#### Answer

$0$

#### Work Step by Step

First, use the fact that $\log a^b= b \log a$ to simplify the expression to $$2 \ln x^{-2} + \ln x^4$$ $$\ln (x^{-2})^2+\ln x^4$$ Since $(a^m)^n=a^{mn}$, $$=\ln (x^{-4})+\ln(x^4)$$ Finally, using the fact that $\ln(a)+\ln(b)=\ln(ab)$ the expression simplifies to $\ln(x^{-4}(x^4))=\ln(x^4/x^4)=\ln(1)=\ln e^0=0$