Rational Exponents Worksheets - Math Monks - Free Printable

Let's solve each problem step by step, simplifying expressions with rational exponents and ensuring all answers contain only positive exponents.

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

We apply the exponent $-1$ to both terms inside the parentheses:

$$
= x^{-\frac{1}{2}} y^{-\frac{1}{2}}
$$

Now convert negative exponents to positive:

$$
= \frac{1}{x^{\frac{1}{2}} y^{\frac{1}{2}}} = \frac{1}{\sqrt{x} \sqrt{y}} = \frac{1}{\sqrt{xy}}
$$

✔ Answer: $\boxed{\dfrac{1}{\sqrt{xy}}}$

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

Any non-zero expression raised to the power of 0 is 1.

Note: $a \cdot ab^2 = a^2b^2$, which is not zero (assuming $a, b \neq 0$), so:

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

✔ Answer: $\boxed{1}$

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

Break it into parts:

$$
= (81)^{\frac{1}{2}} \cdot (m^6)^{\frac{1}{2}} = \sqrt{81} \cdot m^{6 \cdot \frac{1}{2}} = 9 \cdot m^3
$$

✔ Answer: $\boxed{9m^3}$

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

$$
= (64)^{\frac{1}{6}} \cdot (n^{12})^{\frac{1}{6}} = (2^6)^{\frac{1}{6}} \cdot n^{12 \cdot \frac{1}{6}} = 2 \cdot n^2
$$

✔ Answer: $\boxed{2n^2}$

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

First, write $2x$ and $4x$ in terms of primes:

- $2x = 2x$
- $4x = 2^2 x$

So:

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

Denominator: $(4x)^{\frac{4}{3}} = (2^2 x)^{\frac{4}{3}} = 2^{\frac{8}{3}} x^{\frac{4}{3}}$

Now divide:

$$
= \frac{2^{\frac{7}{4}} x^{\frac{7}{4}}}{2^{\frac{8}{3}} x^{\frac{4}{3}}} = 2^{\frac{7}{4} - \frac{8}{3}} \cdot x^{\frac{7}{4} - \frac{4}{3}}
$$

Compute exponents:

- $\frac{7}{4} - \frac{8}{3} = \frac{21 - 32}{12} = -\frac{11}{12}$
- $\frac{7}{4} - \frac{4}{3} = \frac{21 - 16}{12} = \frac{5}{12}$

So:

$$
= 2^{-\frac{11}{12}} x^{\frac{5}{12}} = \frac{x^{\frac{5}{12}}}{2^{\frac{11}{12}}}
$$

We can write this as:

$$
= \frac{x^{\frac{5}{12}}}{\sqrt[12]{2^{11}}}
$$

But better to leave as:

✔ Answer: $\boxed{\dfrac{x^{\frac{5}{12}}}{2^{\frac{11}{12}}}}$

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

Simplify numerator and denominator separately.

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

Now divide:

$$
= \frac{x^{\frac{9}{2}} y^3}{x^{-\frac{1}{4}} y^{-\frac{1}{6}}} = x^{\frac{9}{2} + \frac{1}{4}} y^{3 + \frac{1}{6}}
$$

Add exponents:

- $\frac{9}{2} + \frac{1}{4} = \frac{18 + 1}{4} = \frac{19}{4}$
- $3 + \frac{1}{6} = \frac{18 + 1}{6} = \frac{19}{6}$

✔ Answer: $\boxed{x^{\frac{19}{4}} y^{\frac{19}{6}}}$

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

First simplify numerator:

$$
3x^{\frac{1}{2}} \cdot 3x^{\frac{1}{2}} y^{\frac{1}{3}} = 9x^{\frac{1}{2} + \frac{1}{2}} y^{\frac{1}{3}} = 9x^1 y^{\frac{1}{3}}
$$

Now divide by denominator:

$$
\frac{9x y^{\frac{1}{3}}}{3y^{-\frac{7}{4}}} = \frac{9}{3} \cdot x \cdot y^{\frac{1}{3} - (-\frac{7}{4})} = 3x y^{\frac{1}{3} + \frac{7}{4}}
$$

Compute exponent:

