Nonlinear equation

Problem

Determine the solutions of \((4x-3)\cdot\ln\left(x^2-5x+7\right)=0\) for \(x\in\mathbb{R}\).

Solution

The zeros of the function on the left-hand side are determined by the zeros of the two factors.

Solving the requirement \(4x-3=0\) for \(x\) immediately yields the first zero \(x_1 = 3/4\).

The second factor vanishes provided the argument of the logarithm equals one. One thus needs to find solutions of \(x^2-5x+7=1\). The solutions of the resulting quadratic equation \(x^2-5x+6=0\) are obtained by means of

\[x_{2,3} = \frac{5\pm\sqrt{25-24}}{2}.\]

We thus obtain two more zeros \(x_2=2\) and \(x_3=3\).

These results can easily be checked with Sage: