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Pre-Calculus 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.

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

    and

    solve

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

    ty in advance =)
     
  2. jcsd
  3. Jan 20, 2005 #2
    I tried solving the log equation and got to log(base three)x(x-6) = 3 and then forgot what to do @_@
     
  4. Jan 20, 2005 #3

    learningphysics

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    if
    log(base a) p = q

    then a^q = p

    Can you take it from here?

    Show what you tried for the first problem. I'd rather not just tell you how to do it. Do you have any ideas?
     
    Last edited: Jan 20, 2005
  5. Jan 21, 2005 #4

    Zurtex

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    There are a few ways to solve the 1st problem. Here is the way I like doing it.

    Get it in the form R(x) then inequality then 0, where R(x) is some rational function. You already have it in this form, but remember when you need to get it in to this form you can't multiply or divide by a linear function of x.

    Your critical values are:

    x=-7
    x=2
    x=6

    These are the values on the graph where the function changes sign. So look at less than x=-7, all the 3 factors are negative so the whole function is negative, each time you get to a critical value the sign changes. So for x<7 negative for -7<x<2 positive for 2<x<6 negative and for x>6 negative. Think what you want x to be and it's simple.
     
  6. Jan 21, 2005 #5

    arildno

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    Dearly Missed

    I would like to add to Zurtex' good advice the following graphical procedure:
    1. Draw the real numberline, and mark on it, at the very least, where each of the zeroes of your constituent polynomials are (in your case, your polynomials are all linear)
    2. Directly beneath the drawn numberline, write down the "sign line for (x-6) like this: -----0++++++
    The zero shall lie directly below the "6"-mark on your number line.
    3. Continue to draw beneath your previous sign line the "sign line for (x+7):
    This will look like: --0+++++++, with its zero directly beneath "-7".

    4. You are to do the same with the (x-2) factor, the image you've drawn should look like this:
    ----------------------0+++++++++++++++
    --0+++++++++++++++++++++++++++
    ----------------0+++++++++++++++++++
    Now, you can at each point "multiply the signs", to form the "sign line for your rational function":
    --0+++++++(inf)--0+++++++++++++++
    (Note that as we approach the zero of the denominator, the function will tend to infinity (infinity sign depends on which direction you are approaching the singularity)
     
    Last edited: Jan 21, 2005
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