Solving a Fractional Equation Involving a Logarithm I may be being stupid right now, so I've come to Stack to see if this elementary algebra holds up. 
Suppose I have the equation $$\frac{\ln x}{(1+ \ln x)^2} = \frac{1}{4}$$
My chosen way to solve this would be to cross multiply and expand brackets, solve the quadratic and get the value of $x$. 
However, a student I am helping got this by saying $\ln x = 1$  gives $x = \mathrm{e}$ and at $x= \mathrm{e}$, the denominator $(1+ \ln x)^2 = 4$. 
Hence $x= \mathrm{e}$. 
Is this approach always correct or is it just luck here? 
In general if I have $\frac{f(x)}{g(x)} = \frac{m(x)}{n(x)}$, can I solve it by finding the common solutions of $f(x) = m(x)$ and $g(x) = n(x)$? 
[Edit: clearly not because if I have $\frac{x}{x+2} = \frac{1}{x+3}$, then $x= 1$ and $x+2 = x + 3$ don't give you anything..., so why does it work in this case?]

$$\frac{\ln x}{(1+\ln x)^2}=\frac{1}{4}$$
with $u=\ln(x)$ we get:
$$\frac{u}{(1+u)^2}=\frac{1}{4}$$
$$4u=1+2u+u^2$$
$$u^2-2u+1=0\Rightarrow (u-1)^2=0$$
$$\therefore u=1$$
$$x=e^u\Rightarrow x=e,$$
This appear to be the only solution
