Question

If $5{\log }_{abc}(a^3+b^3+c^3)=3\lambda \left(\frac{1+{\log }_3\left(abc\right)}{{\log }_{3}abc}\right)$ and ${\left(abc\right)}^{a+b+c}=1$ and $\lambda =\frac{m}{n}$, where $m$ and $n$ are relatively prime, then the value of $|m+n|+|m-n|=$

• A. $8$
• B. $10$
• C. $12$
• D. $14$

Answer 1

The correct option is B $10$

$5{\log }_{abc}{a}^3+b^3+c^3=3\lambda \left[\frac{1+{\log }_{3}abc}{{\log }_{3}abc}\right]$

$5{\log }_{abc}{a}^3+b^3+c^3=3\lambda [1+\frac{1}{{\log }_{3}abc}]$

$5{\log }_{abc}{a}^3+b^3+c^3=3\lambda [{\log }_{abc}\left(abc\right)+{\log }_{abc}\left(3\right)]$

$5{\log }_{abc}{a}^3+b^3+c^3=3\lambda {\log }_{abc}\left(3abc\right)$

${\log }_{abc}{(a^3+b^3+c^3)}^5={\log }_{abc}{\left(3abc\right)}^{3\lambda }$

${(a^3+b^3+c^3)}^5={\left(3abc\right)}^{3\lambda }$(equation $1$)

Given, ${\left(abc\right)}^{a+b+c}=1$

Taking log both side.

$(a+b+c)\log \left(abc\right)=\log \left(1\right)$

$(a+b+c)\log \left(abc\right)=0$

$\log \left(abc\right)\ne 0\therefore (a+b+c)=0$

$\Rightarrow (a+b+c{)}^3=0$

$\Rightarrow a^3+b^3+c^3=3abc$(equation$2$)

From equation $1$ and $2$

$(3abc{)}^5=(3abc{)}^{3\lambda }$

$5=3\lambda$

$\lambda =\frac{5}{3}=\frac{m}{n}$

$|m+n|+|m-n|=|5+3|+|5-3|=8+2=10$