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?]
 A: If $f(x) = m(x)$ and $g(x) = n(x)$ do have a common solution, then that solution is also a solution of $\frac{f(x)}{g(x)} = \frac{m(x)}{n(x)}$. The problem is that it is not guaranteed to be the only solution. An example where this is the case would be:
$$\frac{(x-1)}{2x-1} = \frac{(x-1)}{x(3x-2)}$$
The solution $x=1$ can be obtained using the method you proposed, but there is another solution, namely $x=1/3$.
A: Let $u = \ln x$.
Then,
$$ \frac{u}{(u^{2} + 1)^{2}} = \frac{1}{4}$$
from which it follows 
$$ 4u = u^{2} + 2u + 1 $$
that is,
$$ u^{2} - 2u + 1 = (u - 1)^{2} = 0 $$
whence,
$$ u = 1 ; $$
Therefore, $x = e$ is the only solution. 
A: $$\frac{\ln x}{(1+\ln x)^2}=\frac{1}{4}$$
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
A: $\frac ab = \frac cd$ does not mean $a = c$ and $b = d$.
Here are two examples:
$\frac 13 = \frac 26$ but $1 \ne 2$ and $3\ne 6$.  Likewise if $x =2$ we would have $\frac x{x+4} = \frac 13$. That does not mean $x=1$ and $x+4 =3$.
One (not advised) way of doing this is to notice that $\frac ab = \frac {\beta a}{\beta b}$ for all $\beta$ (excepts $0$) so there must be some $\alpha$ where we have $\frac ab = \frac {\alpha a}{\alpha c} = \frac cd$ and $a = \alpha c$ and $b=\alpha d$.  We just don't have any reason to assume that $\alpha$ is equal to $1$.
So when $\frac 13 = \frac 26$ we have $1 = [\frac 12]*2$ and $3 = [\frac 12]*6$.  And for $\frac x{x+4} = \frac 13$ we have $x = \alpha$ and $x+ 4 = 3\alpha$.  (We can solve this as $\alpha = x$ we have $x +4 = 3x$ so $2x =4$ and $x=2$...)
But the better way is to do:
$\frac ab = \frac cd \implies ad = bc$.  For our examples we get $\frac 13 = \frac 26 \implies 1*6 = 2*3$ which it does and $\frac x{x+4} =\frac 13 \implies 3x = x+4\implies x = 2$.
So in your problem you have:
$\frac{\ln x}{(1+ \ln x)^2} = \frac{1}{4}$ so that means
$4\frac {\ln x} = (1 + \ln x)^2$.  It does not mean $\ln x = 1$ or that $(1+\ln x)^2 = 4$.  It does mean that there is some constant $\alpha$ so taht $\ln x = \alpha$ and $1+\ln x)^2 = \alpha *4$ but .... that's just the same thing with a complicated and unnecessary variable thrown in.
To do this I'd just replace $\ln x$ with $w$ and have $4w = (1 + w)^2$ and solve for $w$.
$4w = 1 + 2w + w^2$
$0 = 1-2w + w^2$
$(w-1)^2 = 0$
$w-1 = 0$
$w = 1$ 
So $\ln x = 1$ so $x = e$. 
