What would be the set of all solutions satisfying this logarithmic equation? The question is as follows:-
Find the set of all solutions for the equation:
$$\log_3x \log_4x\log_5x=\log_3x \log_4x+\log_4x \log_5x+\log_3x \log_5x$$
(I)  $\{1\}$
(II)  $\{1,60\}$
(III) $\{1,5,10,60\}$
(IV)  None
I can see that $x=1$ would be a solution to this equation, (right?), because then LHS = RHS = $0$, so option I is probable. But my book attempts the question as follows:-
Dividing throughout by $\log_3x \log_4x\log_5x$ (thus assuming $\log_3x \log_4x\log_5x \neq0$, and thus $x\neq1$ ), we get
$$1=\frac{1}{\log_5x}+\frac{1}{\log_4x}+\frac{1}{\log_3x}$$
$$\implies \log_x3+\log_x4+\log_x5=1$$
$$\implies \log_x60 =1 $$
$$\implies x=60$$
And thus they say that the answer is IV, ie, None of the above.
But my question is, why is it even being assumed that $x\neq1$? How can we assume that if $x=1$ does satisfy it?? Why are we dividing by $\log_3x \log_4x\log_5x$ in the first place if it can be equal to zero? If we do not assume that , $x=1$ would be a solution, but here they assume that $x\neq1$and find $x=60$ to be a solution. 
Can someone explain to me why division by $\log_3x \log_4x\log_5x$ assuming $x\neq1$ is being done here?

write your equation in the form
$$\frac{(\ln(x))^3}{\ln(3)\ln(4)\ln(5)}=\frac{(\ln(x))^2}{\ln(3)\ln(4)}+\frac{(\ln(x))^2}{\ln(4)\ln(5)}+\frac{(\ln(x))^2}{\ln(3)\ln(5)}$$
