Question

The value of $\frac{{\log }_{4}81}{{\log }_{4}3}-\frac{{\log }_{2}16}{{\log }_{2}4}$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

$A=\frac{{\log }_{4}81}{{\log }_{4}3}-\frac{{\log }_{2}16}{{\log }_{2}4}$

$={\log }_{3}81-{\log }_{4}16$....................$[{\log }_{a}N=\frac{\log N}{\log a}]$

$={\log }_3{3}^4-{\log }_4{4}^2=4-2=2$.