Question

The set of all solutions of the equation
${\log }_{3}x.{\log }_{4}x.{\log }_{5}x={\log }_{3}x.{\log }_{4}x+{\log }_{4}x.{\log }_{5}x+{\log }_{5}x.{\log }_{3}x$ is

• A. $\left\{1\right\}$
• B. $\{1,60\}$
• C. $\{1,5,10,60\}$
• D. None of these

Answer 1

Answer: (B)

$\{1,60\}$
${\log }_{3}x.{\log }_{4}x.{\log }_{5}x={\log }_{3}x.{\log }_{4}x+{\log }_{4}x.{\log }_{5}x+{\log }_{5}x.{\log }_{3}x$ ....(1)
For $x=1$, both parts of the equation (1) vanish.
Therefore, $x=1$ is root of the equation (1).
For $x\ne 1$,
${\log }_{3}x.{\log }_{4}x.{\log }_{5}x={\log }_{3}x.{\log }_{4}x+{\log }_{4}x.{\log }_{5}x+{\log }_{5}x.{\log }_{3}x$
Divide both sides by ${\log }_{3}x.{\log }_{4}x.{\log }_{5}x$
$\Rightarrow 1=\frac{1}{{\log }_{5}x}+\frac{1}{{\log }_{3}x}+\frac{1}{{\log }_{4}x}={\log }_{x}5+{\log }_{x}3+{\log }_{x}4[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
$={\log }_{x}60[\because \log a+\log b+\log c=\log (abc\left)\right]$
$\Rightarrow x=60$
Thus, $x=\{1,60\}$

Ans: