Worksheet featuring a cartoon eye with a magnifying glass, illustrating the concept of simplifying expressions with negative exponents.
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Step-by-step solution for: Simplifying Expressions With Negative Exponents Worksheet With ...
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Show Answer Key & Explanations
Step-by-step solution for: Simplifying Expressions With Negative Exponents Worksheet With ...
Absolutely! Let’s solve each problem step by step, applying the negative exponent rule:
> Negative Exponent Rule:
> For any nonzero number \( a \) and integer \( n \),
> \[
a^{-n} = \frac{1}{a^n} \quad \text{or} \quad \frac{1}{a^{-n}} = a^n
\]
Also, we’ll use:
- Power of a power: \( (a^m)^n = a^{m \cdot n} \)
- Product of powers: \( a^m \cdot a^n = a^{m+n} \)
- Quotient of powers: \( \frac{a^m}{a^n} = a^{m-n} \)
- Power of a product/quotient: \( (ab)^n = a^n b^n \), \( \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \)
---
Apply the negative exponent rule to the entire fraction:
\[
\left( \frac{-6}{x} \right)^{-3} = \left( \frac{x}{-6} \right)^3 = \frac{x^3}{(-6)^3} = \frac{x^3}{-216} = -\frac{x^3}{216}
\]
✔ Answer: \( \boxed{-\dfrac{x^3}{216}} \)
---
First, simplify inside the parentheses using negative exponents:
\[
\frac{2x^{-2}}{y^{-3}} = 2 \cdot x^{-2} \cdot y^{3} = \frac{2y^3}{x^2}
\]
Now raise to the 4th power:
\[
\left( \frac{2y^3}{x^2} \right)^4 = \frac{(2)^4 (y^3)^4}{(x^2)^4} = \frac{16 y^{12}}{x^8}
\]
✔ Answer: \( \boxed{\dfrac{16y^{12}}{x^8}} \)
---
This is a binomial raised to a negative exponent. Apply the rule directly:
\[
(18 + a)^{-2} = \frac{1}{(18 + a)^2}
\]
We can leave it like this unless asked to expand the denominator.
✔ Answer: \( \boxed{\dfrac{1}{(18 + a)^2}} \)
---
Apply the outer exponent \(-1\) to both numerator and denominator:
\[
= \frac{(2^{-6})^{-1}}{((x+2)^2)^{-1}} = \frac{2^{6}}{(x+2)^{-2}} = 2^6 \cdot (x+2)^2 = 64(x+2)^2
\]
✔ Answer: \( \boxed{64(x+2)^2} \)
---
First, simplify inside the parentheses:
Note: \( \frac{1}{s^{-4}} = s^4 \)
So,
\[
s t^{-3} \cdot s^4 = s^{1+4} \cdot t^{-3} = s^5 t^{-3}
\]
Now square the result:
\[
(s^5 t^{-3})^2 = s^{10} t^{-6} = \frac{s^{10}}{t^6}
\]
✔ Answer: \( \boxed{\dfrac{s^{10}}{t^6}} \)
---
First, simplify inside the parentheses:
\[
\frac{x y^{-2}}{4x} = \frac{1}{4} \cdot \frac{x}{x} \cdot y^{-2} = \frac{1}{4} y^{-2}
\]
Now apply the exponent \(-2\):
\[
\left( \frac{1}{4} y^{-2} \right)^{-2} = \left( \frac{1}{4} \right)^{-2} \cdot (y^{-2})^{-2} = 4^2 \cdot y^4 = 16 y^4
\]
But don’t forget the negative sign in front:
\[
-16 y^4
\]
✔ Answer: \( \boxed{-16y^4} \)
---
First, square the fraction:
\[
\left( \frac{3}{x^3} \right)^2 = \frac{9}{x^6}
\]
Now multiply by \( x \):
\[
\frac{9}{x^6} \cdot x = \frac{9x}{x^6} = \frac{9}{x^5}
\]
✔ Answer: \( \boxed{\dfrac{9}{x^5}} \)
---
First, simplify inside: \( \frac{1}{(3x)^{-3}} = (3x)^3 \)
Then square it:
\[
\left( (3x)^3 \right)^2 = (3x)^6 = 3^6 \cdot x^6 = 729 x^6
\]
✔ Answer: \( \boxed{729x^6} \)
---
This expression has no negative exponents to simplify. The numerator is a binomial cubed, but unless asked to expand, we leave it as is.
