Question

The number of rational roots of ${\log }_x^{3}10-6{\log }_x^{2}10+11{\log }_{x}10-6=0$, is

Answer 1

Put ${\log }_{x}10=t$ in the given equation, we get
$t^3-{6t}^2+11t-6=0$
$\Rightarrow (t-1)(t-2)(t-3)=0$, then $\{\begin{array}{c}t=1\\ t=2\\ t=3\end{array}$
It follows that
$\{\begin{array}{c}{\log }_{x}10=1\\ {\log }_{x}10=2\\ {\log }_{x}10=3\end{array}\Rightarrow \{\begin{array}{c}x=10\\ x^2=10\\ x^3=10\end{array}\Rightarrow \{\begin{array}{c}x=10\\ x=\sqrt{10}\\ x=\sqrt[3]{310}\end{array}(\because x>0\&\ne 1)$
$\therefore x_1=10,x_2=\sqrt{10}\&x_3=\sqrt[3]{310}$ are the roots of the original equation.
Only $x_1=10$ is a rational root, others are irrational.