Official solution
f(x) = \[{{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}\] = y
\[ \therefore \] \[{{y + 1} \over {y - 1}} = {{{{2.8}^{2x}}} \over { - {{2.8}^{ - 2x}}}}\]
\[ \Rightarrow \] \[{{1 + y} \over {1 - y}}\] = 84x
\[ \Rightarrow \] \[{\log _e}\left( {{{1 + y} \over {1 - y}}} \right)\] = 4x \[{\log _e}8\]
\[ \Rightarrow \] x = \[{1 \over {4{{\log }_e}8}}{\log _e}\left( {{{1 + y} \over {1 - y}}} \right)\]
\[ \therefore \] f-1(x) = \[{1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 + x} \over {1 - x}}} \right)\]