Solve the given inequality

Question

$$\frac{\log^2 x-3\log x+3}{\log x -1}<1$$

If I solve it my usual way, where I consider $\log x = a,$ it is impossible to factorise. Is there any other way to solve it?

Answer 1

Put all terms to one side, then solve $$\frac{a^2-3a+3}{a-1}-1<0.$$ This simplifies to $$\frac{(a-2)^2}{a-1}<0$$ which is only negative, when the denominator is negative.

Answer 2

Hint: With the help of your Substitution we get $$\frac{(a-2)^2}{a-1}<0$$

Answer 3

If $\log x=a$, then if $a>1$, $a^2-3a+3<a-1$, so $a^2-4a+4<0$, i.e. $(a-2)^2<0$ which has no solutions.

If $a<1$, we instead get $a^2-3a+3>a-1$, so $(a-2)^2>0$, which has every $a<1$ as a solution.

Since $a<1$, $\log x<1$ so $0<x<e$ is the solution.