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Math Help - More SAT help

  1. #1
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    More SAT help

    Hello! I bought an SAT practice book to get ready for the big test. It is pretty well formated, except it only has the answers/explanations to random questions. I was just hoping that you guys could help me with a question here or there that I didn't know how to solve. Thanks!


    Solve for x, where 0<x<pi: 6e^x * tan(x) - 3tan(x) + 8e^x -4 = 0



    This is what i did:


    factor: 3 tan(x) [2ex - 1] + 4 [2ex - 1] = 0

    factor further: .[2ex - 1] [3 tan(x) + 4] = 0


    know what? I'm not so familiar with the e's (are they called natural logs).

    Thanks, for your help!
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  2. #2
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    Quote Originally Posted by Mr_Green View Post

    Solve for x, where 0<x<pi: 6e^x * tan(x) - 3tan(x) + 8e^x -4 = 0

    This is what i did:

    factor: 3 tan(x) [2ex - 1] + 4 [2ex - 1] = 0

    factor further: .[2ex - 1] [3 tan(x) + 4] = 0


    know what? I'm not so familiar with the e's (are they called natural logs).

    Thanks, for your help!
    First, I like to mention. Questions on SAT do not EVEN come close to the difficulty you are posting. I think you accidently brought yourself a book called AIME (that is a math competition) and thought it said SAT. But anywats if you can do these tough problems you will kill the SAT math section.

    You did an excellent job factoring.
    Now you set each one equal to zero,
    2e^x-1=0
    e^x=1/2
    Now, by definition of logarithms,
    x=\log_e 1/2 = \ln 1/2=\ln 2^{-1}=-\ln 2.
    (Note the meaning of \log_e is the same as [tex]\ln ).

    For the second part you have,
    3\tan x+4=0
    \tan x = -3/4
    Now you solve it through arc-tangent.
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  3. #3
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    wat about the part of the problem that said, "solve for x, where 0 < x< pi"

    doesnt that have some effect on the answer?

    like the only answer the graphing calc provides in that range is about 2.214
    Last edited by Mr_Green; January 28th 2007 at 11:02 AM.
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  4. #4
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    what does factoring all that down help us if its not in the range?
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  5. #5
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    Quote Originally Posted by Mr_Green View Post
    what does factoring all that down help us if its not in the range?
    Factoring makes us express the equation in factors.
    Now that the original equation is in multiplication, we can set any of the factors to zero and solve for the unknown in that factor.

    The domain 0 < x < pi is for the factor that is in trigonometric function.
    The Problem wants only the angle x that is anywhere between 0 and pi radians.
    x = arctan(-4/3) -----***
    Now there are two such x's in the domain, or you called "range", 0<x<2pi. One in the 2nd Quadrant, and another in the 4th Quadrant. The Problem wants only that one in the 2nd quadrant.

    Using a simple scientific calculator,
    x = arctan(-4/3) = -0.9273 radians
    That's the one in the 4th Quadrant, measured clockwise.
    In the 2nd Quadrant, and measured conterclockwise,
    x = pi -0.9273 = 3.1416 -0.9273 = 2.2143 radians ----------same as what you're getting in your graphing calc in that range or region.
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