Question

Find the value of ${\log }_2\frac{75}{16}-2{\log }_2\frac{5}{9}+{\log }_2\frac{32}{243}$

Answer 1

${\log }_2\frac{75}{16}-2{\log }_2\frac{5}{9}+{\log }_2\frac{32}{243}={\log }_2\left(\frac{75,32,81}{16,243,25}\right)={\log }_{2}2=1$