Question

The value of $x$ that satisfies the expression
${\log }_{2}x\cdot {\log }_{3}x\cdot {\log }_{5}x={\log }_{2}x\cdot {\log }_{3}x+{\log }_{3}x{\log }_{5}x+{\log }_{5}x{\log }_{2}x$, where $x\ne 1$ is

Answer 1

We mus have, $x>0$ to define log
Let ${\log }_{2}x=A,{\log }_{3}x=B,{\log }_{5}x=C$
$\therefore ABC=AB+BC+CA$
$\Rightarrow \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=1$
$\Rightarrow {\log }_{x}2+{\log }_{x}3+{\log }_{x}5=1,[\because {\log }_{b}a=\frac{1}{{\log }_{a}b}]$
$\Rightarrow {\log }_{x}30=1,[\because \log a+\log b+\log c=\log (abc\left)\right]$
$\therefore x=30,[\because {\log }_{a}a=1]$
Ans: 30