Solve for x. (e^x-e^-x)/((e^x+e^-x)=t

2 answers:

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$\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=t\\\\\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=(e^x-e^{-x}):(e^x+e^{-x})=\left(e^x-\dfrac{1}{e^x}\right):\left(e^x+\dfrac{1}{e^x}\right)\\\\=\left(\dfrac{(e^x)^2}{e^x}-\dfrac{1}{e^x}\right):\left(\dfrac{(e^x)^2}{e^x}+\dfrac{1}{e^x}\right)=\dfrac{e^{2x}-1}{e^x}:\dfrac{e^{2x}+1}{e^x}\\\\=\dfrac{e^{2x}-1}{e^x}\cdot\dfrac{e^x}{e^{2x}+1}=\dfrac{e^{2x}-1}{e^{2x}+1}\\\\\text{substitute}\ e^{2x}=s$

$\text{therefore}\\\\\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=t\to\dfrac{s-1}{s+1}=t\ \ \ \ |\cdot(s+1)\\\\s-1=t(s+1)\\\\s-1=ts+t\ \ \ \ |+1\\\\s=ts+t+1\ \ \ |-ts\\\\s-ts=t+1\\\\s(1-t)=t+1\ \ \ \ |:(1-t)\neq0\\\\s=\dfrac{t+1}{1-t}\to e^{2x}=\dfrac{t+1}{1-t}\ \ \ \ |\ln\\\\\ln e^{2x}=\ln\left(\dfrac{t+1}{1-t}\right)\\\\2x\ln e=\ln\left(\dfrac{t+1}{1-t}\right)\\\\2x=\ln\left(\dfrac{t+1}{1-t}\right)\ \ \ \ |:2\\\\\boxed{x=\dfrac{1}{2}\ln\left(\dfrac{t+1}{1-t}\right)}\\\\\text{The domain:}\\\\t\in(-1;\ 1)$