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Thread: Resolving a logarithmic equation

  1. #1
    DCB
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    Resolving a logarithmic equation

    Hey everyone,

    So I've been given the following equation:
    Resolving a logarithmic equation-gif.gif
    and the following interval:
    Resolving a logarithmic equation-gif-1-.gif
    and I have been asked to solve it for a=3 and a=1/3

    I assume to do so I must start by using the log(ab)=log(a)+log(b) identity to make it:
    $\displaystyle log_{a}((2cos(x)+1)(cos(x)))\geq 0$
    but after that I am lost (it is safe to say I suck at logs). Could someone please help me out?

    Much appreciated.
    Last edited by DCB; Jan 12th 2014 at 12:04 PM.
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  2. #2
    MHF Contributor
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    Re: Resolving a logarithmic equation

    $\displaystyle \log_a b=c$ means $\displaystyle a^c = b$. There is not much of a trick. Just take $\displaystyle a$ to the power of each side.
    Thanks from DCB
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  3. #3
    DCB
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    Re: Resolving a logarithmic equation

    Thanks for the help!
    That seems way simpler than I thought this problem would be!
    Could you just tell me if I am doing it correctly?

    $\displaystyle log_{3}((2cos(x))(cos(x)))\geq 0$

    $\displaystyle log_{3}(2cos^{2}(x)+cos(x))\geq 0$

    $\displaystyle 3^{(2cos^{2}(x)+cos(x))}\geq 0$

    $\displaystyle 3^{0}\geq (2cos^{2}(x)+cos(x))$

    $\displaystyle 1\geq (2cos^{2}(x)+cos(x))$

    $\displaystyle 2cos^{2}(x)+cos(x)-1\leq 0$

    $\displaystyle (2cos(x)-1)(cos(x)+1)\leq 0$

    $\displaystyle (2cos(x)-1)\leq 0$

    $\displaystyle cos(x)\leq 1/2$

    OR

    $\displaystyle (cos(x)+1)\leq 0$

    $\displaystyle cos(x)\leq -1$



    And wouldn't that mean that a could equal 3, 1/3, or anything else, and the answer would still be the same?

    thanks for the help

    Spoiler:
    EDIT:
    Actually major brain fart on the line where I apply 3^0. Better redo this.

    EDIT 2:
    Redid the problem. Hope it is right this time:

    $\displaystyle log_{3}((2cos(x))(cos(x)))\geq 0$

    $\displaystyle log_{3}(2cos^{2}(x)+cos(x))\geq 0$

    $\displaystyle 3^{(2cos^{2}(x)+cos(x))}\geq 0$

    $\displaystyle 3\geq (2cos^{2}(x)+cos(x))$

    $\displaystyle 0\geq 2cos^{2}(x)+cos(x)-3$

    $\displaystyle 0\geq (2cos(x)-3)(cos(x)+1)$

    ------------------------------------

    $\displaystyle 0\geq (2cos(x)-13)$

    $\displaystyle \frac{3}{2}\geq cos(x)$

    All values of x are negative


    OR


    $\displaystyle -1\geq cos(x)$

    All values of x are positive except x = pi


    And since a positive x negative equals it will always be negative (or zero) and therefore all values of x in the given interval resolve the equation


    EDIT 3:
    Nope, it turns out I was right the first time. Spoilering my whole little... "episode" up there. Well this problem was a roller coaster.
    Last edited by DCB; Jan 12th 2014 at 02:05 PM.
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  4. #4
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    Re: Resolving a logarithmic equation

    You have:

    $\displaystyle \log_a(2\cos^2 x + \cos x)\ge 0$

    $\displaystyle \dfrac{\ln(2\cos^2 x + \cos x)}{\ln a} \ge 0$

    When $\displaystyle a>1, \ln a >0$, so multiplying both sides by $\displaystyle \ln a$ does not change the inequality. When $\displaystyle a < 1, \ln a < 0$, so multiplying both sides by $\displaystyle \ln a$ flips the inequality (since you are multiplying by a negative number). That is the only difference between when $\displaystyle a=3$ or when $\displaystyle a = \dfrac{1}{3}$
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