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Math Help - Still trigonometry

  1. #1
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    Question Still trigonometry

    22x+1 - 5 * 2x - 12 = 0

    If 2x is y, then it would be like this:

    y2 - 5y - 12 = 0

    Seems simple enough, but:

     \frac{-5 \pm \sqrt{5^{2} - 4 * 1 * -12  }}{2}

    equals:

     \frac{-5 \pm \sqrt{73 }}{2}

    And there is no string square root of 73. Is there a way to get around this, some trick perhaps?


    Next one is:

     5^{x^{2} - 3x + 1} = \frac{1}{5}

    I believe I have to solve the  x^{2} - 3x + 1 part first, and so I tried, but encountered the same problem as above: The square root ends up like a decimal - which shouldn't happen :/

    And neither last nor least, is:

     (\frac{1}{2})^{x} * (\frac{2}{3})^{x+1} * (\frac{3}{4})^{x+2} = 6

    I don't even know how to start this one. I believe that's about it from exponentials, but I have to cover two more topics about trigonometry :O

    Thanks
    Last edited by dirtyharry; December 12th 2012 at 06:11 AM.
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  2. #2
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    Re: Still trigonometry

    You have made an error in your substitution in the first problem. Your first term should be 2y^2. Do you see why?
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    Re: Still trigonometry

    I do now, and it makes sense. idk why I didn't see it before.

    What about the other two?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Still trigonometry

    For the second one, rewrite the right side as 5^{-1}, then equate exponents and solve for x.

    For the third, I would rewrite the equation as:

    \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2\cdot\left(\frac{1}{2} \right)^x\left(\frac{2}{3} \right)^x\left(\frac{3}{4} \right)^x=6

    \left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4} \right)^x=6\cdot\frac{8}{3}

    Can you proceed from here?
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    Re: Still trigonometry

    Quote Originally Posted by MarkFL2 View Post
    For the second one, rewrite the right side as 5^{-1}, then equate exponents and solve for x.

    For the third, I would rewrite the equation as:

    \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2\cdot\left(\frac{1}{2} \right)^x\left(\frac{2}{3} \right)^x\left(\frac{3}{4} \right)^x=6

    \left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4} \right)^x=6\cdot\frac{8}{3}

    Can you proceed from here?
    I believe I can. I'm not familiar with how you got the 8/3. I know it's from \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2 but I can't see how you got there.
    If the right answer is -2, then I got it right.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Still trigonometry

    Quote Originally Posted by dirtyharry View Post
    I believe I can. I'm not familiar with how you got the 8/3. I know it's from \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2 but I can't see how you got there.
    If the right answer is -2, then I got it right.
    \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2=\frac{2}{3}\cdot\frac{9}{16}=\frac{3}{8}

    So, then I multiplied both sides by \frac{8}{3}. And yes, x=-2 is the correct answer, since we find:

    \left(\frac{1}{4} \right)^x=16=\left(\frac{1}{4} \right)^{-2}
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    Re: Still trigonometry

    Thanks. I'm doing logorythms now. Can you help me with this: What's X if:
    X =  X = 5^{2 + 2log_{5}4}
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Still trigonometry

    I would write:

    X=5^{2+2\log_5(4)}=5^2\cdot5^{\log_5(4^2)}=?
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