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Question Number 196683 by MATHEMATICSAM last updated on 29/Aug/23

$$\mathrm{If}y=\frac{e^x-e^{-x}}{e^x+e^{-x}}\mathrm{then}\mathrm{show}\mathrm{that}$$$$x=\frac{1}{2}{\log }_e\left(\frac{1+y}{1-y}\right).$$

Answered by Frix last updated on 29/Aug/23

$$y=\frac{{\mathrm{e}}^{2x}-1}{{\mathrm{e}}^{2x}+1}$$$${\mathrm{e}}^{2x}=\frac{1+y}{1-y}$$$$2x=\ln \frac{1+y}{1-y}$$$$x=\frac{1}{2}\ln \frac{1+y}{1-y}$$

Answered by som(math1967) last updated on 29/Aug/23

$$\frac{1}{y}=\frac{e^x+\frac{1}{e^x}}{e^x-\frac{1}{e^x}}$$$$\Rightarrow \frac{1+y}{1-y}=\frac{2e^x}{\frac{2}{e^x}}$$$$\Rightarrow \frac{1+y}{1-y}=e^{2x}$$$$\Rightarrow {log}_e\left(\frac{1+y}{1-y}\right)={log}_e{e}^{2x}$$$$\therefore 2x={log}_e\left(\frac{1+y}{1-y}\right)$$$$x=\frac{1}{2}{log}_e\left(\frac{1+y}{1-y}\right)$$

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