log_8 (x-3) - log_8 (3x+11) = log_8 [1 / (x+1)] I use the quotient law for the left side and eliminate log_8. And then what? My Work: log_8 [(x-3)/(3x+11)] = log_8 [1 / (x+1)] [(x-3)/(3x+11)] = [1 / (x+1)] Answer: 7
(x-3)(x+1) = 3x+11 x^2 - 2x - 3 = 3x + 11 x^2 - 5x -14 = 0 (x-7)(x+2) = 0 ok?
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