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I know how to find vertical asymptote(s) of rational function, but have not idea how to solve this problem.

Find the vertical asymptote(s) for the graph of $y=log_2x$+$log_2(x-1)$.

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Well, we have the function:

$$\text{y}\left(x\right):=\log_2\left(x\right)+\log_2\left(x-1\right)=\frac{\ln\left(x\right)}{\ln\left(2\right)}+\frac{\ln\left(x-1\right)}{\ln\left(2\right)}=$$ $$\frac{\ln\left(x\right)+\ln\left(x-1\right)}{\ln\left(2\right)}=\frac{\ln\left(x\left(x-1\right)\right)}{\ln\left(2\right)}\tag1$$

Now, what happens when:

  1. $$\text{y}\left(0\right)=\lim_{x\to0}\frac{\ln\left(x\left(x-1\right)\right)}{\ln\left(2\right)}=\frac{1}{\ln\left(2\right)}\lim_{x\to0}\ln\left(x\left(x-1\right)\right)\to-\infty\tag2$$
  2. $$\text{y}\left(1\right)=\lim_{x\to1}\frac{\ln\left(x\left(x-1\right)\right)}{\ln\left(2\right)}=\frac{1}{\ln\left(2\right)}\lim_{x\to1}\ln\left(x\left(x-1\right)\right)\to-\infty\tag3$$
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  • $\begingroup$ Note that the original function is defined only for $x > 1$, but the fraction at the end of (1) is defined both for $x < 0$ and $x > 1$. $\endgroup$
    – Uwe
    Mar 1 '17 at 15:44
  • $\begingroup$ $(x(x-1))$ is positive for $x < 0$ and for $x > 1$, so $\ln(x(x-1))$ is defined there. By contrast, $\log_2(x-1)$ is only defined for $x > 1$. The crucial point is that after the transformation, you get a second vertical asymptote at $x \to 0$, but this is outside of the domain of the original function. $\endgroup$
    – Uwe
    Mar 2 '17 at 12:49

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