Question

The set of all the solutions of the equation
${\log }_{5}x{\log }_{6}x{\log }_{7}x={\log }_{5}x.{\log }_{6}x+{\log }_{6}x.{\log }_{7}x+{\log }_{7}x.{\log }_{5}x$

• A. $\left\{0.1\right\}$
• B. $\{1,210\}$
• C. $\{1,5,6,7,210\}$
• D. $\{1,2\}$

Answer 1

Answer: (B)

$\{1,210\}$

Given ${\log }_{5}x{\log }_{6}x{\log }_{7}x={\log }_{5}x.{\log }_{6}x+{\log }_{6}x.{\log }_{7}x+{\log }_{7}x.{\log }_{5}x$

$1=\frac{{\log }_{5}x.{\log }_{6}x+{\log }_{6}x.{\log }_{7}x+{\log }_{7}x.{\log }_{5}x}{{\log }_{5}x{\log }_{6}x{\log }_{7}x}$

$1=\frac{{\log }_{5}x.{\log }_{6}x}{{\log }_{5}x{\log }_{6}x{\log }_{7}x}+\frac{{\log }_{6}x.{\log }_{7}x}{{\log }_{5}x{\log }_{6}x{\log }_{7}x}+\frac{{\log }_{7}x.{\log }_{5}x}{{\log }_{5}x{\log }_{6}x{\log }_{7}x}$

$1=\frac{1}{{\log }_{5}x}+\frac{1}{{\log }_{6}x}+\frac{1}{{\log }_{7}x}$

$1={\log }_{x}5+{\log }_{x}6+{\log }_{x}7$

$1={\log }_x(5\times 6\times 7)$

$1={\log }_x\left(210\right)$

$x^1=210$

$x=210$

$\therefore$ the set of solutions that satisfy the given equation is $\{1,210\}$