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Pre-Cal help

  1. Jan 20, 2005 #1
    This should be rather easy for the rest of you .. but somehow I can't remember what to do.

    [ (x-6) (x+7) ] / (x-2) is greater than or equal to 0



    log(base three)x + log(base three)(x-6) = 3

    ty in advance =)
  2. jcsd
  3. Jan 20, 2005 #2
    what have you done so far?

    [tex] \frac{(x-6)(x+7)}{x-2} \geq 0 [/tex]
    what are the critical numbers( make numerator 0 or denominator 0)

    Then form your test intervals and you get the answer.

    for [tex] \log_3x + \log_3(x-6) = 3 [/tex] use log rules

    hint: [tex] \log_bxy = log_bx + log_by [/tex]
    Last edited: Jan 20, 2005
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