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Question Number 133893 by mathlove last updated on 25/Feb/21

$$iff\left(x\right)=\frac{{10}^x-{10}^{-x}}{{10}^x+{10}^{-x}}then{f}^{-1}\left(x\right)=?$$

Answered by rs4089 last updated on 25/Feb/21

$$y=\frac{{10}^x-{10}^{-x}}{{10}^x+{10}^{-x}}\Rightarrow \frac{1+y}{1-y}=\frac{{10}^x}{{10}^{-x}}$$$${10}^{2x}=\frac{1+y}{1-y}$$$$x=\frac{1}{2}{log}_{10}\left(\frac{1+y}{1-y}\right)$$$$x\to f^{-1}\left(x\right)andy\to x$$$$f^{-1}\left(x\right)=\frac{1}{2}{log}_{10}\left(\frac{1+x}{1-x}\right)$$

Answered by TheSupreme last updated on 25/Feb/21

$${10}^x=e^{ln\left(10\right)x}$$$$f\left(x\right)=\frac{\frac{e^{ln\left(10\right)x}-e^{ln\left(10\right)x}}{2}}{\frac{e^{ln\left(10\right)x}+e^{ln\left(10\right)x}}{2}}=\frac{sinh\left(ln\left(10\right)x\right)}{cosh\left(ln\left(10\right)x\right)}=tanh\left(ln\left(10\right)x\right)$$$$ln\left(10\right)x={tanh}^{-1}\left(y\right)$$$$x=\frac{{tanh}^{-1}\left(y\right)}{ln\left(10\right)}$$

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