Given x = log₁₀ 12, y = log₄ 2 × log₁₀ 9 and z = log₁₀ 0.4, find :

(i) x - y - z

(ii) 13^{x - y - z}

Given,

x = log₁₀ 12, y = log₄ 2 × log₁₀ 9 and z = log₁₀ 0.4

(i) Substituting value of x, y and z in equation x - y - z, we get :

⇒ x - y - z = log₁₀ 12 - log₄ 2 × log₁₀ 9 - log₁₀ 0.4

= log₁₀ 12 - log(_2²) 2 × log₁₀ 9 - log₁₀ 0.4

= log₁₀ 12 - $\dfrac{1}{2}$ log ₂ 2 × log₁₀ 9 - log₁₀ 0.4

= log₁₀ 12 - $\dfrac{1}{2}$ × 1 × log₁₀ 9 - log₁₀ 0.4

= log₁₀ 12 - $\dfrac{1}{2}$ × log₁₀ 9 - log₁₀ 0.4

= log₁₀12 - log₁₀ $9^{\dfrac{1}{2}}$ - log₁₀ 0.4

= log₁₀12 - log₁₀3 - log₁₀0.4

= log₁₀12 - (log₁₀3 + log₁₀0.4)

= log₁₀ $\dfrac{12}{3 \times 0.4}$

= log₁₀ $\dfrac{12}{1.2}$

= log₁₀ 10

= 1.

Hence, x - y - z = 1.

(ii) Substituting value of x - y - z in 13^{x - y - z}, we get :

⇒ 13¹ = 13.

Hence, 13^{x - y - z} = 13.