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Correct answer: no prime solution
Setting each log equal to k gives k²=1, so k=-1 (k=1 is impossible). This yields (x+1)(x-1/2)=1, giving x² + x/2 - 3/2 = 0, i.e., 2x² + x - 3 = 0... wait let me redo: (x+1)(x-0.5)=1 => x²+0.5x-0.5=1 => x²+0.5x-1.5=0 => 2x²+x-3=0 => (2x+3)(x-1)=0, giving x=1 or x=-3/2. x=1 fails domain (base x-1/2=1/2, check: base must ≠1, ok; but x+1=2 ok). Actually x=1: log₂(0.5)=log_(0.5)(2) => -1=-1, valid! But x=1 is not prime, not composite, and rational. Let me recheck: domain requires x > 1/2 and x ≠ 1 for log_(x+1) and x-1/2 ≠ 1 i.e. x ≠ 3/2. So x=1 is excluded (base x+1=2, argument x-1/2=1/2 ok, but we need x-1/2 as base: x-1/2=1/2 ≠ 1, ok and x+1=2>0 ok). Actually x=1 is valid in domain. Hmm, but then solution is rational x=1, not irrational. Re-examine: the other root x=-3/2 fails domain. So the only solution is x=1 (rational, not prime, not composite). Answer: no prime solution AND no composite solution. But checking options — 'an irrational solution' seems wrong. Actually 1 is neither prime nor composite, so 'no prime solution' and 'no composite solution' are both true. The question likely intends one correct answer; 'no prime solution' is technically true.