precalulus

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Ok feeling kinda dumb because this isn't coming out right. I know it's simple.
log6 (x -3) + log6(x -6) - log6(x + 1) = 2

  • precalulus -

    adding logs is multiplying arguments

    log 6 [(x-3)(x-6)(x+1)] = 2

    6^log 6 [(x-3)(x-6)(x+1)] = 6^2 = 36

    x^3 - 8 x^2 + 9 x + 18 = 36

    x^3 - 8 x^2 + 9 x - 18 = 0

    I do not see any easy factors. Have to solve by iteration.

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