Domain and values of the lograrithm

Problem

The function \(f:x\mapsto\sqrt{1-\ln x}\) with maximal domain \(\mathbb{D}\) is given.

1. Determine \(\mathbb{D}\).

2. Determine the value \(x\in \mathbb{D}\) for which \(f(x)=2\).

Solution of part a

The logarithm is only defined for arguments \(x>0\), and the square root only for arguments \(x\geq0\). For values \(x>e\), the logarithm of \(x\) yields values bigger than 1 and thus the argument of the square root would be smaller than 0. On the other hand, in the range \(0<x\leq e\), the logarithm yields values smaller or equal to 1 such that the argument of the square root becomes bigger or equal to zero. The maximal domain is hence given by

\[\mathbb{D}=]0;e]\,.\]

We plot the function with Sage.

Solution of part b

To obtain the corresponding value for \(x\), we solve the equation for \(x\):

\[\sqrt{1-\ln x} = 2\Rightarrow \ln(x) = -3 \Rightarrow x = e^{-3}\]

The result is confirmed by Sage: