finding all possible values of $a,b,c$ when $f(x)=\frac{\ln^3 x - c \ln x + 30}{(\ln^2 x - b \ln x + 6)(\ln x + a)}$ Given the function $$f(x)=\frac{\ln^3 x - c \ln x + 30}{(\ln^2 x - b \ln x + 6)(\ln x + a)},  \quad (a,b,c)\in\mathbb{R},$$
and vertical asymptote at $x=e^2$, and removable discontinuity at $x=e^3$.
Need to find possible values of $a,b,c$.
I think $a$ has two options but i'm not sure how to get them.
I tried long Division but didn't work for me.
thanks.
 A: If you assume $\ln x$=y and transforming everywhere $x=e^y$ and frame the question again,it will be easy to solve.Your question will become Given the function $$f(e^y)=\frac{y^3 - c y + 30}{(y^2 - by + 6)(y + a)},  \quad (a,b,c)\in\mathbb{R},$$
and horizontal asymptote at y=2, and removable discontinuity at $y=3$.
Need to find possible values of $a,b,c$.
For y=2 function $\to$ $\infty$ and considering removable discontinuity at $y=3$ $\implies$$y-3$ is a factor for numerator and denominator both.:-$\mathbf Case I$: $(y+a)=(y-3)$ in case of removable discontinuity at $y=3$,then a=-3 and $(y-2)$ is root of $(y^2 - by + 6)=(y-2)(y-3) \implies b=5$ But this case is not possible as there will be  $(x-3)^2$ and will lead to discontinuity at $x=3$ as there is only one $x=3$ factor in numerator to cancel out as no value of $c$ satisfies it.$\mathbf Case II$ Let $y-3$ be a factor of $ y^2 - by + 6 \implies b=5$ accordingly as in $\mathbf Case I$.Now for limit to exist at $y-3$ and $y-2$, both terms should be factor of numerator.On solving numerator as $(y-2)(y-3)(y-k)=y^3 - c y + 30 \implies k=-5$ and $c=+19$ and $a$ can assume any value except $a=-3$. $\mathbf SOLUTION$  a $\in$ $R$-{-3}, $b=5$,$c=19$
