solve (e^x+3)/(4e^-x+5)=2

 (e^x+3)/(4e^-x+5)=2

e^x + 3   =  2 (4e^(-x) + 5)

e^x + 3  = 8e^(-x) + 10      

e^x - 8e^(-x) - 7  = 0       multiply through by  e^x

e^(2x) - 7e^x - 8  = 0       factor

(e^x + 1)  (e^x - 8)  = 0           setting both factors to  0

e^x + 1   = 0    has no real solutions

e^x - 8 = 0     add 8 to both sides

e^x = 8         take the ln of both sides

ln e ^x  = ln 8

x * ln e   = ln 8       [ ln e = 1, so we can disregard this ]

x = ln 8 

I am going to use another way to solve this equation.

aftenn provided a good start, Substitute x = ln a.......

$\dfrac{e^x+3}{4e^{-x}+5}=2\\ e^x+3=\dfrac{8}{e^x}+10\\ e^{\ln a}+3=\dfrac{8}{e^{\ln a}}+10\\ a+3 = \dfrac{8}{a}+10\\ a= \dfrac{8}{a}+7\\ a^2-7a-8=0\\ (a+1)(a-8)=0\\ (e^x+1)(e^x-8)=0\\\text{Set each factor to 0}\\ e^x=-1 \text{ have no real solutions}\\\text{PS: }x=\pi i \text{ is a solution but it's not real}\\ e^x=8\\ x = \ln 8$

If you want imaginary solutions too x = pi(i) or x = ln 8