MCQ: Indices and Surds - 1 - SSC CGL MCQ

16^3/2 + 16^-3/2
Apply the law of Fractional Exponents and Laws of Exponents
a^m x aⁿ = a^m+n
a^-m = 1/a^m
⇒16^3/2 + 1/16^3/2
⇒(16^1/2)³ + 1/(16^1/2)³
⇒(4^{2 x 1/2})³ + 1/(4^{2 x 1/2})³
⇒4³ + 1/4³
⇒ 64 + 1/64
⇒ (64 x 64+ 1)/64
= (4096+1)/64
= 4097/64

Given that ,(16)⁹ ÷ (16)⁴ x (16) ³ = (16)^?
Apply the law of Fractional Exponents and Laws of Exponents
(a^m) x (aⁿ) = a^{m + n}-------------------1
a^m ÷ aⁿ = a^{m ? n}----------------2
⇒ (16)⁹ ÷ (16)⁴ x (16) ³ = (16)^?
⇒ (16)^? = (16)^{9 + 3 - 4}
⇒ ? = 12 - 4 = 8

(1/5 )^3a = 0.008 = 8/1000 = 1/125 = (1/5 )³
Apply the Algebra Law and Solve the equation.
⇒ 3a = 3
∴ a = 1
∴ ( 0.25 )^a = ( 0.25 )¹ = 0.25

? = 1/ ( 1 + x^{b - a} + x^{c - a} ) + 1/ ( 1 + x^{a - b} + x^{c - b} ) + 1 / ( 1 + x^{b - c} + x^{a - c} )
Apply the law of algebra and solve the given equation.
⇒ ? =1/ ( 1 + x^b/x^a + x^c/x^a ) + 1/ ( 1+ x^a/x^b + x^c/x^b ) + 1/ ( 1 + x^b/x^c + x^a/x^c )
⇒ ? = x^a/ ( x^a+x^b+x^c ) + x^b/ ( x^b + x^a+ x^c ) + x^c/ ( x^c+x^b+x^a )
⇒ ? = ( x^a + x^b + x^c ) / ( x^a + x^b + x^c ) = 1

[(2a²)³]³ [(3a³)²]² ÷ [(6a⁶)²]²
= (8a⁶)³. (9a⁶)² ÷ (36a¹²)²

6a³ b³ c² ÷ 2ab²c

= 3a^{3 - 1} b^{3 - 2} c^{2 - 1} = 3a²bc

If any given number (suppose x) is a rational number other than zero, then x⁰ = 1.

Given that ,
m = 7 - 4√3
Then I/m = 1/7 - 4√3......................(1)
Now multiply and divide with 7 + 4√3 in Eq. (1)
We will get,
1/m = ( 1/7 - 4√3 ) x ( 7 + 4√3 / 7 + 4√3)
⇒1/m = ( 7 + 4√3 )/ ( 7 - 4√3 ) x ( 7 + 4√3
⇒1/m = ( 7 + 4√3 ) / ( 7² - ( 4√3)² )
⇒1/m = ( 7 + 4√3 ) / ( 49 - 4 x 4 x 3 )
⇒1/m = ( 7 + 4√3 ) /49 - 48
⇒1/m = ( 7 + 4√3 ) / 1
⇒1/m = 7 + 4√3
∴ m + 1/m = 14
⇒ ( m + 1/m ) + 2 = 14 + 2 = 16
∴ (√m + 1/√m)² = ( m + 1/m )+ 2
⇒ (√m + 1/√m)² = 16
⇒ (√m + 1/√m)² = 4²
⇒ (√m + 1/√m) = 4

Given that, a = 2 + √3
Then, 1/a = 2 - √3 [ by conjugate property ]
Conjugate law is below,
if x = p + √q
then 1/x = p - √q
Apply the law of algebra in above equation.
Now, we have, a² + a ^-2
= ( a + 1/a )² - 2
= ( 2 + √3 + 2 - √3)² - 2
= (4)² - 2 = 16 - 2 = 14

Given equation is below
(p/q)^{n - 1} = (q/p)^{n - 3}
Apply the law of Fractional Exponents and Laws of Exponents
a ^{- m} = 1/a ^m
or
a^m = 1/a ^{- m}
(p/q)^{n - 1} = (q) ^{- (n - 3)} x 1/ (p) ^{- (n - 3)}
(p/q)^{n - 1} = (p) ^{3 - n} x 1/ (q) ^{3 - n}
(p/q)^{n - 1} = (p/q) ^{3 - n}
if p^X = p^Y then X will be equal to Y. means X = Y;
⇒ n - 1 = 3 - n
⇒ 2n = 4
⇒ n = 2

? = [(12)^-2]²/[(12)²]^-2
Apply the Algebra Law,
(a^m)ⁿ = a^mn
= (12) ^{- 4}/(12) ^{- 4}
= 1

? = [ (11)³ x (6)² ] x 1/4³
? = [ 11³ x 6² ] x 1/4³
Solve the equation by algebra law.
?= [ 1331 x 36 ] x 1/64 =1331 x 36 / 64
? = 748.6875

(16)^–3/4 + 2^–3 + (8)^–2/3

If m and n are natural numbers, then m√n is irrational unless n is m^th power of an integer.