Nonlinear equation

Problem

State for \(x\in\mathbb{R}^+\) the solutions of the following equation:

\[(\ln x-1) \cdot \left( \mathrm{e}^x -2 \right) \cdot \left( \dfrac{1}{x} -3 \right)=0\]

Solution

The zeros of the function are obtained by determining the zeros of the three factors.

The requirement \(\ln x - 1=0\) implies \(\ln x = 1\). Applying the exponential function to both sides, one finds \(\mathrm{e}^{\ln x} = \mathrm{e}^1 = \mathrm{e}\). The logarithm being the inverse of the exponential function, \(\mathrm{e}^{\ln x}=x\) holds. Therefore, we find as a first zero \(x_1 = \mathrm{e}\).

From the second factor, we find \(\mathrm{e}^x = 2\) to which we apply the logarithm on both sides. In analogy to the previous reasoning, we have \(\ln \mathrm{e}^x = x\). The second zero thus follows as \(x_2=\ln 2\).

By simply solving for \(x\), the last factor yields the zero \(x_3 = \frac{1}{3}\).

This result can easily be verified by means of Sage