Question

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

Answer 1

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

$\log \frac{75}{16}=\log 75-\log 16(\because \log \frac{a}{b}=\log a-\log b;\log ab=\log a+\log b;\log a^n=n\log a)$

$=\log \left(5^2\times 3\right)-4\log 2$

$=2\log 5+\log 3-4\log 2$ ........................(i)

$2\log \frac{5}{9}=2\log 5-2\log 9$

$=2\log 5-4log3\to 2$

$\log \frac{32}{243}=\log 32-\log 243$

$=5\log 2-5\log 3\to 3$

From $1,2,3$

$A=(2\log 5+\log 3-4\log 2)-(2\log 5-4\log 3)+(5\log 2-5\log 3)$

$=\log 2$

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