Solving a Fractional Equation Involving a Logarithm

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 $x=1/3$.

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.

$$\frac{\ln x}{(1+\ln x)^2}=\frac{1}{4}$$ 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