Given log₁₀x = 2a and log₁₀y = $\dfrac{b}{2}$,

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

(ii) write 10^{2b + 1} in terms of y.

(iii) if log₁₀P = 3a - 3b, express P in terms of x and y.

(i) Given,

log₁₀x = 2a

⇒ x = 10^2a

⇒ x = (10^a)²

⇒ 10^a = $\sqrt{x}$.

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

(ii) Given,

log₁₀y = $\dfrac{b}{2}$

$\Rightarrow y = 10^{\dfrac{b}{2}} \\[1em] \Rightarrow y = (10^b)^{\dfrac{1}{2}} \\[1em]$

Squaring both sides we get,

⇒ y² = 10^b

Simplifying 10^{2b + 1} we get,

10^{2b + 1} = 10^2b.10

= (10^b)².10

= (y²)².10

= 10y⁴.

Hence, 10^{2b + 1} = 10y⁴.

(iii) Given,

log₁₀P = 3a - 2b

⇒ P = 10^{3a - 2b}

⇒ P = 10^3a.10^-2b

= (10^a)³.(10^b)^-2

$= (\sqrt{x})^3 \times (y^2)^{-2} \\[1em] = (x^3)^{\dfrac{1}{2}} \times \dfrac{1}{(y^2)^2} \\[1em] = \dfrac{x^{\dfrac{3}{2}}}{y^4}$

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