- $\frac{1}{3} + \frac{7}{4} = \frac{4 + 21}{12} = \frac{25}{12}$

✔ Answer: $\boxed{3x y^{\frac{25}{12}}}$

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

First simplify denominator:

$$
x^{-\frac{5}{4}} y^{-\frac{5}{3}} \cdot x y^{\frac{1}{2}} = x^{-\frac{5}{4} + 1} y^{-\frac{5}{3} + \frac{1}{2}}
$$

Compute exponents:

- $-\frac{5}{4} + 1 = -\frac{1}{4}$
- $-\frac{5}{3} + \frac{1}{2} = \frac{-10 + 3}{6} = -\frac{7}{6}$

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

Now divide:

$$
\frac{2x^{-2} y^{\frac{5}{3}}}{x^{-\frac{1}{4}} y^{-\frac{7}{6}}} = 2 \cdot x^{-2 + \frac{1}{4}} \cdot y^{\frac{5}{3} + \frac{7}{6}}
$$

Compute:

- $-2 + \frac{1}{4} = -\frac{8}{4} + \frac{1}{4} = -\frac{7}{4}$
- $\frac{5}{3} + \frac{7}{6} = \frac{10 + 7}{6} = \frac{17}{6}$

So:

$$
= 2x^{-\frac{7}{4}} y^{\frac{17}{6}} = \frac{2 y^{\frac{17}{6}}}{x^{\frac{7}{4}}}
$$

✔ Answer: $\boxed{\dfrac{2 y^{\frac{17}{6}}}{x^{\frac{7}{4}}}}$

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

First simplify numerator:

Inside: $x^{\frac{4}{3}} y^{\frac{1}{3}} \cdot y = x^{\frac{4}{3}} y^{\frac{1}{3} + 1} = x^{\frac{4}{3}} y^{\frac{4}{3}}$

Now raise to $-1$:

$$
(x^{\frac{4}{3}} y^{\frac{4}{3}})^{-1} = x^{-\frac{4}{3}} y^{-\frac{4}{3}}
$$

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

So:

$$
\frac{x^{-\frac{4}{3}} y^{-\frac{4}{3}}}{x^{\frac{1}{3}} y^{-2}} = x^{-\frac{4}{3} - \frac{1}{3}} y^{-\frac{4}{3} + 2} = x^{-\frac{5}{3}} y^{\frac{2}{3}}
$$

Convert negative exponent:

$$
= \frac{y^{\frac{2}{3}}}{x^{\frac{5}{3}}}
$$

✔ Answer: $\boxed{\dfrac{y^{\frac{2}{3}}}{x^{\frac{5}{3}}}}$

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

Simplify inside first:

Numerator: $x^{\frac{1}{2}} y^{-2}$

Denominator: $y \cdot x^{-\frac{7}{4}} = x^{-\frac{7}{4}} y^1$

Now divide:

$$
\dfrac{x^{\frac{1}{2}} y^{-2}}{x^{-\frac{7}{4}} y} = x^{\frac{1}{2} - (-\frac{7}{4})} y^{-2 - 1} = x^{\frac{1}{2} + \frac{7}{4}} y^{-3}
$$

Compute exponent:

- $\frac{1}{2} + \frac{7}{4} = \frac{2 + 7}{4} = \frac{9}{4}$

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

Now raise to 4th power:

$$
\left(x^{\frac{9}{4}} y^{-3}\right)^4 = x^{\frac{9}{4} \cdot 4} y^{-3 \cdot 4} = x^9 y^{-12}
$$

Convert negative exponent:

$$
= \frac{x^9}{y^{12}}
$$

✔ Answer: $\boxed{\dfrac{x^9}{y^{12}}}$

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $\dfrac{1}{\sqrt{xy}}$ |
| 2 | $1$ |
| 3 | $9m^3$ |
| 4 | $2n^2$ |
| 5 | $\dfrac{x^{\frac{5}{12}}}{2^{\frac{11}{12}}}$ |
| 6 | $x^{\frac{19}{4}} y^{\frac{19}{6}}$ |
| 7 | $3x y^{\frac{25}{12}}$ |
| 8 | $\dfrac{2 y^{\frac{17}{6}}}{x^{\frac{7}{4}}}$ |
| 9 | $\dfrac{y^{\frac{2}{3}}}{x^{\frac{5}{3}}}$ |
| 10 | $\dfrac{x^9}{y^{12}}$ |

Let me know if you'd like these rewritten in radical form or further explained!