# Solutions to Simplify Exponents and Radicals

We present detailed solutions and explanations to the questions in simplify exponents and radicals. The rules for radicals and exponents are used to simplify numerical and algebraic expressions with exponents and radical expressions and therefore need to be reviewed.

## Evaluate Numerical Expressions with Exponents

### Question 1

Evaluate the following expressions:
1.   32 = 3 × 3 = 9
2.   - 34 = - (3 × 3 × 3 × 3 ) = - 81
3.   (- 3)4 = - 3 × - 3 × - 3 × - 3 ) = 81
4.   120 = 1
5.   020 = 0
6.   0- 4 = 1 / 0 4 = 1 / 0 : undefined because division by zero is not allowed in maths.
7.   (-1)7 = - 1
8.   (-1)4 = 1
9.   (-1) - 3 = 1 / (-1) 3 = 1 / (- 1) = - 1
10.   (- 2 / 3) -2 = (- 3 / 2) 2 = (-1) 2 ( 3 / 2) 2 = 9 / 4
11.   (- 2)-2 / 4-2 = 42 / (- 2)2 = 16 / 4 = 4
12.   - 3-3 + (- 2)-2 = - 1 / 33 + 1 / (- 2)2 = - 1 / 27 + 1 / 4 = 23 / 108
13.   x4 for x = - 2
x4 = (-2)4 = 16
14.   - x2 for x = - 3
- x2 = - (-3)2 = - 9
15.   - x5 for x = - 2
- x5 = -( - 2) 5 = 32
16.   x4 for x = - 2 / 3
(- 2 / 3)4 = (-2)4 / (3)4 = 16 / 81
17.   - x-2 for x = - 1 / 2
- x-2 = - (- 1 / 2)-2 = - ( - 1)-2 ( 1 / 2) -2
= - (1) (2 / 1) 2 = - 4 / 1 = - 4
18.   - x5 for x = - 1 / 3
- x5 = - ( - 1 / 3)5 - ( - 1) 5( 1 / 3)5
= - (-1) (1 / 243) = 1 / 243

## Evaluate Numerical Expressions with Radicals and rational exponents

### Question 2

Evaluate the following expressions:
1.   √(64) = √(82) = 8
2.   ∛(-8) = ∛((-2)3) = - 2
3.   43/2 = (√4) 3 = 2 3 = 8
4.   82/3 = (∛(8)) 2 = 2 2 = 4
5.   0.00013/4 = ( 4√(0.0001) )3
= ( 4√(1/10000) )3 = ( 4√(1/104) )3
= ( 1 / 10 )3 = 1 / 103 = 0.001
6.   16- 3/4 = 1 / 16 3/4 = 1 / ( 4√(16)) 3
= 1 / 2 3 = 1 / 8
7.   √2 √8 = √(2 × 8) = √(16) = 4
8.   ∛2 ∛(32) = ∛(2 × 32) = ∛(64) = 4
9.   √2 / √8 = √(2 / 8) = √(1 / 4) = 1 / 2
10.   ∛(-16) / ∛2 = ∛(-16/2) = ∛(- 8) = - 2
11.   - 2 (6√8) (56√8) = (- 2 × 5) 6√(8 × 8)
= - 10 6√(64) = - 10 6√(26) = -10(2) = - 20
12.   - 4 ∛(375) / (2∛3) = (- 4 / 2) ∛ (375 / 3) = - 2 ∛ (375 / 3)
= - 2 ∛ (125) = - 2 × 5 = - 10

## Simplify Algebraic Expressions with Exponents

### Question 3

Simplify the following expressions:
1.   x2 x5 = x2 + 5 = x7
2.   y4 / y2 = y4 - 2 = y2
3.   (x2)-2 = (x2 × (-2)) = x- 4 = 1 / x4

4. \large {3(\dfrac{1}{2} x^4)(\dfrac{1}{3} x^3) = (3 \times \dfrac{1}{2} \times \dfrac{1}{3} ) x^{4 + 3} = \dfrac{1}{2} x^7}

5. \large {(2 x)^4(\dfrac{1}{4} x^{-3}) = 2^4 x^4 (\dfrac{1}{4} x^{-3}) \\\\ = (2^4 \times \dfrac{1}{4}) x^{4-3} = 4 x}

6. \large { \dfrac{(3x)^2(-2x)^3}{(2x)^2} = \dfrac{3^2 x^2 (-2)^3 x^3}{2^2 x^2} \\\\ = \dfrac{3^2 (-2)^3}{2^2} \dfrac{x^3}{x^2} = - 18 x }

7. \large { \dfrac{(4x)^2(100 x)^0}{(3x)^2} = \dfrac{4^2 x^2 \times 1}{3^2 x^2} = 16/9 }
8.   (- 2 x2 y -3)3 = (- 2)3 (x2)3 (y -3) 3 = - 8 x6 y-9 = - 8 x6 / y9
9.   (3 x2 y3) (2 x5 y - 2) = (3 × 2)(x2 x5)(y3 y - 2) = 6 x2 + 5 y3 - 2 = 6 x7 y
10.   (- 2 x2 y3 z4) ( - 4 x3 y z - 8) ( 5 x y2 z2) = (-2 × (-4) × 5)(x2 x3 x)(y3 y y2)(z4 z-8 z2)
= 40 x2 + 3 + 1 y3 + 1 + 2 z4 - 8 + 2 = 40 x6 y6 z- 2 = 40 x6 y6 / z2

11. \large { (-2xy^2)^2 \left (\dfrac{x^6}{(2x)^2} \right)^3 = (-2)^2 x^2 y^4 \left (\dfrac{x^{18}}{(2x)^6} \right) = \dfrac{4x^2 y^4x^{18}}{2^6 x^6}} \\\\ \large { = \dfrac{4}{2^6} \dfrac{x^{2+18}y^4}{x^6} = \dfrac{1}{16} {x^{14} y^4} }

