For what value of $x$ does \[\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?\]
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 $\fbox{256}$.