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Math Help - An logarithmic problem and an trigonometrical problem

  1. #1
    Junior Member
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    An logarithmic problem and an trigonometrical problem

    Hello

    I'd like an check on my answers for these two equations:

    1 2ln(x) = ln(2 - x) + ln(1)

    2 cos(-x) = sin (-x) | -2\pi \leq x \leq 2\pi
    Note: We're working in radians here

    ----

    Ok the first one:

    2ln(x) = ln(2 - x) + ln(1)

    ln(x^2) = ln(2 - x) + ln(1)

    ln(x^2) = ln(2 - x)

    x^2 = 2 - x

    x^2 +x - 2 = 0

    x = - \frac{1}{2} \pm \sqrt{(\frac{1}{2})^2 + 2 }

    x = - \frac{1}{2} \pm \sqrt{\frac{9}{4}}

    x = - \frac{1}{2} \pm \frac{3}{2}

    x_{1} = \frac{2}{2}

    x_{2} = -\frac{4}{2}

    A check of these answers seems to give good results.

    ------
    2

    cos(-x) = sin (-x)

    1 = \frac{sin (-x)}{cos(-x)}

    1 = - \frac{sin (x)}{cos(x)}

    1 = -tan(x)

    -1 = tan(x)

    x = tan^{-1} (-1)

    x = - \frac{\pi}{4}

    x = - \frac{\pi}{4} + n \cdot \pi

    Thanks for replies

    Edit: Is the period exresssion for 2 correct? I'm not too familiar with radians.
    Last edited by λιεҗąиđŗ; March 1st 2008 at 06:30 AM.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, λιεҗąиđŗ!

    A few minor things . . .


    I'd like an check on my answers for these two equations:

    1)\;\;2\ln(x) \:= \:\ln(2 - x) + \ln(1)

    My work: . 2\ln(x) \:= \:\ln(2 - x) + \ln(1) \quad\Rightarrow\quad \ln(x^2) \:= \:\ln(2 - x) + 0

    . . \ln(x^2) \:= \:\ln(2 - x) \quad\Rightarrow\quad x^2 \:= \:2 - x \quad\Rightarrow\quad x^2 +x - 2 \:= \:0 . . . . Right!
    Factor: . (x-1)(x+2) \:=\:0

    Hence: . x \:=\:1,-2

    But x = -2 is extraneous.

    Solution: . x \:= \:1



    2)\;\;\cos(-x) \:= \sin(-x),\quad -2\pi \leq x \leq 2\pi

    My work: . \cos(-x) \:= \:\sin (-x)\quad\Rightarrow\quad 1 \:= \:\frac{\sin (-x)}{\cos(-x)} \quad\Rightarrow\quad 1 \:=\: -\frac{\sin (x)}{\cos(x)}

    . . 1 \:= \:-\tan(x) \quad\Rightarrow\quad -1 \:=\: \tan(x)\quad\Rightarrow\quad x \:= \:\tan^{-1}(-1) . . . . Correct!
    x \;=\;-\frac{5\pi}{4},\;-\frac{\pi}{4},\;\frac{3\pi}{4},\;\frac{7\pi}{4} . . . . on the interval they gave us.

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