Video narration
Did taking logarithms keep every real case, or should you check the original equation? Set a equal to x plus one. For positive a, taking ln, the natural logarithm, gives: a times ln of a equals minus ln of a over a. So ln of a times the quantity a plus one over a equals zero. That second factor cannot be zero, so a equals one and x equals zero. But logarithms require positive a. The negative case was never checked. At a equals minus one, both exponents are minus one, so each side of the original equation equals minus one. Thus x equals minus two works. Could another negative a work? If both powers are real, compare magnitudes. Let r be the magnitude of a. Equality requires r to the minus r to equal r to the one over r, forcing r to equal one. So a equals minus one is the only negative case. No: the real solutions are x equals zero and x equals minus two.