Question

Answer this with explanation

Answer 1

Given : a^x = (a/k)^y = k^m.

Apply 'log' on both sides, we get

⇒ log(a^x) = log(a/k)^y = log(k)^m

We know that log(a)^n = n log a

⇒ x log a = y log(a/k) = m log k.

We know that log(a/b) = log a - log b

⇒ x log a = y(log a - log k) = m log k

Now,

Equating like terms, we get

⇒ x = (m log k)/(log a)

⇒ y = (m log k)/(log a - log k).

Given:

⇒ (1/x) - (1/y)

⇒ (log a/m log k) - (log a - log k/m log k)

⇒ (log a - log a + log k/m log k)

⇒ (log k)/(m log k)

⇒ 1/m.

Therefore, 1/x - 1/y = 1/m.

Hope it helps!

Answer 2

Solution for your question is given in the attached picture.

Value of $\dfrac{1}{x}-\dfrac{1}{y} \:is\:\dfrac{1}{m}$.

Three terms are given,  

Let 1st term = $a^{x}$  

     2nd term = $\bigg(\dfrac{a}{k}\bigg)^{y}$

     3rd term = $k^{m}$

First compare 1st and 3rd terms and then compare 2nd and 3rd terms.