Results 1 to 5 of 5

Math Help - Please help

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    6

    Please help

    Hello. I'm having trouble with the 2 last calculus exercises witch I just can't solve. If someone would please tell me how to solve these two:

    1) 1-sinx -cosx +sin2x+cos2x=0

    2)log(2 below) (4^(x+1)+4) * log(2below) (4^(x)+1)=log(1/square root of 2 below) (square root of 1/8)

    Thanks to everyone who can help me!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Sep 2008
    Posts
    6
    Does anyone have any ideas or tips how to solve these two, I'm really running out of time
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    769
    Hello, MadMan!

    2)\;\;\log_2\left(4^{x+1}+4\right)\cdot \log_2\left(4^x+1\right) \;=\;\log_{\frac{1}{\sqrt{2}}}\left(\sqrt{\frac{1}  {8}}\right)

    The right side is: . \log_{\frac{1}{\sqrt{2}}}\left(\sqrt{\frac{1}{8}}\  right) \:=\:p \quad\Rightarrow\quad \left(\frac{1}{\sqrt{2}}\right)^p \:=\:\frac{1}{\sqrt{8}} \quad\Rightarrow\quad p \:=\:3

    The left side is: . \log_2\bigg[4(4^x+1)\bigg]\cdot\log_2(4^x+1) \;=\;\bigg[\log_2(4) + \log_2(4^x + 1)\bigg]\cdot\log_2(4^x+1)

    . . \;=\bigg[2 + \log_2(4^x+1)\bigg]\cdot\log_2(4^x+1) \;=\;2\cdot\log_2(4^x+1) + \bigg[\log_2(4^x + 1)\bigg]^2


    The equation becomes: . \bigg[\log_2(4^x+1)\bigg]^2 + 2\!\cdot\!\bigg[\log_2(4^x+1)\bigg]\:=\:3
    . . . . \bigg[\log_2(4^x+1)\bigg]^2 + 2\bigg[\log_2(4^x+1)\bigg] - 3 \;=\;0

    Factor: . \bigg[\log_2(4^x+1) +3\bigg]\cdot\bigg[\log_2(4^x+1) -1\bigg] \;=\;0


    (1)\;\;\log_2(4^x+1) + 3 \:=\:0\quad\Rightarrow\quad \log_2(4^x+1) \:=\:-3 \quad\Rightarrow\quad  4^x + 1 \:=\:2^{-3}

    . . . . 4^x + 1 \:=\:\frac{1}{8} \quad\Rightarrow\quad 4^x \:=\:-\frac{7}{8} . . . no real roots


    (2)\;\;\log_2(4^x+1) - 1 \:=\:0 \quad\Rightarrow\quad \log_2(4^x+1) \:=\:1 \quad\Rightarrow\quad 4^x + 1 \:=\:2^1

    . . . . 4^x \:=\:1 \quad\Rightarrow\quad\boxed{ x \:=\:0}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    769
    Hello again, MadMan!

    I'll assume the answers are on the interval [ 0,\:2\pi)


    1)\;\;1-\sin x -\cos x +\sin2x+\cos2x\:=\:0
    Double-angle identities: . \begin{array}{ccc}\sin2A &=& 2\sin A\cos A \\ \cos2A &=& 2\cos^2\!A - 1 \end{array}


    The equation becomes: . 1 - \sin x - \cos x + 2\sin x\cos x + 2\cos^2\!x - 1 \;=\;0

    . . and we have: . 2\sin x\cos x + 2\cos^2\!x - \sin x - \cos x \;=\;0


    Factor: . 2\cos x(\sin x + \cos x) - (\sin x + \cos x) \;=\;0

    Factor: . (\sin x + \cos x)(2\cos x - 1) \;=\;0


    (1)\;\;\sin x + \cos x \:=\:0 \quad\Rightarrow\quad \sin x \:=\:-\cos x \quad\Rightarrow\quad \frac{\sin x}{\cos x} \:=\:-1

    . . . . \tan x \:=\:-1 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{3\pi}{4},\:\frac{7\pi}{4}}


    (2)\;\;2\cos x - 1 \:=\:0\quad\Rightarrow\quad \cos x \:=\:\frac{1}{2} \quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{3},\:\frac{5\pi}{3}}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2008
    Posts
    6
    Soroban, thank you so much, you just made my day!! After being absent from school because of an illness I fell behind and I absolutely could not figure those two out by myself, so A really big thanks to you for helping me out!!
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum