Given log₁₀ x = 2a and log₁₀ y = $\dfrac{b}{2}$ .

(i) Write 10^a in terms of x .

(ii) Write 10^2b+1 in terms of y .

(iii) If log₁₀ P = 3a - 2b, express P in terms of x and y.

(i) Given: log₁₀ x = 2a

⇒ $10^{2a}$ = x

⇒ $(10^a)^2$ = x

⇒ $10^a = \sqrt{x}$

Hence, the value of $10^a = \sqrt{x}$.

(ii) Given, log₁₀ y = $\dfrac{b}{2}$

⇒ $10^{\dfrac{b}{2}}$ = y

Squaring both sides, we get:

⇒ $10^{b} = y^2$

Again squaring both sides, we get:

⇒ $10^{2b} = y^4$

Multiplying both sides by 10,

⇒ $10^{2b} \times 10 = y^4 \times 10$

⇒ $10^{2b + 1} = 10y^4$

Hence, the value of $10^{2b + 1} = 10y^4$.

(iii) Given: log₁₀ P = 3a - 2b

⇒ P = $10^{3a - 2b}$

⇒ P = $10^{3a} \div 10^{2b}$

⇒ P = $(\sqrt{x})^3 \div y^4$

⇒ P = $(x)^{\dfrac{3}{2}} \div y^4$

⇒ P = $\dfrac{x^{\dfrac{3}{2}}}{y^4}$

Hence, the value of P = $\dfrac{x^{\dfrac{3}{2}}}{y^4}$.