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Math Help - 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:
    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; January 12th 2014 at 12:04 PM.
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  2. #2
    MHF Contributor
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    Re: Resolving a logarithmic equation

    \log_a b=c means a^c = b. There is not much of a trick. Just take 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?

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

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

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

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

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

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

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

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

    cos(x)\leq  1/2

    OR

    (cos(x)+1)\leq  0

    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:

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

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

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

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

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

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

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

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

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

    All values of x are negative


    OR


    -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; January 12th 2014 at 02:05 PM.
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  4. #4
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    Re: Resolving a logarithmic equation

    You have:

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

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

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