1. ## 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.

2. 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.

3. Hello,

I am wondering how to solve for the horizontal asymptote

I found the vertical asymptote, but am stuck for the horizontal =(

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

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