'$|x|$ to power of $x$' inequality with polynomial exponent

Question

I have found this in a old math book and I completly don't know how to deal with it: $$|x-1|^{x^4-4x^3+3x^2}<1$$ I've looked for similar problems on internet but I couldn't find anything. You can factor the exponent but it isn't any better. The book says solution is $x\in(0,1)\cup(2,3)$ and WolframAlpha agrees with it.

Do you have any idea?

Answer 1

Taking logarithms, we find that $$(x^4-4x^3+3x^2)\ln|x-1|<\ln 1 \implies x^2(x-1)(x-3)\ln|x-1|<0$$ We know that the logarithmic term can never equal $0$, so the inequality can be simplified to $$x^2(x-1)(x-3)<0$$ Thus the solution is $$x\in(0,1)\cup(2,3)$$ since when $x \in [1, 2]$, both $\ln|x-1|$ and $x-3$ are less than or equal to $0$ which makes the LHS positive.

Answer 2

Hint: $k^n<1$ if $0<k<1,n>0$ or $k>1,n<0$.

What values give $|x-1|<1$ or $|x-1|>1$?

Factorize the exponent. What values give positive/negative exponent?

Which regions satisfy both conditions?