Given :

log₃ m = x and log₃ n = y

(i) Express 3^{2x - 3} in terms of m.

(ii) Write down 3^{1 - 2y + 3x} in terms of m and n.

(iii) If 2 log₃ A = 5x - 3y; find A in terms of m and n.

Given,

1st equation :

⇒ log₃ m = x

⇒ m = 3^x ……(1)

2nd equation :

⇒ log₃ n = y

⇒ n = 3^y ……(2)

(i) Given,

3^{2x - 3}

⇒ 3^2x.3^-3

⇒ (3^x)².3^-3

Substituting value of 3^x from equation (1) in above equation, we get :

⇒ m².3^-3

⇒ $\dfrac{m^2}{3^3}$

⇒ $\dfrac{m^2}{27}$.

Hence, 3^{2x - 3} = $\dfrac{m^2}{27}$.

(ii) Simplifying the expression,

⇒ 3^{1 - 2y + 3x}

⇒ 3¹.3^-2y.3^3x

⇒ 3.(3^y)^-2.(3^x)³

Substituting value of 3^x and 3^y from equation (1) and (2) in above equation, we get :

⇒ 3.n^-2.m³

⇒ $\dfrac{3m^3}{n^2}$.

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

(iii) Given,

⇒ 2log₃ A = 5x - 3y

⇒ log₃ A² = 5x - 3y

⇒ A² = 3^{5x - 3y}

⇒ A² = 3^5x.3^-3y

⇒ A² = (3^x)⁵.(3^y)^-3

Substituting value of 3^x and 3^y from equation (1) and (2) in above equation, we get :

⇒ A² = m⁵.n^-3

⇒ A² = $\dfrac{m^5}{n^3}$

⇒ A = $\sqrt{\dfrac{m^5}{n^3}}$.

Hence, A = $\sqrt{\dfrac{m^5}{n^3}}$.