Answer
$ln[16(\frac{x^2+1}{x(x-4)})^{\frac{1}{2}}]$
Work Step by Step
$\frac{1}{2}ln(x^2+1)-4ln\frac{1}{2}-\frac{1}{2}[ln(x-4)+lnx]=ln(x^2+1)^{\frac{1}{2}}+4ln2-ln(x(x-4))^{\frac{1}{2}}=ln[(x^2+1)^{\frac{1}{2}}\times2^4\times(\frac{1}{x(x-4)})^{\frac{1}{2}}]=ln[16(\frac{x^2+1}{x(x-4)})^{\frac{1}{2}}]$