12. \large { \left (\dfrac{4 x^6 y^3}{-8x^3y^{-2}} \right) = \dfrac{4}{-8} \; \dfrac{x^6 y^3}{x^3y^{-2}} = -\dfrac{1}{2} x^{6-3} y^{3-(-2)} = -\dfrac{1}{2} x^3 y^5 }

13. \large { \left (\dfrac{- 4 x^3 y^2}{3 x^3 y^3} \right) \left (\dfrac{3 x^2 y^5}{-6x^3y^2} \right) = \dfrac{(- 4 \times 3)}{(3 \times (-6))} \dfrac{(x^3 y^2)(x^2 y^5)}{(x^3 y^3)(x^3y^2)} }\\\\\\ \large { = \dfrac{2}{3} \dfrac{x^5 y^7}{x^6 y^5} = \dfrac{2}{3} x^{5-6} y^{7-5}= \dfrac{2}{3} \dfrac{y^2}{x} }

14. \large { \dfrac{1}{6}(x^2 y z^3)\left (\dfrac{4 x^2 y^3}{3 x^2 y^{-3}z^2} \right) \left (\dfrac{3 x^2 y^5 z}{-x^3 y^2 z^3} \right) = \dfrac{1 \times 4 \times 3}{6 \times 3 \times (-1)} \dfrac{(x^2 y z^3)(x^2 y^3)(x^2 y^5 z)}{(x^2 y^{-3}z^2)(-x^3 y^2 z^3)} } \\\\\\ \large { = (-\dfrac{2}{3}) \dfrac{x^{2+2+2}y^{1+3+5}z^{3+1}}{x^{2+3} y^{-3+2} z^{2+3} } = - \dfrac{2}{3} \dfrac{x^6 y^9 z^4}{x^5 y z^5} = - \dfrac{2}{3} \; \dfrac{x y^{10} }{z} }

15. \large { (4 x^{2/3})(-8x^{1/2}) = (4 \times; (-8))(x^{2/3+1/2}) = - 32 x^{7/6} }

16. \large { (4 x)^{3/2}(9x)^{1/2} = 4^{3/2} x^{3/2} 9^{1/2} x^{1/2} = (\sqrt 4)^3 (\sqrt 9) x^{3/2+1/2} = 8 \times 3 x^2 = 24 x^2}

17. \large { (- 27 x^{3/2})^{1/3} = (- 27)^{1/3} (x^{3/2})^{1/3} = \sqrt[3]{-27} x^{3/2 \times 1/3} = - 3 x^{1/2} = - 3 \sqrt x }

18. \large {\left (\dfrac{-8x^3}{y^{-6}} \right)^{2/3} = \dfrac{(-8x^3)^{2/3}} {(y^{-6})^{2/3}} = \dfrac{(-8)^{2/3} (x^3)^{2/3}} {(y^{-6 \times (2/3)}}} \\\\\\ \large { = \dfrac{(\sqrt[3]{-8})^2 (x^{3 \times (2/3)})} {(y^{-4})} = 4 x^2 y^4 }

19. \large { x^{5/3} x^{1/6} x ^{-11/6} = x^{5/3+1/6 -11/6} = x^0 = 1 , x \ne 0 }

## Simplify Algebraic Expressions with Radicals

### Question 4

1. \large { \sqrt{x^2} = | x |}

2. \large { \sqrt[4]{16x^4} = \sqrt[4]{16} \sqrt[4]{x^4} = 4 | x | }

3. \large { \sqrt{(2x - 1)^2} = |2x - 1| }

4. \large { \sqrt[3]{-27x^3} = \sqrt[3]{-27} \sqrt[3]{x^3} = \sqrt[3]{(-3)^3} \sqrt[3]{x^3} = - 3 x }

5. \large { \sqrt[3]{8 x^6 y^3} = \sqrt[3]{8} \sqrt[3]{ x^6} \sqrt[3]{y^3} = 2 x^{6/3} y = 2 x^2 y}

6. \large { \sqrt{x^3} \sqrt{x^2} = \sqrt{x^3 x^2} = \sqrt{x^{3+2}} = \sqrt{x^5} = \sqrt{x^4 \times x} = x^2 \sqrt x }

7. \large { \sqrt[3]{\left (\dfrac{-8x^6}{y^{-3}} \right)} = \dfrac{\sqrt[3]{-8x^6}}{\sqrt[3]{y^{-3}}} = \dfrac{\sqrt[3]{-8} \sqrt[3]{x^6}} {\sqrt[3]{y^{-3} }} }\\\\\\ \large { = \dfrac{\sqrt[3]{(-2)^3} \sqrt[3]{(x^2)^3}} {\sqrt[3]{(y^{-1})^3 }} = \dfrac{-2 x^2}{y^{-1}} = -2 x^2 y}

8. \large { \dfrac{ \sqrt[5]{64x^9 y^7}}{ \sqrt[5]{2 x^4 y^2}} = \sqrt[5] {\dfrac{64x^9 y^7}{2 x^4 y^2}}} \\\\\\ \large { = \sqrt[5] {32 x^{9-4} y^{7-2}} = \sqrt[5] {2^5} \sqrt[5] x^{5} \sqrt[5] y^{5} = 2 x y }

9. \large { (4\sqrt[8]{b^2})( 5\sqrt[8]{b^3})( \sqrt[8]{b^3}) = (4 \times 5) \sqrt[8]{b^2 b^3 b^3 } = 20 \sqrt[8]{b^8 } = 20 |b| = 20 b }
Note: In the last question; since 8√(b3) is real then b is greater than or equal to zero. Hence |b| = b.