∵y=lg[x+√(x²-1)] ==>x+√(x²-1)=10^y.(1)
==>[x-√(x²-1)]/{[x+√(x²-1)][x-√(x²-1)]}=10^y
==>1/[x-√(x²-1)]=10^y
==>x-√(x²-1)=10^(-y).(2)
∴由(1)式+(2)得 2x=10^y+10^(-y) ==>x=[10^y+10^(-y)]/2
故所求反函数是 x=[10^y+10^(-y)]/2.