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Math Help - Logarithm

  1. #1
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    Logarithm

    For which value of x does the following inequality hold?
    \log_{x^2-3}\left(4x+2\right)\ge1
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  2. #2
    Moo
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    Hello,

    If there is no mistake in the subject, i'll write this :

    \log_{x} y = \frac{\ln(y)}{ln(x)}

    So you have to solve : \frac{\ln(4x+2)}{\ln(x^2 -3)} \geq 1

    Differentiate the case when ln(x-3) is positive (ie x-3 > 1) and the case when ln(x-3) is negative (ie x-3 < 1).
    Multiply the two sides with ln(x-3), change the sign when it's negative.


    Then, you may have ln(4x+2) > (or <) ln(x-3), which is ln(4x+2) - ln(x-3) > (or <) 0=ln(1)

    <=> \ln(\frac{4x+2}{x^2-3}) \geq ln(1) (or <)

    Ln is an increasing function, so a<b <=> ln(a)<ln(b)
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  3. #3
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    Hello, james_bond!

    This one is trickier than I thought . . .


    Solve for x\!:\;\;\log_{x^2-3}(4x+2)\:\ge\:1

    We know that: . \log_b(N) \:\geq \:1 means: N \:\geq \:b\quad\hdots .
    (Think about it!)

    So we have: . 4x + 2 \:\geq \:x^2-3 \quad\Rightarrow\quad 5 + 4x - x^2 \:\geq \:0

    This is a down-opening parabola which is positive between its x-intercepts.
    Its x-intercepts are: (-1,0) and (5,0).
    . . Hence, the inequality holds on the interval: . (-1,\,5)


    But the base of a logarithm must be positive.
    . . So we have: . x^2-3 \:>\:0\quad\Rightarrow\quad |x| \:>\: \sqrt{3}

    Moreover, the base of a logarithm cannot be 1.
    . . So we have: . x^2-3 \:\neq \:1\quad\Rightarrow\quad x \neq \pm 2


    Therefore: . (\sqrt{3},\,2) \:\cup \:(2,\,5)

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  4. #4
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    It is indeed tricky.
    1.8 \in \left( {\sqrt 3 ,2} \right)\quad  \Rightarrow \quad \log _{1.8} \left( {4(1.8) + 2} \right) =  - 1.555
    Recall that we need x^2  - 3 > 1
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  5. #5
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    Hi Soroban!

    I can't see a mistake in your solution but Wolfram Mathematica says that only (2,5) is the solution. Am I wrong?

    EDIT: Yeah I didn't recognize Plato's post... Sorry.

    I now think that the problem might be with the 0<x^2-3<1 case...
    Attached Thumbnails Attached Thumbnails Logarithm-logarithm.png  
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