Given log x = m + n and log y = m - n, express the value of log $\dfrac{10x}{y^2}$ in terms of m and n.

Given,

⇒ log x = m + n

⇒ x = 10^{m + n} ……..(1)

Given,

⇒ log y = m - n

⇒ y = 10^{m - n} ……..(2)

Substituting value of x and y from equation (1) and equation (2) in log $\dfrac{10x}{y^2}$, we get :

$\Rightarrow \text{log} \dfrac{10x}{y^2} = \text{log } \dfrac{10 \times 10^{m + n}}{(10^{m - n})^2} \\[1em] = \text{log } \dfrac{10^{m + n + 1}}{10^{2(m - n)}} \\[1em] = \text{log } 10^{m + n + 1 - 2(m - n)} \\[1em] = \text{log } 10^{m + n + 1 - 2m + 2n} \\[1em] = \text{log } 10^{3n - m + 1} \\[1em] = (3n - m + 1) \text{ log} 10 \\[1em] = (3n - m + 1) \times 1 \\[1em] = 3n - m + 1 \\[1em] = 1 - m + 3n$

Hence, log $\dfrac{10x}{y^2}$ = 1 - m + 3n.