Solution of $x$ for $(e^{-x}) -x = 0$?
This was my attempt to solve it:
$(e^{-x}) - x = 0$
$e^{-x} = x$
$\ln(e^{-x}) = \ln(x)$
$-x\times \ln(e) = \ln(x)$
$-x = \ln(x)$
But I'm stuck, how can I obtain $x$?
There is no exact algebraic solution to this equation. You can rewrite it as $-1=\frac{\ln(x)}{x}$ and then plot the function $f(x)=\frac{\ln(x)}{x}$. It takes the value $-1$ exactly once at approximately $x \sim 0.5$. You could compute a numeric approximation as well but you can't solve it explicitly in a $x= $ some number way.