how do I solve x in
$\displaystyle 4+7 \ln e^{x^2} = 2 e^{4 \ln x}$
Use these identities: $\displaystyle e^{\ln x}=\ln e^x=x$
Then, $\displaystyle \ln e^{x^2}=x^2$ and $\displaystyle e^{4\ln x}=e^{\ln x^4}=x^4$
The equation becomes $\displaystyle 4+7x^2=2x^4, \ x>0$
Let $\displaystyle x^2=t, \ t>0\Rightarrow 2t^2-7t-4=0\Rightarrow t_1=4, \ t_2=-\frac{1}{2}$
Now replace t with 4: $\displaystyle x^2=4\Rightarrow x=2$