Question

If $7{\log }_{abc}\left(a^3+b^3+c^3\right)=3k\left(\frac{1+{\log }_3(abc)}{{\log }_3(abc)}\right)$ and $(abc{)}^{a+b+c}=1$ and $k=\frac{m}{n}$ in lowest form, where $m,n\in N$ then find the value $(m+n)$.

Answer 1

Answer: 0010

$a+b+c=0$
$\Rightarrow a^3+b^3+c^3=3abc$
$\text{ L.H.S. }=7{\log }_{abc}(3abc)$
$\text{ R.H.S. }=3k\frac{{\log }_3(3abc)}{{\log }_3(abc)}$
$=3k{\log }_{abc}(3abc)$
So $7{\log }_{abc}(3abc)=3k{\log }_{abc}(3abc)$
$\therefore k=\frac{7}{3}\equiv \frac{m}{n}$
$\therefore (m+n)=7+3=10$