✔ Answer: \( \boxed{\dfrac{(1 - y)^3}{y^3}} \)
*(Note: If you were to expand, it would be \( \frac{1 - 3y + 3y^2 - y^3}{y^3} \), but that’s not required here.)*
---
Simplify each part:
First term: \( \frac{1}{(t^3)^2} = \frac{1}{t^6} \)
Second term: \( \frac{t^5}{2t} = \frac{1}{2} \cdot t^{5-1} = \frac{1}{2} t^4 \)
Now multiply:
\[
\frac{1}{t^6} \cdot \frac{1}{2} t^4 = \frac{1}{2} \cdot \frac{t^4}{t^6} = \frac{1}{2} t^{-2} = \frac{1}{2t^2}
\]
✔ Answer: \( \boxed{\dfrac{1}{2t^2}} \)
---
## ✔ Final Answers Summary:
1. \( \boxed{-\dfrac{x^3}{216}} \)
2. \( \boxed{\dfrac{16y^{12}}{x^8}} \)
3. \( \boxed{\dfrac{1}{(18 + a)^2}} \)
4. \( \boxed{64(x+2)^2} \)
5. \( \boxed{\dfrac{s^{10}}{t^6}} \)
6. \( \boxed{-16y^4} \)
7. \( \boxed{\dfrac{9}{x^5}} \)
8. \( \boxed{729x^6} \)
9. \( \boxed{\dfrac{(1 - y)^3}{y^3}} \)
10. \( \boxed{\dfrac{1}{2t^2}} \)
Let me know if you’d like to see any expanded forms or step-by-step breakdowns for specific problems!
> Negative Exponent Rule:
> For any nonzero number \( a \) and integer \( n \),
> \[
a^{-n} = \frac{1}{a^n} \quad \text{or} \quad \frac{1}{a^{-n}} = a^n
\]
Also, we’ll use:
- Power of a power: \( (a^m)^n = a^{m \cdot n} \)
- Product of powers: \( a^m \cdot a^n = a^{m+n} \)
- Quotient of powers: \( \frac{a^m}{a^n} = a^{m-n} \)
- Power of a product/quotient: \( (ab)^n = a^n b^n \), \( \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \)
---
Problem 1: \( \left( \frac{-6}{x} \right)^{-3} \)
Apply the negative exponent rule to the entire fraction:
\[
\left( \frac{-6}{x} \right)^{-3} = \left( \frac{x}{-6} \right)^3 = \frac{x^3}{(-6)^3} = \frac{x^3}{-216} = -\frac{x^3}{216}
\]
✔ Answer: \( \boxed{-\dfrac{x^3}{216}} \)
---
Problem 2: \( \left( \frac{2x^{-2}}{y^{-3}} \right)^4 \)
First, simplify inside the parentheses using negative exponents:
\[
\frac{2x^{-2}}{y^{-3}} = 2 \cdot x^{-2} \cdot y^{3} = \frac{2y^3}{x^2}
\]
Now raise to the 4th power:
\[
\left( \frac{2y^3}{x^2} \right)^4 = \frac{(2)^4 (y^3)^4}{(x^2)^4} = \frac{16 y^{12}}{x^8}
\]
✔ Answer: \( \boxed{\dfrac{16y^{12}}{x^8}} \)
---
Problem 3: \( (18 + a)^{-2} \)
This is a binomial raised to a negative exponent. Apply the rule directly:
\[
(18 + a)^{-2} = \frac{1}{(18 + a)^2}
\]
We can leave it like this unless asked to expand the denominator.
