Question

If $\log \frac{75}{16}-2\log \frac{5}{9}+\log \frac{32}{243}=\log p$. Find the value of ${}^{\prime }{p}^{\prime }$

Answer 1

$\log \frac{75}{16}-2\log \frac{5}{9}+\log \frac{32}{243}=\log P$

$\Rightarrow \log \frac{75}{16}-\log {\left(\frac{5}{9}\right)}^2+\log \frac{32}{243}$

$\Rightarrow \log [\frac{\frac{75}{16}}{\frac{25}{81}}\times \frac{32}{243}]$

$\Rightarrow \log [\frac{75}{16}\times \frac{81}{25}\times \frac{32}{243}]$

$\therefore \log 2=\log P$

$\Rightarrow P=2$