Question

The value of $\frac{1}{2}\log 9+2\log 6+\frac{1}{4}\log 81-\log 12$ in terms of $\log 3$ is

• A. $3\log 3$
• B. $4\log 3$
• C. $2\log 3$
• D. $\log 3$

Answer 1

The correct option is A $3\log 3$

$\frac{1}{2}\log 9+2\log 6+\frac{1}{4}\log 81-\log 12$

$=\log 9^{\frac{1}{2}}+\log 6^2+\log {81}^{\frac{1}{4}}-\log 12$

$=\log 3+\log 36+\log 3-\log 12$

$=\log \left(\frac{3\times 36\times 3}{12}\right)=\log 27=\log 3^3=3\log 3$