Assuming that x, y, z are positive real numbers, simplify each of the following:
(i)(√x⁻³)³ (ii) √x³y⁻² (iii) x⁻²/³ y⁻¹/²)² (iv) (√x)⁻²/³ √y⁴÷√xy⁻¹/² (v) s√243X¹⁰Y⁵Z¹⁰ (VI) ( X⁻⁴/Y⁻¹⁰)⁵/⁴
Answers
By using these law of exponents : [a⁻¹ = 1/a] , [a^-p = 1/ a^p], (a^p )^q = a^pq
(i) (√x⁻³)⁵
( √1/√x³)⁵
= (1 / x³/²)⁵
= (1/x³/² × ⁵)
= (1 / x¹⁵/²)
(ii) √x³y⁻²
= (x³/y²)½
= (x³×¹/² /y²×¹/²)
= x³/²/y
(iii) (x⁻²/³ y⁻¹/²)²
= 1/(x²/³ × y¹/²)²
= 1/(x²/³ × ² × y¹/² × ²)
= 1/(x⁴/³y)
(iv) (√x)⁻²/³ √y⁴ ÷ √xy⁻¹/²
= (x¹/²)⁻²/³ y⁴×¹/² ÷ (xy-¹/²)¹/²
= X⁻¹/³ y² ÷ x¹/²y⁻¹/⁴
= x⁻¹/³ ⁻¹/² y² ⁺¹/⁴
= x( ⁻²⁻³)/⁶ y(⁸⁺ ¹)/⁴
= (x⁻⁵/⁶) (y⁹/⁴)
= (y⁹/⁴) / (x⁵/⁶)
(v) ⁵√243x¹⁰y⁵z¹⁰
= (243x¹⁰y⁵z¹⁰)¹/⁵
= 3⁵ˣ¹/⁵ x¹⁰ ˣ¹/⁵y⁵×¹/⁵ z¹⁰×¹/⁵
= 3x²yz²
(vi) (x⁻⁴/y⁻¹⁰)⁵/⁴
= (y¹⁰/x⁴)⁵/⁴
= (y¹⁰×⁵/⁴ /x⁴×⁵/⁴)
= y²⁵/² / x⁵
HOPE THIS ANSWER WILL HELP YOU…..
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Answer:
Step-by-step explanation:
(i) (√x⁻³)⁵
( √1/√x³)⁵
= (1 / x³/²)⁵
= (1/x³/² × ⁵)
= (1 / x¹⁵/²)
(ii) √x³y⁻²
= (x³/y²)½
= (x³×¹/² /y²×¹/²)
= x³/²/y
(iii) (x⁻²/³ y⁻¹/²)²
= 1/(x²/³ × y¹/²)²
= 1/(x²/³ × ² × y¹/² × ²)
= 1/(x⁴/³y)
(iv) (√x)⁻²/³ √y⁴ ÷ √xy⁻¹/²
= (x¹/²)⁻²/³ y⁴×¹/² ÷ (xy-¹/²)¹/²
= X⁻¹/³ y² ÷ x¹/²y⁻¹/⁴
= x⁻¹/³ ⁻¹/² y² ⁺¹/⁴
= x( ⁻²⁻³)/⁶ y(⁸⁺ ¹)/⁴
= (x⁻⁵/⁶) (y⁹/⁴)
= (y⁹/⁴) / (x⁵/⁶)
(v) ⁵√243x¹⁰y⁵z¹⁰
= (243x¹⁰y⁵z¹⁰)¹/⁵
= 3⁵ˣ¹/⁵ x¹⁰ ˣ¹/⁵y⁵×¹/⁵ z¹⁰×¹/⁵
= 3x²yz²