If log₁₀ 8 = 0.90; find the value of :

(i) log₁₀ 4

(ii) log $\sqrt{32}$

(iii) log 0.125

Given,

⇒ log₁₀ 8 = 0.90

⇒ log₁₀ 2³ = 0.90

⇒ 3log₁₀ 2 = 0.90

⇒ log₁₀ 2 = $\dfrac{0.90}{3}$

⇒ log₁₀ 2 = 0.30 ………(1)

(i) Given,

⇒ log₁₀ 4

⇒ log₁₀ 2²

⇒ 2 log₁₀ 2

Substituting value of log₁₀ 2 from equation (1) in above equation, we get :

⇒ 2 × 0.30

⇒ 0.60

Hence, log₁₀ 4 = 0.60

(ii) Given,

$\Rightarrow \text{log } \sqrt{32} \\[1em] \Rightarrow \text{log } 4\sqrt{2} \\[1em] \Rightarrow \text{log } 4 + \text{log } \sqrt{2} \\[1em] \Rightarrow \text{log } 2^2 + \text{log } 2^{\dfrac{1}{2}} \\[1em] \Rightarrow 2 \times \text{log 2} + \dfrac{1}{2} \times \text{log 2}$

Substituting value of log 2 from equation (1), in above equation, we get :

$\Rightarrow 2 \times 0.30 + \dfrac{1}{2} \times 0.30 \\[1em] \Rightarrow 0.60 + 0.15 \\[1em] \Rightarrow 0.75$

Hence, log $\sqrt{32}$ = 0.75

(iii) Given,

$\Rightarrow \text{log } 0.125 \\[1em] \Rightarrow \text{log } \dfrac{125}{1000} \\[1em] \Rightarrow \text{log } \dfrac{1}{8} \\[1em] \Rightarrow \text{log } \dfrac{1}{2^3} \\[1em] \Rightarrow \text{log } 2^{-3} \\[1em] \Rightarrow -3 \times \text{log 2} \\[1em] \Rightarrow -3 \times 0.30 \\[1em] \Rightarrow -0.90$

Hence, log 0.125 = -0.90