✔ Answer: \( \boxed{\dfrac{1}{(18 + a)^2}} \)
---
Problem 4: \( \left( \frac{2^{-6}}{(x+2)^2} \right)^{-1} \)
Apply the outer exponent \(-1\) to both numerator and denominator:
\[
= \frac{(2^{-6})^{-1}}{((x+2)^2)^{-1}} = \frac{2^{6}}{(x+2)^{-2}} = 2^6 \cdot (x+2)^2 = 64(x+2)^2
\]
✔ Answer: \( \boxed{64(x+2)^2} \)
---
Problem 5: \( \left( s t^{-3} \cdot \frac{1}{s^{-4}} \right)^2 \)
First, simplify inside the parentheses:
Note: \( \frac{1}{s^{-4}} = s^4 \)
So,
\[
s t^{-3} \cdot s^4 = s^{1+4} \cdot t^{-3} = s^5 t^{-3}
\]
Now square the result:
\[
(s^5 t^{-3})^2 = s^{10} t^{-6} = \frac{s^{10}}{t^6}
\]
✔ Answer: \( \boxed{\dfrac{s^{10}}{t^6}} \)
---
Problem 6: \( -\left( \frac{x y^{-2}}{4x} \right)^{-2} \)
First, simplify inside the parentheses:
\[
\frac{x y^{-2}}{4x} = \frac{1}{4} \cdot \frac{x}{x} \cdot y^{-2} = \frac{1}{4} y^{-2}
\]
Now apply the exponent \(-2\):
\[
\left( \frac{1}{4} y^{-2} \right)^{-2} = \left( \frac{1}{4} \right)^{-2} \cdot (y^{-2})^{-2} = 4^2 \cdot y^4 = 16 y^4
\]
But don’t forget the negative sign in front:
\[
-16 y^4
\]
✔ Answer: \( \boxed{-16y^4} \)
---
Problem 7: \( \left( \frac{3}{x^3} \right)^2 \cdot x \)
First, square the fraction:
\[
\left( \frac{3}{x^3} \right)^2 = \frac{9}{x^6}
\]
Now multiply by \( x \):
\[
\frac{9}{x^6} \cdot x = \frac{9x}{x^6} = \frac{9}{x^5}
\]
✔ Answer: \( \boxed{\dfrac{9}{x^5}} \)
---
Problem 8: \( \left( \frac{1}{(3x)^{-3}} \right)^2 \)
First, simplify inside: \( \frac{1}{(3x)^{-3}} = (3x)^3 \)
Then square it:
\[
\left( (3x)^3 \right)^2 = (3x)^6 = 3^6 \cdot x^6 = 729 x^6
\]
✔ Answer: \( \boxed{729x^6} \)
---
Problem 9: \( \frac{(1 - y)^3}{y^3} \)
This expression has no negative exponents to simplify. The numerator is a binomial cubed, but unless asked to expand, we leave it as is.
✔ Answer: \( \boxed{\dfrac{(1 - y)^3}{y^3}} \)
*(Note: If you were to expand, it would be \( \frac{1 - 3y + 3y^2 - y^3}{y^3} \), but that’s not required here.)*
---
Problem 10: \( \frac{1}{(t^3)^2} \cdot \frac{t^5}{2t} \)
Simplify each part:
First term: \( \frac{1}{(t^3)^2} = \frac{1}{t^6} \)
Second term: \( \frac{t^5}{2t} = \frac{1}{2} \cdot t^{5-1} = \frac{1}{2} t^4 \)
Now multiply:
\[
\frac{1}{t^6} \cdot \frac{1}{2} t^4 = \frac{1}{2} \cdot \frac{t^4}{t^6} = \frac{1}{2} t^{-2} = \frac{1}{2t^2}
\]
✔ Answer: \( \boxed{\dfrac{1}{2t^2}} \)
---
## ✔ Final Answers Summary:
1. \( \boxed{-\dfrac{x^3}{216}} \)
2. \( \boxed{\dfrac{16y^{12}}{x^8}} \)
3. \( \boxed{\dfrac{1}{(18 + a)^2}} \)
4. \( \boxed{64(x+2)^2} \)
5. \( \boxed{\dfrac{s^{10}}{t^6}} \)
6. \( \boxed{-16y^4} \)
7. \( \boxed{\dfrac{9}{x^5}} \)
8. \( \boxed{729x^6} \)
9. \( \boxed{\dfrac{(1 - y)^3}{y^3}} \)
10. \( \boxed{\dfrac{1}{2t^2}} \)
Let me know if you’d like to see any expanded forms or step-by-step breakdowns for specific problems!
Parent Tip: Review the logic above to help your child master the concept of negative exponents worksheet printable.