The value of 2 log 3 - $\dfrac{1}{3}$ log 64 + log 12 is :

1. log 27

2. 27

3. -27

4. -log 27

Given,

$\Rightarrow \text{2 log 3} - \dfrac{1}{3} \text{log 64 + log 12} \\[1em] \Rightarrow \text{log 3}^2 - \text{log 64}^{\dfrac{1}{3}} + \text{log 12} \\[1em] \Rightarrow \text{log 3}^2 - (\text{log 4}^3)^{\dfrac{1}{3}} + \text{log 12} \\[1em] \Rightarrow \text{log 9} - (\text{log 4})^{3 \times \dfrac{1}{3}} + \text{log 12} \\[1em] \Rightarrow \text{log 9 - log 4 + log 12} \\[1em] \Rightarrow \text{log 9 + log 12 - log 4} \\[1em] \Rightarrow \text{log } \dfrac{9 \times 12}{4} \\[1em] \Rightarrow \text{log } \dfrac{108}{4} \\[1em] \Rightarrow \text{log } 27.$

Hence, Option 1 is the correct option.