11}{12}} x^{\frac{5}{12}}$

- Convert negative exponent to positive: $\frac{x^{\frac{5}{12}}}{2^{\frac{11}{12}}}$

Final answer: $\frac{x^{\frac{5}{12}}}{2^{\frac{11}{12}}}$

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Problem 6:
$\frac{(x^3 y^2)^{\frac{3}{2}}}{(x^{-1} y^{-\frac{2}{3}})^{\frac{1}{4}}}$

- First, simplify numerator:
$(x^3 y^2)^{\frac{3}{2}} = x^{3 \cdot \frac{3}{2}} y^{2 \cdot \frac{3}{2}} = x^{\frac{9}{2}} y^3$

- Denominator:
$(x^{-1} y^{-\frac{2}{3}})^{\frac{1}{4}} = x^{-1 \cdot \frac{1}{4}} y^{-\frac{2}{3} \cdot \frac{1}{4}} = x^{-\frac{1}{4}} y^{-\frac{1}{6}}$

- Now divide:
$\frac{x^{\frac{9}{2}} y^3}{x^{-\frac{1}{4}} y^{-\frac{1}{6}}}$

- Use $\frac{a^m}{a^n} = a^{m-n}$

- For x: $x^{\frac{9}{2} - (-\frac{1}{4})} = x^{\frac{9}{2} + \frac{1}{4}} = x^{\frac{18 + 1}{4}} = x^{\frac{19}{4}}$

- For y: $y^{3 - (-\frac{1}{6})} = y^{3 + \frac{1}{6}} = y^{\frac{18 + 1}{6}} = y^{\frac{19}{6}}$

Final answer: $x^{\frac{19}{4}} y^{\frac{19}{6}}$

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Problem 7:
$\frac{3x^{\frac{1}{2}} \cdot 3x^{\frac{1}{2}} y^{-\frac{1}{3}}}{3y^{-\frac{7}{4}}}$

- First, multiply the numerators:
$3x^{\frac{1}{2}} \cdot 3x^{\frac{1}{2}} = 9x^{\frac{1}{2} + \frac{1}{2}} = 9x^1 = 9x$

- So numerator: $9x \cdot y^{-\frac{1}{3}}$

- Denominator: $3y^{-\frac{7}{4}}$

- Now divide: $\frac{9x y^{-\frac{1}{3}}}{3 y^{-\frac{7}{4}}}$

- Simplify coefficients: $9/3 = 3$

- For x: $x^1$ (no change)

- For y: $y^{-\frac{1}{3} - (-\frac{7}{4})} = y^{-\frac{1}{3} + \frac{7}{4}}$

- Compute: $-\frac{1}{3} + \frac{7}{4} = \frac{-4 + 21}{12} = \frac{17}{12}$

- So we have: $3x y^{\frac{17}{12}}$

Final answer: $3x y^{\frac{17}{12}}$

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Problem 8:
$\frac{2x^{-2} y^{\frac{5}{3}}}{x^{-\frac{5}{4}} y^{-\frac{5}{3}} \cdot x y^{\frac{1}{2}}}$

- First, simplify the denominator:
Multiply the terms: $x^{-\frac{5}{4}} \cdot x^1 = x^{-\frac{5}{4} + 1} = x^{-\frac{1}{4}}$

- $y^{-\frac{5}{3}} \cdot y^{\frac{1}{2}} = y^{-\frac{5}{3} + \frac{1}{2}} = y^{-\frac{10}{6} + \frac{3}{6}} = y^{-\frac{7}{6}}$

- So denominator: $x^{-\frac{1}{4}} y^{-\frac{7}{6}}$

- Numerator: $2x^{-2} y^{\frac{5}{3}}$

- Now divide: $\frac{2x^{-2} y^{\frac{5}{3}}}{x^{-\frac{1}{4}} y^{-\frac{7}{6}}}$

- For x: $x^{-2 - (-\frac{1}{4})} = x^{-2 + \frac{1}{4}} = x^{-\frac{8}{4} + \frac{1}{4}} = x^{-\frac{7}{4}}$

- For y: $y^{\frac{5}{3} - (-\frac{7}{6})} = y^{\frac{5}{3} + \frac{7}{6}} = y^{\frac{10}{6} + \frac{7}{6}} = y^{\frac{17}{6}}$

- So we have: $2 x^{-\frac{7}{4}} y^{\frac{17}{6}}$

- Convert negative exponent to positive: $\frac{2 y^{\frac{17}{6}}}{x^{\frac{7}{4}}}$

Final answer: $\frac{2 y^{\frac{17}{6}}}{x^{\frac{7}{4}}}$

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Problem 9:
$\frac{(x^{\frac{4}{3}} y^{-\frac{1}{3}} \cdot y)^{-1}}{x^{\frac{1}{3}} y^{-2}}$

- First, simplify inside the parentheses:
$x^{\frac{4}{3}} y^{-\frac{1}{3}} \cdot y = x^{\frac{4}{3}} y^{-\frac{1}{3} + 1} = x^{\frac{4}{3}} y^{\frac{2}{3}}$

- Now raise to the power of -1: $(x^{\frac{4}{3}} y^{\frac{2}{3}})^{-1} = x^{-\frac{4}{3}} y^{-\frac{2}{3}}$

- Now divide by denominator: $\frac{x^{-\frac{4}{3}} y^{-\frac{2}{3}}}{x^{\frac{1}{3}} y^{-2}}$

- For x: $x^{-\frac{4}{3} - \frac{1}{3}} = x^{-\frac{5}{3}}$

- For y: $y^{-\frac{2}{3} - (-2)} = y^{-\frac{2}{3} + 2} = y^{-\frac{2}{3} + \frac{6}{3}} = y^{\frac{4}{3}}$

- So: $x^{-\frac{5}{3}} y^{\frac{4}{3}}$

- Convert negative exponent: $\frac{y^{\frac{4}{3}}}{x^{\frac{5}{3}}}$

Final answer: $\frac{y^{\frac{4}{3}}}{x^{\frac{5}{3}}}$

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Problem 10:
$\left(\frac{x^{\frac{1}{2}} \cdot y^{-2}}{y x^{-\frac{7}{4}}}\right)^4$

- First, simplify inside the parentheses:

- $x^{\frac{1}{2}} / x^{-\frac{7}{4}} = x^{\frac{1}{2} - (-\frac{7}{4})} = x^{\frac{1}{2} + \frac{7}{4}} = x^{\frac{2}{4} + \frac{7}{4}} = x^{\frac{9}{4}}$

- $y^{-2} / y^1 = y^{-2 - 1} = y^{-3}$

- So inside: $x^{\frac{9}{4}} y^{-3}$

- Now raise to the 4th power: $(x^{\frac{9}{4}} y^{-3})^4 = x^{\frac{9}{4} \cdot 4} y^{-3 \cdot 4} = x^9 y^{-12}$

- Convert negative exponent: $\frac{x^9}{y^{12}}$

Final answer: $\frac{x^9}{y^{12}}$

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Final Answer:
1. $\frac{1}{x^{\frac{1}{2}} y^{\frac{1}{2}}}$
2. $1$
3. $9m^3$
4. $2n^2$
5. $\frac{x^{\frac{5}{12}}}{2^{\frac{11}{12}}}$
6. $x^{\frac{19}{4}} y^{\frac{19}{6}}$
7. $3x y^{\frac{17}{12}}$
8. $\frac{2 y^{\frac{17}{6}}}{x^{\frac{7}{4}}}$
9. $\frac{y^{\frac{4}{3}}}{x^{\frac{5}{3}}}$
10. $\frac{x^9}{y^{12}}$