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Math Help - Finding asymptotes:

  1. #1
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    Cool Finding asymptotes:

    Given:

    f(x) = log(base 4)(x-4) / 3 - 2log(base 4)x

    I think whats throwing me off is the logs. I have no problem finding asymptotes for a rational function with simple variables. But this is confusing me.
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  2. #2
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    Assuming \displaystyle f(x) = \frac{\log_4(x-4)}{3-2\log_4x}, x\neq 8 \cap (-\infty,5)

    Sketch the graph using a table of values or technology for a better understanding.
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  3. #3
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    Hello,

    I am wondering how to solve for the horizontal asymptote

    I found the vertical asymptote, but am stuck for the horizontal =(
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  4. #4
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    Hello, FallenStar117!

    f(x) \:=\: \dfrac{\log_4(x-4)}{3 - 2\log_4x}

    Vertical asymptotes occur where the denominator is zero.

    . . 3-2\log_4x \:=\:0 \quad\Rightarrow\quad \log_4x \:=\:\frac{3}{2} \quad\Rightarrow\quad x \:=\:4^{\frac{3}{2}}

    Therefore, there is a vertical asymptote at x\,=\,8.



    I suggest that you re-write the function . . .
    The denominator is: . 3 - 2\log_4x \;=\;\log_4(64) - \log_4(x^2) \;=\;\log_4\left(\dfrac{64}{x^2}\right)
    The function becomes: . f(x) \;=\;\dfrac{\log_4(x-4)}{\log_4\left(\frac{64}{x^2}\right)}

    Using the Base-change Formula, we can convert to any base.

    So we have: . f(x) \;=\;\dfrac{\ln(x-4)}{\ln(\frac{64}{x^2})}

    \displaystyle \text{Now determine: }\;\lim_{x\to\infty}\dfrac{\ln(x-4)}{\ln(\frac{64}{x^2})}

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