Question
The inverse of $$f\left( x \right) = \frac{2}{3}\frac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}}$$ is :
A.
$$\frac{1}{3}\,{\log _{10}}\frac{{1 + x}}{{1 - x}}$$
B.
$$\frac{1}{2}\,{\log _{10}}\frac{{2 + 3x}}{{2 - 3x}}$$
C.
$$\frac{1}{3}\,{\log _{10}}\frac{{2 + 3x}}{{2 - 3x}}$$
D.
$$\frac{1}{6}\,{\log _{10}}\frac{{2 - 3x}}{{2 + 3x}}$$
Answer :
$$\frac{1}{2}\,{\log _{10}}\frac{{2 + 3x}}{{2 - 3x}}$$
Solution :
$$\eqalign{
& {\text{If }}y = \frac{2}{3}\frac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}},\,\,\,{10^{2x}} = \frac{{3y + 2}}{{2 - 3y}} \cr
& {\text{or }}x = \frac{1}{2}\,{\log _{10}}\frac{{2 + 3y}}{{2 - 3y}} \cr
& \therefore \,{f^{ - 1}}\left( x \right) = \frac{1}{2}\,{\log _{10}}\frac{{2 + 3x}}{{2 - 3x}}\, \cr} $$