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 log₁₀P = 3a - 2b, express P in terms of x and y.

(i) Given,

⇒ log₁₀ x = 2a

⇒ x = 10^2a

⇒ x = (10^a)²

Taking square root on both sides, we get :

$\Rightarrow \sqrt{x} = 10^a$

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

(ii) Given,

⇒ 10^{2b + 1}

⇒ 10^2b.10¹ ………(1)

Given,

$\Rightarrow \text{log}_{10}y = \dfrac{b}{2} \\[1em] \Rightarrow y = 10^{\dfrac{b}{2}} \\[1em] \Rightarrow y^4 = (10^{\dfrac{b}{2}})^{4} \\[1em] \Rightarrow y^4 = 10^{2b} \text{ ……(2)}$

Substituting value of 10^2b from equation (2) in equation (1), we get :

⇒ y⁴.10¹

⇒ 10y⁴.

Hence, 10^{2b + 1} = 10y⁴.

(iii) Given,

⇒ log₁₀ P = 3a - 2b

⇒ P = 10^{3a - 2b}

⇒ P = 10^3a.10^-2b

⇒ P = $\dfrac{10^{3a}}{10^{2b}}$

We know that: (shown above)

y⁴ = 10^2b

and

$\sqrt{x}$ = 10^a

∴ P = $\dfrac{10^{3a}}{10^{2b}}$ = $\dfrac{(10^{a})^3}{y^4} = \dfrac{(\sqrt{x})^3}{y^4} = \dfrac{x^{\dfrac{3}{2}}}{y^4}$.

Hence, P = $\dfrac{x^{\dfrac{3}{2}}}{y^4}$