Given log x = 2m - n, log y = n - 2m and log z = 3m - 2n, find in terms of m and n, the value of log $\dfrac{x^2y^3}{z^4}$.

Given,

1st equation :

⇒ log x = 2m - n

⇒ x = 10^{2m - n} …….(1)

2nd equation :

⇒ log y = n - 2m

⇒ y = 10^{n - 2m} …….(2)

3rd equation :

⇒ log z = 3m - 2n

⇒ z = 10^{3m - 2n} …….(3)

Substituting value of x, y and z in log $\dfrac{x^2y^3}{z^4}$, we get :

$\Rightarrow \text{log } \dfrac{(10^{2m - n})^2(10^{n - 2m})^3}{(10^{3m - 2n})^4} \\[1em] \Rightarrow \text{log } \dfrac{10^{2(2m - n)}.10^{3(n - 2m)}}{10^{4(3m - 2n)}} \\[1em] \Rightarrow \text{log } \dfrac{10^{4m - 2n}.10^{3n - 6m}}{10^{12m - 8n}} \\[1em] \Rightarrow \text{log } 10^{4m - 2n + 3n - 6m - (12m - 8n)} \\[1em] \Rightarrow \text{log } 10^{n - 2m - 12m + 8n} \\[1em] \Rightarrow \text{log } 10^{9n - 14m} \\[1em] \Rightarrow (9n - 14m) \text{ log 10} \\[1em] \Rightarrow (9n - 14m) \times 1 \\[1em] \Rightarrow 9n - 14m.$

Hence, log $\dfrac{x^2y^3}{z^4}$ = 9n - 14m.