I have no clue how to do this...
\(\log16x^2+2\log\dfrac{1}{x}=\log(4x)^2+2\log\dfrac{1}{x}=2\log4x+2\log\dfrac{1}{x}\\\\=2\left(\log4x+\log\dfrac{1}{x}\right)=2\log\left(4x\cdot\dfrac{1}{x}\right)=2\log4=\log4^2=\boxed{\log16}\)
