Indices and Surds - 2 - Free MCQ Practice Test with solutions, SSC CGL

Given that, a^{2x + 2} = 1
Apply the algebra Law,
p⁰ = 1
⇒ a^{2x + 2} = a⁰
Apply the algebra law
p^X = p^Y then X will be equal to Y. means X = Y;
⇒ 2x + 2 = 0
⇒ x = -2/2 = -1

Given that, (2^{4x 1/2})^? = 256
Apply the law of Fractional Exponents and Laws of Exponents
⇒ (2²)^? = 2⁸
⇒ 2^{2 x ?} = 2⁸
⇒ 2 x ? = 8
∴ ? = 8/2 = 4

Third power of 4 = 4³ = 4 × 4 × 4 = 64

Third power of 2 = 2³ = 2 × 2 × 2 = 8
And second power of 3 = 3² = 3 × 3 = 9
∴ Difference = 9 – 8 = 1

( 9ⁿ x 3² x (3^-n/2)^-2 - (27)ⁿ ) / ( 3^3m x 2³ ) = 1/27
⇒ ( 3²ⁿ x 3² x 3ⁿ - 3³ⁿ) / ( 3^3m x 2³ ) = 1/27
⇒ ( 3^{2n + n} x 3² - 3³ⁿ) / ( 3^3m x 2³ )= 1/27
⇒ ( 3³ⁿ x 3² - 3³ⁿ) / ( 3^3m x 2³ ) = 1/27
⇒ 3³ⁿ( 3² - 1) / ( 3^3m x 8 ) = 1/27
⇒ 3³ⁿ( 9 - 1) / (3^3m x 8 ) = 1/27
⇒ ( 3³ⁿ x 8 ) / (3^3m x 8 ) = 1/27
⇒ 3³ⁿ / 3^3m = 1/27
⇒ 3^{3n - 3m} = 1/3³
⇒ 3^{3( n - m )} = 3^-3
3(n - m) = -3
or n - m = -1
or m - n = 1

Given that 16 x 8^{n + 2} = 2^m

Apply the law of Fractional Exponents and Laws of Exponents
if a multiply three times a x a x a then
a x a x a = a³
if a multiply two times a x a then
a x a = a²
if a multiply n times a x a x a x....up to n times, then
a x a x a x a ......up to n times = aⁿ
⇒ (2)⁴ x 2^{3 x (n+2)} = 2^m
⇒ (2)⁴ x 2³ⁿ⁺⁶ = 2^m
a^maⁿ = a^m+n
⇒ (2)^{(4 + 3n + 6)} = 2^m
⇒ (2)^{(3n + 10)} = 2^m
if p^X = p^Y then X will be equal to Y. means X = Y;
On comparing, we get
3n + 10 = m
⇒ m = 3n + 10;

∵ 2^{x - 1} + 2^{x + 1} = 320
Apply the law of Algebra
⇒ 2^{x - 1}(1 + 2 ² ) = 320
⇒ 2^{x - 1}(1 + 4 ) = 320
⇒ 2^{x - 1} x 5 = 320
⇒ 2^{x - 1} = 64 = 2 x 2 x 2 x 2 x 2 x 2
⇒ 2^{x - 1} = 2⁶
if p^X = p^Y then X will be equal to Y. means X = Y;
⇒ x - 1 = 6
∴ x = 7

Given equation is
(10)²⁰⁰ ÷ (10)¹⁹⁶
Apply the law of Algebra
a^m ÷ aⁿ = a^{m ? n}
= (10)^{200 - 196}
= 10⁴
= 10000

Given that
(0.00032)^2/5 = (32/100000)^2/5
Solve the equation by algebra Law
= (2⁵/10⁵)^2/5
= {(2/10)⁵}^2/5
= (2/10)^5x2/5
= (1/5)² = 1/25

7^8.9 ÷ (343)^1.7 x (49)^4.8 = 7^?
Apply the law of Fractional Exponents and Laws of Exponents
(a^m)(aⁿ) = a^m+n
a^m÷aⁿ=a^m?n
a^m/aⁿ=a^m?n
(a^m)ⁿ = a^mn
7^8.9 ÷ (343)^1.7 x (49)^4.8 = 7^?
⇒ 7^8.9 ÷ (7³)^1.7 x (7²)^4.8 = 7^?
⇒ 7^8.9 ÷ 7^5.1 x 7^9.6 = 7^?
⇒ 7^{8.9 - 5.1 + 9.6} = 7^?
⇒ 7^{18.5 - 5.1} = 7^?
∴ ? = 13.4

∵ 3^x - 3^{x - 1} = 18
⇒ 3^{x - 1}(3 - 1) = 18
⇒ 3^{x - 1}(2) = 18
⇒ 3^{x - 1} = 18/2
⇒ 3^{x - 1} = 9
⇒ 3^{x - 1} = 3²
Apply the Algebra law,
If a^X = a^Y then X will be equal to Y.
means X = Y;
⇒ x - 1 = 2
⇒ x = 3
Then x^x = (3)³ = 27

2^{x - 1} + 2^{x + 1} = 2560
2^{x - 1} + 2^{x - 1 + 2} = 2560
Apply the Law of Algebra
2^{x - 1} + 2^{x - 1} x 2 ² = 2560
⇒ 2^{x - 1} ( 1 + 2² ) = 2560
⇒ 2^{x - 1} = 2560/5 = 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
⇒ 2^{x - 1} = 2⁹
⇒ x - 1 = 9
∴ x = 9 + 1 = 10

Given, xyz = 1, a^x = b, b^y = c
Now, b = a^x
Apply the law of algebra .
Take the power y of both side equation.
⇒ b^y= a^xy
Take the power z of both side equation.
⇒ b^yz = a^xyz
⇒ ( b^y )^z = a^xyz
Replace the b^y with c and xyz with 1 as per given equation,
⇒ c^z = a

Given equation is
a^-1 - 1
Apply the law of Fractional Exponents and Laws of Exponents
x^-y = 1/x^y
⇒ a^-1 - 1 = 1/a - 1
⇒ a^-1 - 1 = (1 - a)/a
Now divide by a - 1 in above equation,
⇒ ( a^-1 - 1 ) ÷ (a - 1) = (1 - a)/a ÷ (a - 1)
⇒ ( a^-1 - 1 ) ÷ (a - 1) = (1 - a)/a x 1/ (a - 1)
⇒ ( a^-1 - 1 ) ÷ (a - 1) = -1x (a - 1)/a x 1/ (a - 1)
⇒ ( a^-1 - 1 ) ÷ (a - 1) = -1/a
∴ Required quotient = -1/a.

a⁵ × a⁷ = a^{5 + 7} = a¹²