Gordon Young

2023-02-25

How to find the vertical asymptote of a logarithmic function?

Roy Mcclain

Beginner2023-02-26Added 9 answers

$f\left(x\right)={\mathrm{log}}_{b}\left(\text{argument}\right)$ has vertical aymptotes at $\text{argument}=0$

Example $f\left(x\right)=\mathrm{ln}({x}^{2}-3x-4).$ has asymptotes in the vertical

$x=4$ and $x=-1$

graph{y=ln(x^2-3x-4) [-5.18, 8.87, -4.09, 2.934]}

Example $f\left(x\right)=\mathrm{ln}\left(\frac{1}{x}\right)$ has vertical asymptote

$x=0$

graph{ln(1/x) [-5.18, 8.87, -4.09, 2.934]}

Example $f\left(x\right)=\mathrm{ln}\left(\frac{1}{{x}^{2}}\right)$ alsohas vertical asymptote

$x=0$

graph{ln(1/x^2) [-5.18, 8.87, -4.09, 2.934]}

Example $f\left(x\right)=\mathrm{ln}({x}^{2}-3x-4).$ has asymptotes in the vertical

$x=4$ and $x=-1$

graph{y=ln(x^2-3x-4) [-5.18, 8.87, -4.09, 2.934]}

Example $f\left(x\right)=\mathrm{ln}\left(\frac{1}{x}\right)$ has vertical asymptote

$x=0$

graph{ln(1/x) [-5.18, 8.87, -4.09, 2.934]}

Example $f\left(x\right)=\mathrm{ln}\left(\frac{1}{{x}^{2}}\right)$ alsohas vertical asymptote

$x=0$

graph{ln(1/x^2) [-5.18, 8.87, -4.09, 2.934]}