help plzzz

roughly speaking find x in terms of y ,

then u get f^-1(y) ,

and hence to get f^-1(x) , replace y be x

here y-1 = \frac{10^{2x}-1}{10^{2x}+1}

(y-1 )(10^{2x}+1)= 10^{2x}-1

(y-2)10^{2x}=-y

10^{2x}=\frac{y}{2-y}

x=\frac{1 }{2} log_{10}(\frac{y}{2-y})=f^{-1}(y)

\Rightarrow f^{-1}(x)= \frac{1 }{2} log_{10}(\frac{x}{2-x})

@qwerty you are bang on target dude
i too solve it in the same manner

its a standard method , isnt it ??
but still i hav my own confusions :P,....hav a look

http://targetiit.com/iit-jee-forum/posts/functionssss-13844.html