Prove that:
(i) √3 x5⁻³ ÷ 3√3⁻¹√5x6√3 x5⁶= 3/5
(ii) 9³/² -3× 5⁰ -(1/81)⁻¹/²= 15
(iii) (1/4 )⁻² -3×8²/³ ×4⁰+(9/16)⁻¹/²=16/3
(iv) 2¹/²x3¹/³x4¹/⁴ /10⁻¹/⁵x5³/⁵ ÷3⁴/³x5⁻⁷/⁵ / 4⁻³/⁵x6 = 10
(v) √1/4+ (0.01)⁻¹/² – (27)²/³ = 3/2
(vi) 2 ⁿ⁻¹/2ⁿ⁻¹ - 2ⁿ= 3/2
(vii) (64/125)⁻²/³ + 1/(256/625)¹/⁴ + (√25/3√64) = 65/16
(viii) 3⁻³x6²x√98 / 5²x3√1/25x(15)⁻⁴/³ x3¹/³ = 28√2
(ix) (0.6)⁰ - (0.1)⁻¹ / (3/8)⁻¹ (3/2)³ + (-1/3)⁻¹ = - 3/2
Answers
By using these law of exponents : [a⁻¹ = 1/a] , [a^-p = 1/ a^p], (a^p )^q = a^pq , a⁰ = 1, (a × b)ⁿ = aⁿ × bⁿ
(i) √3 x 5⁻³ ÷ ³√3⁻¹√5 x ⁶√3 x 5⁶ = ⅗
L.H.S : √3 x 5⁻³ ÷ ³√3⁻¹√5 x ⁶√3 x 5⁶
= ((3 x 5⁻³)¹/² ÷ (3⁻¹)¹/³(5)¹/²) x (3 x 5⁶)¹/⁶
= ((3)¹/² x (5⁻³)¹/² ÷ (3⁻¹)¹/³(5)¹/² ) x (3¹/⁶ x 5⁶ ×¹/⁶)
= (3)¹/² x (5)⁻³/² ÷ (3⁻¹/³)(5)¹/² ) x ((3)¹/⁶ x 5⁶/⁶)
= (3)¹/² - (-¹/³) x (5)⁻³/² - ¹/²) x ((3)¹/⁶ x 5¹)
= (3)(³+²)/⁶ x (5)(⁻³ - ¹)/² x ((3)¹/⁶ x 5¹)
= (3)⁵/⁶ x (5)⁻⁴/² x ((3)¹/⁶ x 5¹)
= (3)⁵/⁶ x (5)⁻² x ((3)¹/⁶ x 5¹)
= (3)⁵/⁶+¹/⁶ x (5)⁻²+¹
= 3⁶/⁶ × 5⁻¹
= 3¹ × ⅕
= ⅗
= R.H.S
(ix) (0.6)⁰ - (0.1)⁻¹ / (3/8)⁻¹ (3/2)³ - (-1/3)⁻¹ = - 3/2
LH.S - (0.6)⁰ - (0.1)⁻¹ / (3/8)⁻¹ (3/2)³ - (1/3)⁻¹
= (1 - 1/0.1 ) /( 8/3× 3³/2³ - 3/1)
= (1 - 1/0.1 ) /( 8/3 × 27/8 - 3/1)
= ( 1 - 10) / (9 - 3)
= - 9/6
= - 3/2
= RHS
Some solutions are in the attachment below.
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