Question

$2\log 3-\frac{1}{2}\log 16+\log 12$

Answer 1

Given,

$2\log 3-\frac{1}{2}\log 16+\log 12$

$=\log 3^2-\frac{1}{2}\log 4^2+\log 12$

$=\log 9-\frac{2}{2}\log 4+\log 12$

$=\log 9-\log 4+\log 12$

$=\log \left(\frac{12\times 9}{4}\right)$

$=\log 27$.