Results 1 to 8 of 8

Thread: Still trigonometry

  1. #1
    Newbie
    Joined
    Dec 2012
    From
    at Home
    Posts
    9

    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:

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

    equals:

    $\displaystyle \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:

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

    I believe I have to solve the $\displaystyle 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:

    $\displaystyle (\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; Dec 12th 2012 at 05:11 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,023
    Thanks
    758

    Re: Still trigonometry

    You have made an error in your substitution in the first problem. Your first term should be $\displaystyle 2y^2$. Do you see why?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2012
    From
    at Home
    Posts
    9

    Re: Still trigonometry

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

    What about the other two?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,023
    Thanks
    758

    Re: Still trigonometry

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

    For the third, I would rewrite the equation as:

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

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

    Can you proceed from here?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2012
    From
    at Home
    Posts
    9

    Re: Still trigonometry

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

    For the third, I would rewrite the equation as:

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

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

    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 $\displaystyle \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2$ but I can't see how you got there.
    If the right answer is -2, then I got it right.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,023
    Thanks
    758

    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 $\displaystyle \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2$ but I can't see how you got there.
    If the right answer is -2, then I got it right.
    $\displaystyle \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2=\frac{2}{3}\cdot\frac{9}{16}=\frac{3}{8}$

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

    $\displaystyle \left(\frac{1}{4} \right)^x=16=\left(\frac{1}{4} \right)^{-2}$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2012
    From
    at Home
    Posts
    9

    Re: Still trigonometry

    Thanks. I'm doing logorythms now. Can you help me with this: What's X if:
    X = $\displaystyle X = 5^{2 + 2log_{5}4} $
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,023
    Thanks
    758

    Re: Still trigonometry

    I would write:

    $\displaystyle X=5^{2+2\log_5(4)}=5^2\cdot5^{\log_5(4^2)}=?$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometry to Memorize, and Trigonometry to Derive
    Posted in the Trigonometry Forum
    Replies: 12
    Last Post: Feb 23rd 2017, 09:35 AM
  2. Trigonometry help plz!!!
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Nov 10th 2009, 10:39 AM
  3. trigonometry
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 25th 2009, 02:36 AM
  4. Trigonometry
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Oct 15th 2009, 12:19 AM
  5. Trigonometry
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Aug 31st 2008, 09:12 AM

Search Tags


/mathhelpforum @mathhelpforum