How do i solve (e^x+7)/(3e^-x+5)=3 ? The solution should be in the form of: x = ln a

$$\\\frac{(e^x+7)}{(3e^{-x}+5)}=3\\\\ (e^x+7)\div (3e^{-x}+5)=3\\\\ (e^x+7)\div (\frac{3}{e^{x}}+5)=3\\\\ (e^x+7)\div (\frac{3+5e^{x}}{e^{x}})=3\\\\ (e^x+7)\times (\frac{e^{x}}{3+5e^{x}})=3\\\\ (e^x+7)\times e^{x}=3(3+5e^{x})\\\\ (e^x)^2+7e^{x}=9+15e^{x}\\\\ (e^x)^2-8e^{x}-9=0\\\\$$

$$\\let\;y=e^x\\\\ y^2-8y-9=0\\\\ (y-9)(y+1)=0\\\\ y=9\;\;or\;\;y=-1\\\\ e^x=9\;\;or\;\;e^x=-1\\\\ e^x>0\;\;$for all real x so $ \\\\ e^x=9\\\\ x=ln9\\\\ x=ln3^2\\\\ x=2ln3$$