If 21 , 168 = x 4 × y 3 × z 2 , then find ( x + y + z ) y + z x + y , where x , y and z are positive integers.

We have, 21 , 168 = x 4 × y 3 × z 2 ⇒ 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 = x 4 × y 3 × z 2 ⇒ 2 4 × 3 3 × 7 2 = x 4 × y 3 × z 2 Comparing the bases and the exponents of both sides, we obtain, x = 2 , y = 3 and z = 4 . Substituting these values in the given expression, we obtain, ( x + y + z ) y + z x + y = ( 2 + 3 + 4 ) 3 + 4 2 + 3 = 9 × 7 5 = 63 5 = 12 3 5 . Therefore, the required answer is 12 3 5 .

If 21 , 168 = x 4 × y 3 × z 2 , then find ( x + y + z ) y + z x + y , where x , y and z are positive integers.

We have, 21 , 168 = x 4 × y 3 × z 2 ⇒ 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 = x 4 × y 3 × z 2 ⇒ 2 4 × 3 3 × 7 2 = x 4 × y 3 × z 2 Comparing the bases and the exponents of both sides, we obtain, x = 2 , y = 3 and z = 4 . Substituting these values in the given expression, we obtain, ( x + y + z ) y + z x + y = ( 2 + 3 + 4 ) 3 + 4 2 + 3 = 9 × 7 5 = 63 5 = 12 3 5 . Therefore, the required answer is 12 3 5 .

Find the value of 3 4 n + 1 ( 4 ) + ( 81 ) n 3 4 n - 1 ( 5 ) + 3 ( 1 + 4 n ) .

We have, 3 4 n + 1 ( 4 ) + ( 81 ) n 3 4 n - 1 ( 5 ) + 3 ( 1 + 4 n ) = 3 4 n × 3 1 × 4 + ( 3 × 3 × 3 × 3 ) n 3 4 n × 3 - 1 × 5 + 3 1 × 3 4 n ∵ a m × a n = a m + n = 3 4 n × 3 × 4 + ( 3 × 3 × 3 × 3 ) n 3 4 n × 1 3 × 5 + 3 × 3 4 n ∵ 1 a m = a - m = 3 4 n × 12 + ( 3 4 ) n 3 4 n × 5 3 + 3 × 3 4 n ∵ a × a × a × . . . . p times = a k = 3 4 n × 12 + 3 4 n 3 4 n × 5 3 + 3 × 3 4 n ∵ a p q = a p × q = 3 4 n 12 + 1 3 4 n 5 3 + 3 = 13 5 + 9 3 = 13 14 3 = 13 × 3 14 = 39 14 = 2 11 14 . Therefore, the required answer is 2 11 14 .

If 5 n + 2 = 625 , then find n + 3 1 3

Given: 5 n + 2 = 625 We know, if a m = a n , then m = n , if a ≠ 0 , a ≠ 1 and a ≠ - 1 So, 5 n + 2 = 5 4 ⇒ n + 2 = 4 ⇒ n = 4 - 2 ⇒ n = 2 So, n + 3 1 3 = 2 + 3 1 3 = 5 1 3 The value of expression is 5 1 3

Simplify the following: 2 2 3 - 2 - 2 3 2 4 3 + 1 + 2 - 4 3 .

We have, 2 2 3 - 2 - 2 3 2 4 3 + 1 + 2 - 4 3 = 2 2 3 - 2 - 2 3 2 2 3 2 + 2 0 + 2 - 2 3 2 ∵ a m n = a m × n , a 0 = 1 = 2 2 3 - 2 - 2 3 2 2 3 2 + 2 2 3 + - 2 3 + 2 - 2 3 2 = 2 2 3 - 2 - 2 3 2 2 3 2 + 2 2 3 × 2 - 2 3 + 2 - 2 3 2 ∵ a m × a n = a m + n = 2 2 3 3 - 2 - 2 3 3 ∵ a - b a 2 + a b + b 2 = a 3 - b 3 = 2 2 - 2 - 2 ∵ a m n = a m × n = 4 - 1 2 2 ∵ 1 a m = a - m = 4 - 1 4 = 16 - 1 4 = 15 4 = 3 3 4 Therefore, the required answer is 3 3 4 .

Simplify the following: ( 7 6 ) 2 + ( 7 6 ) - 2 ( 7 6 ) 4 - 1 + ( 7 6 ) - 4

We have, ( 7 6 ) 2 + ( 7 6 ) - 2 ( 7 6 ) 4 - 1 + ( 7 6 ) - 4 = 7 1 6 2 + 7 1 6 - 2 7 1 6 4 - 7 0 + 7 1 6 - 4 ∵ x n = x 1 n , x 0 = 1 = 7 2 6 + 7 - 2 6 7 2 6 2 - 7 2 6 + - 2 6 + 7 - 2 6 2 ∵ a m n = a m × n = 7 2 6 + 7 - 2 6 7 2 6 2 - 7 2 6 × 7 - 2 6 + 7 - 2 6 2 ∵ a m × a n = a m + n = 7 2 6 3 + 7 - 2 6 3 ∵ ( a + b ) a 2 - a b + b 2 = a 3 + b 3 = 7 1 + 7 - 1 ∵ a m n = a m × n = 7 + 1 7 ∵ 1 a m = a - m = 49 + 1 7 = 50 7 = 7 1 7 . Therefore, the required answer is 7 1 7 .

If 16 , 200 = 2 a × 3 b × 5 c , then find a , b and c .

We have, 16 , 200 = 2 a × 3 b × 5 c ⇒ 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 = 2 a × 3 b × 5 c ⇒ 2 3 × 3 4 × 5 2 = 2 a × 3 b × 5 c Comparing the bases and exponents of both sides of the equation, we obtain, a = 3 , b = 4 and c = 2 .

Show that a x a γ x 2 + x y + y 2 × a γ a z y 2 + y z + z 2 × a z a x z 2 + x z + x 2 = 1 . where a ≠ 0 .

We have, LHS = a x a γ x 2 + x y + y 2 × a γ a z y 2 + y z + z 2 × a z a x z 2 + x z + x 2 = a x - y x 2 + x y + y 2 × a y - z y 2 + y z + z 2 × a z - x z 2 + x z + x 2 ∵ p m p n = p m - n = a x - y x 2 + x y + y 2 × a y - z y 2 + y z + z 2 × a z - x z 2 + x z + x 2 ∵ p m n = p m × n = a x 3 - y 3 × a y 3 - z 3 × a z 3 - x 3 ∵ ( p - q ) p 2 + p q + q 2 = p 3 - q 3 = a x 3 - y 3 + y 3 - z 3 + z 3 - x 3 ∵ p m × p n = p m + n = a 0 = 1 ∵ p 0 = 1 Hence, proved.