I need this equation to be solved for $x$. I know that I have to get rid of this natural logarithm in the exponent but I don't know how, nothing seems to work.
Is it possible to find $x$? Could you give me some hints? Is there something I'm missing out on?
$$203=412\,x^{1,265-0,0954\ln x}$$
As suggested in the comments by Donald Splutterwit, noting also that $\log$ function is injective, we have that
$$203=412\,x^{1,265-0,0954\ln x} \iff \ln (203)=\ln \left(412\,x^{1,265-0,0954\ln x}\right)$$
then refer to the algebraic foundamental properties of logarithm
$\log (AB) = \log A+\log B$
$\log (A^B) = B\log A$
and solve the new expression obtained for $t=\ln x$ with the condition $x>0$.