Question

The value of $\frac{2\log 2+\log 3}{\log 48-\log 4}$ is

• A. $1$
• B. $2$
• C. $3$
• D. $-1$

Answer 1

The correct option is A $1$

$\frac{2\log 2+\log 3}{\log 48-\log 4}=\frac{\log 2+\log 2+\log 3}{\log 48-\log 4}=\frac{\log (2\times 2\times 3)}{\log \left(\frac{48}{4}\right)}=\frac{\log 12}{\log 12}=1$