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Math Help - Factoring to reduce Complex Fractions. Exponents laws.

  1. #1
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    Factoring to reduce Complex Fractions. Exponents laws.

    It's a bit odd...how I am in Math 3 and still have trouble with factoring! (Please don't tell my math teacher...she won't take it too kindly...)

    "Express in simplest form"
    \frac{a^2-9}{a^2-3a} \bullet \frac{a^2+a}{a+3}
    Work done so far: (Unsure about accuracy...)
    \frac{(a+3)(a-3)}{(a+1)(a-3a)}\bullet\frac{(a+1)(a+a)}{(a+3)} = \frac{(a-3)}{(a-3a)}\bullet\frac{(a+a)}{1}
    "Solve for x: 2^x+2 = 4^x-1"
    Don't even know how to begin this one...set it equal to 0?
    Last edited by MrOats; May 22nd 2008 at 08:14 PM.
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  2. #2
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    Difference of squares; Undistribute.

    "Express in simplest form"


    ----------------------------------------
    Difference of squares, and undistribution:
    [(a+3)(a-3)(a)(a+1)] / [(a)(a-3)(a+3)]
    Cancellation:
    a+1 is what remains, I think.

    Take a look, I will return soon.
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  3. #3
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    "Solve for x: "
    Don't even know how to begin this one...set it equal to 0

    x log 2 + log 2 = x log 4 - log 1
    or this, if you like:

    2^x - 4^x = -3
    x log 2 - x log 4 = log (-3)
    x (log 2 - log 4) = log (-3)
    Take it from there on your scientific calculator.

    I shall return soon...
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  4. #4
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    No more left...

    I thought this mess was another problem but I see now that it is part of a solution:
    ------------------------------------------------
    Work done so far: (Unsure about accuracy...)
    \underline{(a+3)(a-3)}(a+1)(a-3a) \bullet \underline{a+1)(a+a)}(a+3) = \underline{(a-3)}(a-3a) \bullet \underline{(a+a)}1
    ------------------------------------------------
    I don't use LaTex or whatever this mess started out as.

    What is Math 3?
    Looks a lot like what was called Algebra 2 where I went to school.

    Bye.
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  5. #5
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    Thanks for your help Bradley! Yes, I am using LaTex.

    Math 3 is an integrated math course focusing on Algebra II, Geometry, and some Trig. (Basically)...

    Any further help is appreciated!
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by MrOats View Post
    It's a bit odd...how I am in Math 3 and still have trouble with factoring! (Please don't tell my math teacher...she won't take it too kindly...)

    "Express in simplest form"
    \frac{a^2-9}{a^2-3a} \bullet \frac{a^2+a}{a+3}
    Work done so far: (Unsure about accuracy...)
    \frac{(a+3)(a-3)}{(a+1)(a-3a)}\bullet\frac{(a+1)(a+a)}{(a+3)} = \frac{(a-3)}{(a-3a)}\bullet\frac{(a+a)}{1}
    "Solve for x: 2^x+2 = 4^x-1"
    Don't even know how to begin this one...set it equal to 0?
    \frac{a^2-9}{a^2-3a}\cdot\frac{a^2+a}{a+3} \implies \frac{(a+3)(a-3)}{a(a-3)}\cdot\frac{a(a+1)}{a+3}\implies a+1

    For the second question, I will think outside the box:

    Since 4=2^2, then

    2^x+2=2^{2x}-1\implies 2^{2x}-2^x-3=0

    Making a substitution u=2^x, we get the quadratic:

    u^2-u-3=0 \implies u=\frac{1\pm\sqrt{1+12}}{2}.

    Thus,

    2^x=\frac{1\pm\sqrt{1+12}}{2}. We ignore the negative answer, because 2^x can NEVER be negative.

    Thus, 2^x=\frac{1+\sqrt{13}}{2} \implies x=\frac{1+\sqrt{13}}{2\log 2}

    Hope this helped you out!!
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Bradley View Post
    "Solve for x: "
    Don't even know how to begin this one...set it equal to 0

    x log 2 + log 2 = x log 4 - log 1
    or this, if you like:

    2^x - 4^x = -3
    x log 2 - x log 4 = {\color{red}log (-3)}
    x (log 2 - log 4) = {\color{red}log (-3)}
    Take it from there on your scientific calculator.

    I shall return soon...
    Can't take the log of something negative!!
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  8. #8
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    THANK YOU, Chris, for the informative and extensive, detailed help!
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  9. #9
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    \frac{a^2-9}{a^2-3a}\cdot\frac{a^2+a}{a+3} \implies \frac{(a+3)(a-3)}{a(a-3)}\cdot\frac{a(a+1)}{a+3}\implies a+1

    For the second question, I will think outside the box:

    Since 4=2^2, then

    2^x+2=2^{2x}-1\implies 2^{2x}-2^x-3=0

    Making a substitution u=2^x, we get the quadratic:

    u^2-u-3=0 \implies u=\frac{1\pm\sqrt{1+12}}{2}.

    Thus,

    2^x=\frac{1\pm\sqrt{1+12}}{2}. We ignore the negative answer, because 2^x can NEVER be negative.

    Thus, 2^x=\frac{1+\sqrt{13}}{2} \implies x=\frac{1+\sqrt{13}}{2\log 2}

    Hope this helped you out!!
    I caught a typo on my part!

    2^x=\frac{1+\sqrt{13}}{2}\implies x\log2=\log\left(\frac{1+\sqrt{13}}{2}\right)\impl  ies x=\frac{\log\left(\frac{1+\sqrt{13}}{2}\right)}{\l  og2}\approx 1.20337

    Now its the right answer...sorry!
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  10. #10
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    Quote Originally Posted by Chris
    Now its the right answer...sorry!
    Hey, even your wrong answer made a LOT more sense than the crap that I wrote down!

    In all seriousness, THANK YOU very much... =)
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