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
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