Question

If $\log \frac{75}{16}-2\log \frac{5}{9}+\log \frac{32}{243}=\log m$, then value of $m$ is

• A. $6$
• B. $4$
• C. $2$
• D. $8$

Answer 1

Answer: (C)

$2$

$\log \frac{75}{16}-2\log \frac{5}{9}+\log \frac{32}{243}=\log \frac{75}{16}-\log \frac{5}{9}\times \frac{5}{9}+\log \frac{32}{243}=\log (\frac{75}{16}\times \frac{9}{5}\times \frac{9}{5}\times \frac{32}{243})=\log (\frac{3}{16}\times \frac{32}{3})=\log 2$