Simplifying Rational Exponents worksheet with ten problems requiring simplification to positive exponents.

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

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

Now simplify $\sqrt[6]{64}$:

Since $64 = 2^6$, then:

$$
\sqrt[6]{2^6} = 2
$$

So:

$$
= 2n^2
$$

✔ Answer: $\boxed{2n^2}$

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

First, write numerator and denominator separately:

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

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

Now divide:

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

Compute exponents:

- For base 2: $ \frac{7}{4} - \frac{8}{3} = \frac{21 - 32}{12} = -\frac{11}{12} $
- For base x: $ \frac{7}{4} - \frac{4}{3} = \frac{21 - 16}{12} = \frac{5}{12} $

So:

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

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

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

Simplify numerator and denominator separately.

Numerator: $(x^3 y^2)^{3/2} = x^{3 \cdot \frac{3}{2}} y^{2 \cdot \frac{3}{2}} = x^{9/2} y^3$

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

Now divide:

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

Compute:

- $9/2 + 1/4 = 18/4 + 1/4 = 19/4$
- $3 + 1/6 = 18/6 + 1/6 = 19/6$

So:

$$
= x^{19/4} y^{19/6}
$$

✔ Answer: $\boxed{x^{19/4} y^{19/6}}$

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

First, simplify numerator:

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

Now divide by denominator: $3y^{-7/4}$

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

Compute exponent:

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

So:

$$
= 3x y^{25/12}
$$

✔ Answer: $\boxed{3x y^{25/12}}$

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

First, simplify denominator:

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

Compute:

- $-5/4 + 1 = -5/4 + 4/4 = -1/4$
- $-5/3 + 1/2 = -10/6 + 3/6 = -7/6$

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

Now the whole expression:

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

Compute exponents:

- $-2 + 1/4 = -8/4 + 1/4 = -7/4$
- $5/3 + 7/6 = 10/6 + 7/6 = 17/6$

So:

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

Convert negative exponent:

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

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

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

First simplify numerator:

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

Now raise to $-1$: $(x^{4/3} y^{4/3})^{-1} = x^{-4/3} y^{-4/3}$

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

So:

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

Now convert negative exponent:

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

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

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

Simplify inside the parentheses first.

Numerator: $x^{1/2} y^{-2}$

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

So:

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

Compute exponent for x:

$1/2 + 7/4 = 2/4 + 7/4 = 9/4$

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

Now raise to the 4th power:

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

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 | $\dfrac{x^{5/12}}{2^{11/12}}$ |
| 6 | $x^{19/4} y^{19/6}$ |
| 7 | $3x y^{25/12}$ |
| 8 | $\dfrac{2 y^{17/6}}{x^{7/4}}$ |
| 9 | $\dfrac{y^{2/3}}{x^{5/3}}$ |
| 10 | $\dfrac{x^9}{y^{12}}$ |

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