# Math Help - Solving an equation with logarithms in it.

1. ## Solving an equation with logarithms in it.

how do I solve x in

$4+7 \ln e^{x^2} = 2 e^{4 \ln x}$

2. Use these identities: $e^{\ln x}=\ln e^x=x$

Then, $\ln e^{x^2}=x^2$ and $e^{4\ln x}=e^{\ln x^4}=x^4$

The equation becomes $4+7x^2=2x^4, \ x>0$

Let $x^2=t, \ t>0\Rightarrow 2t^2-7t-4=0\Rightarrow t_1=4, \ t_2=-\frac{1}{2}$

Now replace t with 4: $x^2=4\Rightarrow x=2$