Question

Find square of sum of the roots of the equation
${\log }_{3}x{\log }_{4}x{\log }_{5}x={\log }_{3}x{\log }_{4}x+{\log }_{4}x{\log }_{5}x+{\log }_{5}x{\log }_{3}x$.

Answer 1

Let ${\log }_{2}x=A,{\log }_{3}x=B,{\log }_{5}x=C$
$\Rightarrow ABC=AB+BC+CA$
$=ABC(\frac{1}{A}+\frac{1}{B}+\frac{1}{C})$
$\Rightarrow ABC(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}-1)=0$
$\Rightarrow A=0,B=0,C=0$
$\Rightarrow {\log }_{2}x=0,{\log }_{3}x=0,{\log }_{5}x=0$
$\Rightarrow x=(2^0,3^0,5^0)$
$\therefore x=1$
Again $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}-1=0$
Using base change theorem, we have
$\Rightarrow {\log }_{x}2+{\log }_{x}3+{\log }_{x}5-1=0$
$\Rightarrow {\log }_{x}2+{\log }_{x}3+{\log }_{x}5=1$
By using the product rule of logarithms, we have
${\log }_{x}30=1$
$\therefore x=30$
Hence $x=1,30$
$\therefore$ Roots are $1$ and $30$
Hence, the required value $={(1+30)}^2={31}^2=961$.