Be Positive!

$\begin{aligned} (\frac{-2}{5})^{-3}&=(\frac{2}{5})^{2x-1}\\ -1\times(\frac{2}{5})^{-3}&=(\frac{2}{5})^{2x-1}\end{aligned}$

taking Log base $\dfrac{2}{5}$ on both sides

$-3+\log_{\tfrac{2}{5}}(-1)=2x-1$

we know that $e^{i\pi}=-1$ or $\log_{e}(-1)=i\pi$ (considering principal root)

Thus the expression can be simplified to

$-3+\dfrac{i\pi}{\ln(\dfrac{2}{5})}=2x-1$ (changing base to e)

or

$\begin{aligned} x &=-1+\dfrac{i\pi}{2 \ln(\dfrac{2}{5})}\\ x&=-1-1.7143i \hspace{5mm}(approx)\end{aligned}$

$|x|=\sqrt{(-1)^2+(-1.7143)^2}=1.9846$