Simplifying Exponential Expressions worksheet with 18 problems for algebra practice.

Sure! Let’s go through each problem one by one, simplifying the exponential expressions so that the final answer contains only positive exponents and no fractional exponents in the denominator.

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1) $3v^3 \cdot 2v^3$

Multiply coefficients: $3 \cdot 2 = 6$

Add exponents of same base: $v^{3+3} = v^6$

✔ Answer: $6v^6$

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2) $2m^3 \cdot 3m^2$

Coefficients: $2 \cdot 3 = 6$

Exponents: $m^{3+2} = m^5$

✔ Answer: $6m^5$

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3) $2v^2 \cdot v^3$

Coefficient is 2 (implicit 1 for second term)

Exponents: $v^{2+3} = v^5$

✔ Answer: $2v^5$

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4) $3x \cdot x$

Same as $3x^1 \cdot x^1 = 3x^{2}$

✔ Answer: $3x^2$

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5) $(-x^3 y^5 \cdot x^{-5})^2 \cdot (-y x^{-2})$

First, simplify inside the parentheses:

$-x^3 \cdot x^{-5} = -x^{3-5} = -x^{-2}$, so we have:

$(-x^{-2} y^5)^2 \cdot (-y x^{-2})$

Now apply exponent to each part inside:

$(-1)^2 \cdot (x^{-2})^2 \cdot (y^5)^2 = 1 \cdot x^{-4} \cdot y^{10}$

So now: $x^{-4} y^{10} \cdot (-y x^{-2})$

Multiply: $-x^{-4 + (-2)} y^{10 + 1} = -x^{-6} y^{11}$

Convert to positive exponents: $-\frac{y^{11}}{x^6}$

✔ Answer: $-\dfrac{y^{11}}{x^6}$

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6) $(x^{-5} y^3)^4 \cdot (-x^0 y^5)^5$

Note: $x^0 = 1$, so second term: $(-1 \cdot y^5)^5 = (-1)^5 y^{25} = -y^{25}$

First term: $(x^{-5})^4 (y^3)^4 = x^{-20} y^{12}$

Now multiply: $x^{-20} y^{12} \cdot (-y^{25}) = -x^{-20} y^{37}$

Convert to positive exponents: $-\dfrac{y^{37}}{x^{20}}$

✔ Answer: $-\dfrac{y^{37}}{x^{20}}$

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7) $(n^5 \cdot m^0 n^{-5})^2$

$m^0 = 1$, so inside: $n^5 \cdot n^{-5} = n^{0} = 1$

Then $(1)^2 = 1$

✔ Answer: $1$

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8) $-x^4 y^{-5} \cdot -x^{-1} y^5 \cdot (x^{-5} y^{-4})^{-1}$

First, multiply first two terms:

$(-x^4 y^{-5}) \cdot (-x^{-1} y^5) = (+1) x^{4-1} y^{-5+5} = x^3 y^0 = x^3$

Now handle third term: $(x^{-5} y^{-4})^{-1} = x^{5} y^{4}$

Now multiply: $x^3 \cdot x^5 y^4 = x^{8} y^4$

✔ Answer: $x^8 y^4$

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9) $(x^{-3} y^{-5})^4 \cdot x^3 y^{-4}$

First: $(x^{-3})^4 (y^{-5})^4 = x^{-12} y^{-20}$

Multiply by $x^3 y^{-4}$: $x^{-12+3} y^{-20-4} = x^{-9} y^{-24}$

Convert to positive exponents: $\dfrac{1}{x^9 y^{24}}$

✔ Answer: $\dfrac{1}{x^9 y^{24}}$

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10) $(-x^{-5} y^{-5})^4 \cdot (x^{-3})^4$

First term: $(-1)^4 (x^{-5})^4 (y^{-5})^4 = 1 \cdot x^{-20} y^{-20}$

Second term: $(x^{-3})^4 = x^{-12}$

Multiply: $x^{-20} y^{-20} \cdot x^{-12} = x^{-32} y^{-20}$

Convert: $\dfrac{1}{x^{32} y^{20}}$

✔ Answer: $\dfrac{1}{x^{32} y^{20}}$

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11) $y^{-4} \cdot x^4 y^{-2} \cdot (-x^3 y^3)^2$

First, simplify last term: $(-x^3 y^3)^2 = x^6 y^6$

Now multiply all:

$x^4 y^{-4} \cdot y^{-2} \cdot x^6 y^6 = x^{4+6} y^{-4-2+6} = x^{10} y^{0} = x^{10}$

