ln(1/x) + lnx = 0 lnx = -ln(1/x) Replace 1/x by x-1 lnx = -ln(x-1) Use rule lnBA = AlnB lnx = -(-1)lnx lnx = lnx Thus this is an identity for all values of x for which lnx is defined. lnx is defined for all positive x, so the solution set is {x|x > 0} or (0, ¥) Edwin AnlytcPhil@aol.com