Question

Find the value of ${\log }_{2}3+{\log }_{2}3+{\log }_{2}4-{\log }_{2}6$

Answer 1

${\log }_{2}3+{\log }_{2}3+{\log }_2{\left(2\right)}^2-{\log }_2\left(2\times 3\right)$

$={\log }_{2}3+{\log }_{2}3+2{\log }_{2}2-{\log }_{2}2-{\log }_{2}3$ (since $\log a^n=n\log a$ and $\log ab=\log a+\log b$)

$={\log }_{2}3+2-1$ (since ${\log }_{a}a=1$)

$=1+{\log }_{2}3$