✔ Answer: $x^{10}$

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12) $(-x^{-2} y^{-1} \cdot x^{-1} y^3)^{-3}$

First, simplify inside: $-x^{-2-1} y^{-1+3} = -x^{-3} y^{2}$

Now raise to power -3: $(-1)^{-3} (x^{-3})^{-3} (y^2)^{-3} = -1 \cdot x^{9} \cdot y^{-6}$

So: $-x^9 y^{-6} = -\dfrac{x^9}{y^6}$

✔ Answer: $-\dfrac{x^9}{y^6}$

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13) $\dfrac{x^{-3} y^0 \cdot (x y^{-3})^2}{2x^0}$

Numerator:

First, $y^0 = 1$, $x^0 = 1$ → denominator is 2

$(x y^{-3})^2 = x^2 y^{-6}$

So numerator: $x^{-3} \cdot x^2 y^{-6} = x^{-1} y^{-6}$

Overall: $\dfrac{x^{-1} y^{-6}}{2} = \dfrac{1}{2 x y^6}$

✔ Answer: $\dfrac{1}{2 x y^6}$

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14) $\dfrac{b}{(a^2 b^3)^2 \cdot 2b^4}$

Denominator: $(a^2 b^3)^2 = a^4 b^6$, then times $2b^4$: $2 a^4 b^{10}$

So expression: $\dfrac{b}{2 a^4 b^{10}} = \dfrac{1}{2 a^4 b^{9}}$

✔ Answer: $\dfrac{1}{2 a^4 b^9}$

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15) $\dfrac{2x^4 y^2 \cdot y x^{-3}}{(2x^2 y^2 \cdot y x^{-1})^3}$

Numerator: $2x^4 y^2 \cdot y x^{-3} = 2x^{4-3} y^{2+1} = 2x y^3$

Denominator: Inside: $2x^2 y^2 \cdot y x^{-1} = 2x^{2-1} y^{2+1} = 2x y^3$

Then cube: $(2x y^3)^3 = 8 x^3 y^9$

So overall: $\dfrac{2x y^3}{8 x^3 y^9} = \dfrac{1}{4 x^{2} y^{6}}$

✔ Answer: $\dfrac{1}{4 x^2 y^6}$

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16) $\left(\dfrac{y x^{-3} \cdot 2x^0 y^3}{2x y^3}\right)^2$

Simplify inside:

Numerator: $y x^{-3} \cdot 2 \cdot 1 \cdot y^3 = 2 x^{-3} y^{4}$

Denominator: $2x y^3$

So fraction: $\dfrac{2 x^{-3} y^{4}}{2x y^3} = x^{-3-1} y^{4-3} = x^{-4} y$

Now square: $(x^{-4} y)^2 = x^{-8} y^2 = \dfrac{y^2}{x^8}$

✔ Answer: $\dfrac{y^2}{x^8}$

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17) $\dfrac{(2xy)^0}{y^3 \cdot 2x^4 y^3}$

Any non-zero expression to power 0 is 1 → numerator = 1

Denominator: $2x^4 y^{3+3} = 2x^4 y^6$

So: $\dfrac{1}{2x^4 y^6}$

✔ Answer: $\dfrac{1}{2x^4 y^6}$

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18) $\dfrac{(x^0 y^2)^3}{2x y^2 \cdot y^{-3}}$

Numerator: $(1 \cdot y^2)^3 = y^6$

Denominator: $2x y^{2 + (-3)} = 2x y^{-1}$

So: $\dfrac{y^6}{2x y^{-1}} = \dfrac{y^{6 - (-1)}}{2x} = \dfrac{y^7}{2x}$

✔ Answer: $\dfrac{y^7}{2x}$

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## ✔ Final Answers Summary:

1. $6v^6$
2. $6m^5$
3. $2v^5$
4. $3x^2$
5. $-\dfrac{y^{11}}{x^6}$
6. $-\dfrac{y^{37}}{x^{20}}$
7. $1$
8. $x^8 y^4$
9. $\dfrac{1}{x^9 y^{24}}$
10. $\dfrac{1}{x^{32} y^{20}}$
11. $x^{10}$
12. $-\dfrac{x^9}{y^6}$
13. $\dfrac{1}{2 x y^6}$
14. $\dfrac{1}{2 a^4 b^9}$
15. $\dfrac{1}{4 x^2 y^6}$
16. $\dfrac{y^2}{x^8}$
17. $\dfrac{1}{2x^4 y^6}$
18. $\dfrac{y^7}{2x}$

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