Answer to Question #223288 in Calculus for Vanel

Question

The domain of the function define by f(x) = √[Inx/ (Inx-1)] is?

Answer 1

The first restriction:

$$x>0$$

since it is not possible to have $x\le0$ under the $\ln$.

The second restriction:

$$\ln x - 1\ne0\Rightarrow x\ne e$$

since the denominator can not be zero.

The third restriction:

$$\dfrac{\ln x}{\ln x - 1}\ge0$$

since there can not be a negative quantity under the square root sign. The last restriction is satisfied if:

$$\ln x\ge0\space\text{and}\space \ln x-1>0\\ \ln x\ge0\space\text{and}\space\ln x>1\\ x\ge1 \space\text{and}\space x>e\\ x>e$$

or

$$\ln x\le0\space\text{and}\space \ln x-1<0\\ \ln x\le0\space\text{and}\space\ln x<1\\ 0<x\le1 \space\text{and}\space 0<x<e\\ 0<x\le1$$

Finaly, for the third restriction have:

$$x\in (0;1]\cup(e;+\infty)$$

Combining these three restrictions together, find the domain:

$$x\in (0;1]\cup(e;+\infty)$$

It is so, since the third restriction already contains the first and the second ones.

Answer. $x\in (0;1]\cup(e;+\infty)$.