#### Answer

$17920$

#### Work Step by Step

We find the term without $x$ in the expansion of $(8x+\frac{1}{2x})^{8}$ by using Binomial Theorem coefficients.
To figure out when this occurs, we need to find out which term has no $x$ (e.g. has $x^0$). We know the general form of the terms is:
$\displaystyle \left(\begin{array}{l}
8\\r\end{array}\right)(8x)^{r}(\frac{1}{2x})^{8-r}=\left(\begin{array}{l}
8\\r\end{array}\right)\frac{8^r}{2^{8-r}}*\frac{x^r}{x^{8-r}}$
We need to have the power of $x$ be zero, so:
$\frac{x^r}{x^{8-r}}=x^0$
$x^{r-(8-r)}=x^{0}$
$x^{2r-8}=x^{0}$
$2r-8=0$
$r=8/2=4$
Thus we find the 4th term:
$\displaystyle \left(\begin{array}{l}
8\\4\end{array}\right)(8x)^{4}(\frac{1}{2x})^{4}=70\frac{8^{4}x^4}{2^{4}x^4}=70*256=17920$