If 2 x = 3 y = 12 z , show that 1 z = 1 y + 2 x .

Given: 2 x = 3 y = 12 z . We have to prove that 1 z = 1 y + 2 x . Let 2 x = 3 y = 12 z = k . ⇒ 2 = k 1 x , 3 = k 1 y , 12 = k 1 z . Now, 12 = k 1 z ⇒ 2 2 × 3 = k 1 z ⇒ k 1 x 2 × k 1 y = k 1 z ⇒ k 2 x × k 1 y = k 1 z ∵ a m n = a m × n ⇒ k 2 x + 1 y = k 1 z ∵ a m × a n = a m + n ⇒ 2 x + 1 y = 1 z . [ ∵ x p = x q ⇒ p = q ] Hence proved.

If 2 x = 3 y = 12 z , show that 1 z = 1 y + 2 x .

Given: 2 x = 3 y = 12 z . We have to prove that 1 z = 1 y + 2 x . Let 2 x = 3 y = 12 z = k . ⇒ 2 = k 1 x , 3 = k 1 y , 12 = k 1 z . Now, 12 = k 1 z ⇒ 2 2 × 3 = k 1 z ⇒ k 1 x 2 × k 1 y = k 1 z ⇒ k 2 x × k 1 y = k 1 z ∵ a m n = a m × n ⇒ k 2 x + 1 y = k 1 z ∵ a m × a n = a m + n ⇒ 2 x + 1 y = 1 z . [ ∵ x p = x q ⇒ p = q ] Hence proved.

If 2 x = 3 y = 6 - z , show that 1 x + 1 y + 1 z = 0

Given: 2 x = 3 y = 6 - z . We have to prove that 1 x + 1 y + 1 z = 0 . Let 2 x = 3 y = 6 - z = k . ⇒ 2 = k 1 x , 3 = k 1 y , 6 = k 1 - z Now, 6 = 2 × 3 = k 1 - z ⇒ k 1 x × k 1 y = k 1 - z ⇒ k 1 x + 1 y = k 1 - z ∵ a m × a n = a m + n ⇒ 1 x + 1 y = 1 - z ∵ x p = x q ⇒ p = q ⇒ 1 x + 1 y + 1 z = 0 . Hence proved.

If a x = b y = c z and b 2 = a c , then show that y = 2 z x z + x .

Given: a x = b y = c z and b 2 = a c . We have to prove that y = 2 z x z + x . Let a x = b y = c z = k So, a = k 1 x , b = k 1 y , c = k 1 z . Thus, b 2 = a c ⇒ k 1 y 2 = k 1 x k 1 z ⇒ k 2 y = k 1 x + 1 z ∵ a m n = a m × n ; a m × a n = a m + n ⇒ 2 y = 1 x + 1 z ∵ x p = x q ⇒ p = q ⇒ 2 y = z + x x z ⇒ y = 2 z x z + x . Hence proved.

If 3 x = 5 y = ( 75 ) z , show that z = x y 2 x + y .

Given: 3 x = 5 y = ( 75 ) z . We have to show that z = x y 2 x + y . So, let, 3 x = 5 y = ( 75 ) z = k . ⇒ 3 = k 1 x , 5 = k 1 y , 75 = k 1 z . Now, 75 = k 1 2 ⇒ 5 2 × 3 = k 1 z ⇒ k 1 y 2 × k 1 x = k 1 z ⇒ k 2 y × k 1 x = k 1 z [ ∵ a m n = a m × n ] ⇒ k 2 y + 1 x = k 1 z [ ∵ a m × a n = a m + n ] ⇒ 2 y + 1 x = 1 z [ ∵ x p = x q ⇒ p = q ] ⇒ 2 x + y x y = 1 z ⇒ z = x y 2 x + y . Hence proved.