<s>[INST] Logarithmic inequality, can't define the scope of $x$ I'm solving and getting answer that $x >1$.
\begin{align*}
\ln(x)-\ln(2-x)&>0\\
\implies \ln\left(\frac{x}{2-x}\right)&>0\\
\implies e^{\ln(x/(2-x))}&>e^0\\
\implies \frac{x}{2-x}&>1\\
\implies x &> 2-x\\
\implies 2x &> 2\\
\implies x &> 1
\end{align*}
But when I assign a value of $e$ to $x$, which is greater than $1$, I get an error, because I get $\ln(2-e)$.
 [/INST] You have $\dfrac x {2-x} > 1.$
You cannot go from there to $x>2-x$ unless you somehow know that $2-x$ is positive.
There are at least two ways to deal with that:


*

*First of course you rule out $x=2$ as a possible solution because it makes the denominator zero. Then consider what happens when $x>2.$ Then $2-x$ is negative, so from $\dfrac x {2-x}>1$ you conclude $x< 2-x,$ so $2x<2$, so $x<1,$ but $x$ cannot be $<1$ if $x>2.$ Therefore there are no solutions greater than $2.$

*From $\dfrac 2 {2-x}>1$ deduce that $\dfrac 2 {2-x} - 1 > 0.$ The common denominator is $2-x$, so you have $$ \frac 2 {2-x} - \frac{2-x}{2-x}>0 $$ $$ \frac{x}{2-x}>0 $$ The fraction $x/(2-x)$ changes signs at $0$ and and $2$. So consider the three intervals $(-\infty,0),$ $(0,2),$ and $(2,\infty).$ You will find that the fraction is positive only if $0<x<2.$
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