g(x) = e^(x-1) + x - 6 Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1, where x<6

0 = e^x-1+ x - 6 e^x-1 = 6-x x-1 = ln (6-x) -> here we have taken the natural log of both sides, but it only shows on one side as the natural log of e is 1.x = ln (6-x) + 1Question taken from Edexcel 2013 C3 past paper, with my own adapted answer.