Question

If $3\log 2-2\log 3$ can be expressed as $\log \frac{8}{k}$. then $k$=?

Answer 1

:Let $x=3\log 2-2\log 3$

$\therefore x=\log 2^3-\log 3^2$ ....($\log a^b=b\log a$)

$\therefore x=\log \frac{8}{9}$ ....($\log \frac{a}{b}=\log a-\log b$)

comparing it with $\log \frac{8}{k}$

$\therefore k=9$