Answer
$\ln \left(16\sqrt{\dfrac{x^2+1}{x^2-4x}}\right)$
Work Step by Step
We can write the expression as a single logarithm using the rules:
$\log_b(m)+\log_b(n)=\log_b(m\cdot n)$
and
$\log_b(m)+\log_b(n)=\log_b(\frac{m}{n})$
First, we solve the expression inside the square brackets:
$\dfrac{1}{2}\ln \left(x^2+1\right)-4\ln\left(\dfrac{1}{2}\right)-\dfrac{1}{2}\left[\ln(x-4)+\ln x\right]=$
$\dfrac{1}{2}\ln \left(x^2+1\right)-4\ln\left(\dfrac{1}{2}\right)-\dfrac{1}{2}\left[\ln((x-4)\cdot x)\right]=$
$\dfrac{1}{2}\ln \left(x^2+1\right)-4\ln\left(\dfrac{1}{2}\right)-\dfrac{1}{2}\ln(x^2-4x)$
Now, we use the rule $a\log_bm=\log_bm^a$:
$\ln \left(x^2+1\right)^{\frac{1}{2}}-\ln\left(\dfrac{1}{2}\right)^4-\ln(x^2-4x)^{\frac{1}{2}}=$
$\ln \left((x^2+1)^{\frac{1}{2}}\div\dfrac{1}{2^4}\div(x^2-4x)^{\frac{1}{2}}\right)$
Simplify:
$\ln \left((x^2+1)^{\frac{1}{2}}\cdot16\cdot \dfrac{1}{(x^2-4x)^{\frac{1}{2}}}\right)=$
$\ln \left(16\sqrt{\dfrac{x^2+1}{x^2-4x}}\right)$
