# What are the asymptote(s) and hole(s), if any, of  f(x) =1/(e^x-2)?

##### 1 Answer
Jun 13, 2018

There is a asymptote at $x = \ln 2 \approx 0.6931$

#### Explanation:

There are no holes because no factors will cancel out of the numerator and denominator, the asymptote(s) are when we have $\frac{1}{0}$

${e}^{x} - 2 = 0$

${e}^{x} = 2$

$\ln \left({e}^{x}\right) = \ln 2$

$x = \ln 2$

graph{1/(e^x-2) [-10, 10, -5, 5]}