for the first one:
lnx = x-1,
x = e^(x-1) -->taking base e of both sides
ex = e^x -->multiply by e on both sides
by inspection, x = 1.
to prove that there are no further solutions:
d/dx{lnx} = 1/x. therefore, for x>1, the gradient of lnx is less than 1.
d/dx{x-1} = 1, which is always 1...