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

User Jamie Wong
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2 Answers

3 votes

Answer:

its d on edge ;)

Explanation:

User Pavan P
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4 votes


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



\text{therefore}\\\\(e^x-e^(-x))/(e^x+e^(-x))=t\to(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)\\eq0\\\\s=(t+1)/(1-t)\to e^(2x)=(t+1)/(1-t)\ \ \ \ |\ln\\\\\ln e^(2x)=\ln\left((t+1)/(1-t)\right)\\\\2x\ln e=\ln\left((t+1)/(1-t)\right)\\\\2x=\ln\left((t+1)/(1-t)\right)\ \ \ \ |:2\\\\\boxed{x=(1)/(2)\ln\left((t+1)/(1-t)\right)}\\\\\text{The domain:}\\\\t\in(-1;\ 1)



\text{Used:}\\\\a^(-n)=(1)/(a^n)\\\\(a^n)^m=a^(nm)\\\\\log_ab^n=n\cdot\log_ab\\\\\log_aa=1

User Akshay I
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