Question

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

• A. $7$
• B. $9$
• C. $3$
• D. $2$

Answer 1

The correct option is D $2$

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