smc

Using the fact that ${\log }_{x^n}{y}^n={\log }_{x}y$, we see that the equation becomes ${\log }_{2}x+{\log }_{2}x+{\log }_{2}x+{\log }_{2}x+{\log }_{2}x=40$. Thus, $5{\log }_{2}x=40$ and ${\log }_{2}x=8$, so $x=2^8=256$, or $\boxed{256}$.