If a = log₁₀x, find the following in terms of a:

(i) x

(ii) log₁₀$\sqrt[5]{x^2}$

(iii) log₁₀3x

(i) Given,

⇒ a = log₁₀x

⇒ x = 10^a.

Hence, x = 10^a.

(ii) Given,

$\Rightarrow \text{log}${10}\sqrt[5]{x^2} \\[1em] \Rightarrow \text{log}{10}(x^2)^{\dfrac{1}{5}} \\[1em] \Rightarrow \dfrac{1}{5}\text{ log}{10}(x^2) \\[1em] \Rightarrow \dfrac{1}{5} \times 2\text{log}{10}x \\[1em] \Rightarrow \dfrac{1}{5} \times 2\text{log}{10}10^a \\[1em] \Rightarrow \dfrac{2a}{5}\text{log}{10}10 \\[1em] \Rightarrow \dfrac{2a}{5} \times 1 \\[1em] \Rightarrow \dfrac{2a}{5}.

Hence, log₁₀$\sqrt[5]{x^2} = \dfrac{2a}{5}$.

(iii) Given,

⇒ log₁₀3x

⇒ log₁₀3 + log₁₀x

⇒ log₁₀3 + log₁₀10^a

⇒ log₁₀3 + alog₁₀10

⇒ log₁₀3 + a.

Hence, log₁₀3x = log₁₀3 + a.