MCQ: Indices and Surds - 3 - SSC CGL MCQ

Given that , 289 = 17^x/5
Use the formula if a^x = a ^y then x = y will be true.
⇒ 17 ² = 17^x/5
⇒ x /5 = 2
⇒ x = 10 .

Apply the law of Fractional Exponents and Laws of Exponents
(a^m) x (aⁿ) = a^{m + n}
a^-m = 1/a^m
p⁰ = 1
and if p^X = p^Y then X will be equal to Y. means X = Y;
(a^m)ⁿ = a^{m x n}
(a^m) x (aⁿ) = a^{m + n}
a^m ÷ aⁿ = a^{m ? n}
Or
a^m/aⁿ = a^{m ? n}

Given equation is
2^p + 3^q = 17 ----------------(i)
2^{p + 2} - 3^{q + 1} = 5
⇒ 2^p x 2² - 3^q x 3 = 5
⇒ 2^p x 4 - 3q x 3 = 5
⇒ 4 x 2^p - 3 x 3q = 5---------------(ii)
On multiplying Eq. (i) by 3 and adding it, with Eq. (ii), we get

3 x 2^p + 3 x 3^q = 51
4 x 2^p - 3 x 3^q = 5
_____________________________
7 x 2^p = 56

⇒ 7 x 2^p = 7 x 8
⇒ 2^p = 8
⇒ 2^p = 2³
⇒ p = 3

On substituting the value of p in Eq. (i) . we get
2³ + 3^q = 17
⇒ 8 + 3^q = 17
⇒ 3^q = 17 - 8
⇒ 3^q = 9
⇒ 3^q = 3²
⇒ q = 2

81^2.5 X 9^4.5 ÷ 3^4.8 = 9^?
⇒ 81^2.5 x 9^4.5 ÷ 3^4.8 = 9^?
Apply the law of Fractional Exponents and Laws of Exponents
(a^m)ⁿ = a^{m x n} ----------------1
(a^m) x (aⁿ) = a^{m + n}-------------2
a^m ÷ aⁿ = a^{m ? n}----------------3
and if p^X = p^Y then X will be equal to Y. means X = Y;----------------4

Apply Law 1
⇒ (9²)^2.5 x 9^4.5 ÷ 3^{2 x 2.4} = 9^?
⇒ 9⁵ x 9^4.5 ÷ 9^2.4 = 9^?
Apply law 2 and 3,
⇒ 9^{(5 + 4.5 - 2.4)} = 9^?
⇒ 9^? = 9^7.1
Apply the Law 4
∴ ? = 7.1

Given equation is,
( 5 + 2√3 ) / (7 + 4√3 )
Now Multiply and divide by ( 7 - 4√3 ) in above given equation.
By Rationalization, we will get,
[ ( 5 + 2√3 ) / ( 7 + 4√3 ) ] x [ ( 7 - 4√3 ) / ( 7 - 4√3 ) ]
[ ( 5 + 2√3 ) x ( 7 - 4√3 ) ] / [ ( 7 + 4√3 ) x ( 7 - 4√3 ) ]
Apply the formula and multiplication rule of algebra,
( P + Q ) x (R - S ) = P x R - P x S + Q x R - Q x S .................(1)
( a + b ) x ( a - b) = a² - b².......................................................(2)
= ( 5 x 7 - 5 x 4√3 + 7 x 2√3 - 2√3 x 4√3 )/ ( 7² - (4√3)² )
= 35 - 20√3 + 14√3 - 8 x 3 / ( 49 - 16 x 3 )
= ( 35 - 24 - 6√3 )/(49 - 48 )
= 11 - 6√3
∵ ( 5 + 2√3 ) /( 7 + 4√3 ) = a + b√3
∴ a + b√3 = 11 - 6√3
on equating the above equation and we get,
or a = 11 and b = - 6

Given equation is,
( 6a^-2bc^-3/4ab^-3c² ) ÷ ( 5a^-3 b²c^-1/3ab^-2c³ )
Apply the algebra law a ÷ b = a x 1/b
= ( 6a^-2bc^-3/4ab^-3c² ) x ( 3ab^-2c³/5a^-3 b²c^-1 )
= ( 6a^-2 bc^-3 x 3ab^-2c³) / ( 4ab^-3c² x 5a^-3b²c^-1 )
Apply the law of Fractional Exponents and Laws of Exponents
(a^m)(aⁿ) = a^m+n
a^m÷aⁿ=a^m-n
Or
a^m/aⁿ=a^m-n
= 18a^{-2 + 1} b^{1 - 2} c ^{-3 + 3} / 20a^{1 - 3} b^{-3 + 2} c^{2 - 1}
= 9a^-1b^-1c⁰/ 10a ^{- 2} b^-1c¹
= ( 9/10 ) a^{-1 + 2}b^{-1 + 1} c^{0 - 1}
= ( 9/10 ) a¹b⁰ c^-1
= ( 9/10 ) ac^-1 [∵ b⁰ = 1]

27³ × 3⁴ ÷ 3¹⁰

= 3^{13 - 10} = 3³
= 3 × 3 × 3 = 27

∵ 10^2/5 × 10^8/5 = 10ⁿ
⇒ 10^{2/5 + 8/5} = 10ⁿ
⇒ 10^10/5 = 10ⁿ ⇒ 10² = 10ⁿ ⇒ n = 2
Hence, the value of n is 2.

12x²y⁴z³ ÷ (2xy² × 3yz²)

√4 × 4^3/2 ÷ 4^-3 = 2^x

⇒ 4²⁺³ = 2^x
⇒ ( 2² )⁵ = 2^x ⇒ 2¹⁰ = 2^x ⇒ x = 10
∴ Value of x is 10.

17^3.5 x 17^7.3 ÷ 17^4.2 = 17^?
Apply the Laws of Exponents ::
(a^m) x (aⁿ) = a^m+n
Fractional Exponents ::
a^m÷ a ⁿ=a^m?n
⇒ 17^{3.5 + 7.3 - 4.2} = 17^?
⇒ 17^6.6 = 17^?
∴ ? = 6.6

^L√3 = M^1/L = Surd of L^th order.

Given that, (42 x 229) ÷ (9261)^1/3 = ?
As we know that (9621)^1/3 = (21 x 21 x 21)^1/3 = 21^3x1/3
∴ 42 x 229 ÷ (21^{3 x} )^1/3 = ?
∴ 42 x 229 ÷ 21 = ?
∴ 42 x 229 x 1/ 21 = ?
∴ 2 x 229 = ?
∴ ? = 2 x 229
∴ ? = 458

Given expression = [(√2)^√2]^√2
= (√2)^(2)√2/2
= (√2)^(2)1/√2
= (2)^1/2 x 2^1/√2
= 2^(2/21/√2)
= (2)^{(2)(1/√2 - 1)}
which denotes a real number but not a rational number .