Question

ln( e^x -1)+ ln( e^x +1) =1

Answer 1

$ln(e^x-1)+ln(e^x+1) = 1 \Rightarrow ln\Big((e^x-1)(e^x+1)\Big) = 1 \Rightarrow \\ \Rightarrow ln\Big((e^x)^2-1^2\Big) = 1 \Rightarrow ln(e^{2x}-1) = 1 \Rightarrow ln(e^{2x}-1) = ln(e) \Rightarrow \\ \\ \Rightarrow e^{2x}-1 = e \Rightarrow e^{2x} = e + 1 \Rightarrow 2x = ln(e+1) \Rightarrow \boxed{x = \dfrac{ln(e+1)}{2} }$

$- la primul rand m-am folosit de formula: ln(a)+ln(b) = ln(a\cdot b) \\ - la al doilea rand m-am folosit de formula: (a-b)(a+b) = a^2-b^2$

Answer 2

ln(e^x-1)(e^x+1)=1=lne
ln(e^2x-1)=lne

e^2x-1=e
e^2x=1+e
2x=ln(1+e)
x= (1/2)*ln(1+e)

verificare ln (e^ln(1+e)-1)=ln(1+e-1)=lne=1