How do i solve (e^x+7)/(3e^-x+5)=3 ? The solution should be in the form of: x = ln a
e$${\frac{\left({{\mathtt{e}}}^{{\mathtt{x}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}\right)}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{{\mathtt{\,-\,}}{\mathtt{x}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}} = {\mathtt{3}}$$
+118608 $$\\\frac{(e^x+7)}{(3e^{-x}+5)}=3\\\\
(e^x+7)\div (3e^{-x}+5)=3\\\\
(e^x+7)\div (\frac{3}{e^{x}}+5)=3\\\\
(e^x+7)\div (\frac{3+5e^{x}}{e^{x}})=3\\\\
(e^x+7)\times (\frac{e^{x}}{3+5e^{x}})=3\\\\
(e^x+7)\times e^{x}=3(3+5e^{x})\\\\
(e^x)^2+7e^{x}=9+15e^{x}\\\\
(e^x)^2-8e^{x}-9=0\\\\$$
$$\\let\;y=e^x\\\\
y^2-8y-9=0\\\\
(y-9)(y+1)=0\\\\
y=9\;\;or\;\;y=-1\\\\
e^x=9\;\;or\;\;e^x=-1\\\\
e^x>0\;\;$for all real x so $ \\\\
e^x=9\\\\
x=ln9\\\\
x=ln3^2\\\\
x=2ln3$$
+118608 $$\\\frac{(e^x+7)}{(3e^{-x}+5)}=3\\\\
(e^x+7)\div (3e^{-x}+5)=3\\\\
(e^x+7)\div (\frac{3}{e^{x}}+5)=3\\\\
(e^x+7)\div (\frac{3+5e^{x}}{e^{x}})=3\\\\
(e^x+7)\times (\frac{e^{x}}{3+5e^{x}})=3\\\\
(e^x+7)\times e^{x}=3(3+5e^{x})\\\\
(e^x)^2+7e^{x}=9+15e^{x}\\\\
(e^x)^2-8e^{x}-9=0\\\\$$
$$\\let\;y=e^x\\\\
y^2-8y-9=0\\\\
(y-9)(y+1)=0\\\\
y=9\;\;or\;\;y=-1\\\\
e^x=9\;\;or\;\;e^x=-1\\\\
e^x>0\;\;$for all real x so $ \\\\
e^x=9\\\\
x=ln9\\\\
x=ln3^2\\\\
x=2ln3$$
