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Solution

Convert $(12x)_3$ to decimal: $1\cdot 3^2 + 2\cdot 3 + x = 9 + 6 + x = 15 + x$ Convert $(123)_x$ to decimal: $1\cdot x^2 + 2\cdot x + 3$ Equate: $15 + x = x^2 + 2x + 3$ Bring all to one side: $x^2 + x - 12 = 0$ Factor: $(x + 4)(x - 3) = 0$ So $x = 3$ or $x = -4$ But base $x$ must be > 3 (because digits 1,2,3 appear). Allowed value = 3 (but digit '3' cannot appear in base 3). So no valid base exists.