1) log3 (x^2-11x+27)=2 2) logx (2x^2+x-2)=3

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1) Since log3(x^2 - 11x + 27) = 2, this can be rewritten in exponential form as 3^2 = x^2 - 11x + 27. Therefore, 9 = x^2 - 11x + 27. Rearranging this equation, we get x^2 - 11x + 18 = 0. This quadratic can be factored as (x - 2)(x - 9) = 0. Therefore, x = 2 or x = 9.

2) Since logx(2x^2 + x - 2) = 3, this can be rewritten in exponential form as x^3 = 2x^2 + x - 2. Rearranging this equation, we get x^3 - 2x^2 - x + 2 = 0. This equation can be factored as (x - 1)(x^2 - x - 2) = 0. Factoring further, we get (x - 1)(x + 1)(x - 2) = 0. Therefore, x = 1, x = -1, or x = 2. But since the base of the logarithm cannot be negative or 1, the only valid solution is x = 2.