Express the following as a single logarithm:

$\dfrac{1}{2}$log 9 + 2log 3 - log 6 + log 2 - 2

Given,

$\Rightarrow \dfrac{1}{2}\text{log 9 + 2log 3 - log 6 + log 2 - 2} \\[1em] \Rightarrow \dfrac{1}{2}\text{log 3}^2 + \text{log 3}^2 - \text{log 6 + log 2 - 2log 10} \\[1em] \Rightarrow \dfrac{1}{2} \times 2 \times \text{log 3 + log 9 - log 6 + log 2 - log 10}^2 \\[1em] \Rightarrow \text{log 3 + log 9 + log 2 - log 6 - log 100} \\[1em] \Rightarrow \text{log }\dfrac{3 \times 9 \times 2}{6 \times 100} \\[1em] \Rightarrow \text{log } \dfrac{9}{100}.$

Hence, $\dfrac{1}{2}\text{log 9 + 2log 3 - log 6 + log 2 - 2} = \text{log }\dfrac{9}{100}$.