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Thread: An logarithmic problem and an trigonometrical problem

  1. #1
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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 $\displaystyle 2ln(x) = ln(2 - x) + ln(1)$

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

    ----

    Ok the first one:

    $\displaystyle 2ln(x) = ln(2 - x) + ln(1)$

    $\displaystyle ln(x^2) = ln(2 - x) + ln(1)$

    $\displaystyle ln(x^2) = ln(2 - x)$

    $\displaystyle x^2 = 2 - x$

    $\displaystyle x^2 +x - 2 = 0$

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

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

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

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

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

    A check of these answers seems to give good results.

    ------
    2

    $\displaystyle cos(-x) = sin (-x)$

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

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

    $\displaystyle 1 = -tan(x)$

    $\displaystyle -1 = tan(x)$

    $\displaystyle x = tan^{-1} (-1)$

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

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

    Thanks for replies

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

    A few minor things . . .


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

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

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

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

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

    But $\displaystyle x = -2$ is extraneous.

    Solution: .$\displaystyle x \:= \:1$



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

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

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

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