Question Number 133893 by mathlove last updated on 25/Feb/21
$$\:\:{if}\:\:\:\:\:\:\:{f}\left({x}\right)=\frac{\mathrm{10}^{{x}} −\mathrm{10}^{−{x}} }{\mathrm{10}^{{x}} +\mathrm{10}^{−{x}} }\:\:\:\:\:\:{then}\:\:{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$
Answered by rs4089 last updated on 25/Feb/21
$${y}=\frac{\mathrm{10}^{{x}} −\mathrm{10}^{−{x}} }{\mathrm{10}^{{x}} +\mathrm{10}^{−{x}} }\:\:\Rightarrow\:\frac{\mathrm{1}+{y}}{\mathrm{1}−{y}}=\frac{\mathrm{10}^{{x}} }{\mathrm{10}^{−{x}} } \\ $$$$\mathrm{10}^{\mathrm{2}{x}} =\frac{\mathrm{1}+{y}}{\mathrm{1}−{y}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}{log}_{\mathrm{10}} \left(\frac{\mathrm{1}+{y}}{\mathrm{1}−{y}}\right) \\ $$$${x}\rightarrow{f}^{−\mathrm{1}} \left({x}\right)\:{and}\:\:{y}\rightarrow{x} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{log}_{\mathrm{10}} \left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$
Answered by TheSupreme last updated on 25/Feb/21
$$\mathrm{10}^{{x}} ={e}^{{ln}\left(\mathrm{10}\right){x}} \\ $$$${f}\left({x}\right)=\frac{\frac{{e}^{{ln}\left(\mathrm{10}\right){x}} −{e}^{{ln}\left(\mathrm{10}\right){x}} }{\mathrm{2}}}{\frac{{e}^{{ln}\left(\mathrm{10}\right){x}} +{e}^{{ln}\left(\mathrm{10}\right){x}} }{\mathrm{2}}}=\frac{{sinh}\left({ln}\left(\mathrm{10}\right){x}\right)}{{cosh}\left({ln}\left(\mathrm{10}\right){x}\right)}={tanh}\left({ln}\left(\mathrm{10}\right){x}\right) \\ $$$${ln}\left(\mathrm{10}\right){x}={tanh}^{−\mathrm{1}} \left({y}\right) \\ $$$${x}=\frac{{tanh}^{−\mathrm{1}} \left({y}\right)}{{ln}\left(\mathrm{10}\right)} \\ $$