Question

If $l=\log \frac{a^2}{bc}$, $m=\log \frac{b^2}{ca}$ and $n=\log \frac{c^2}{ab}$, find the value of $l+m+n$

• A. $0$
• B. $1$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A $0$

$l+m+n=\log \frac{a^2}{bc}+\log \frac{b^2}{ca}+\log \frac{c^2}{ab}$

$\therefore l+m+n=\log (\frac{a^2}{bc}\times \frac{b^2}{ca}\times \frac{c^2}{ab})$

$\therefore l+m+n=\log 1=0$