If log 4 = 0.602 and log 27 = 1.431; find :

(i) log 8

(ii) log 12

Given, log 4 = 0.602

⇒ log $(2^2)$ = 0.602

⇒ 2log 2 = 0.602

⇒ log 2 = $\dfrac{0.602}{2}$

⇒ log 2 = 0.301

And, log 27 = 1.431

⇒ log $(3^3)$ = 1.431

⇒ 3log 3 = 1.431

⇒ log 3 = $\dfrac{1.431}{3}$

⇒ log 3 = 0.477

(i) log 8

= log $(2^3)$

= 3 x log 2

= 3 x 0.301 (∵ Using log 2 = 0.301)

= 0.903

Hence, the value of log 8 = 0.903.

(ii) log 12

= log $(4 \times 3)$

= log 4 + log 3

= 0.602 + 0.477 (∵ Using log 4 = 0.602 and log 3 = 0.477)

= 1.079

Hence, the value of log 12 = 1.079.