the solution of $|x|^{x^2-x-2} < 1$

Question

I have tried it using log of $|x|$ which gives RHS value 0 and LHS gives quadratic equation of $x^2-x-2$ and solving it i get value of x between - 1 to 2 EXCEPT 0 but answer is $x$ between (1,2).

Answer 1

If $|x|>1\iff x>1\ \ \ \ (1)$ or $x<-1\ \ \ \ (2)$

we need $0>x^2-x-2=(x-2)(x+1)\iff -1<x<2\ \ \ \ (3)$

Find the intersection of $(1),(3)$ and $(2),(3)$

If $|x|<1\iff-1<x<1\ \ \ \ (4)$

we need $x^2-x-2>0\implies$ either $x>2\ \ \ \ (5)$ or $x<-1\ \ \ \ (6)$

Find the intersection of $(4),(5)$ and $(4),(6)$

Answer 2

It is convenient covert the inequality by logarithm (since log function is monotonically increasing):

$$|x|^{x^2-x-2} < 1\iff (x^2-x-2)\log |x|<0$$

Then it is convenient eliminate the absolute value considering 2 cases:

1) $x>0$ $\implies(x^2-x-2)\log x<0$

2) $x<0$ $\implies(x^2-x-2)\log (-x)<0 \implies y^2+y-2 \log y <0 \quad y=-x>0$

You can find the solution for each case by studing the sign of $x^2-x-2$, $x^2+x-2$ and $\log x$.

Answer 3

It is equivalent to: $$|x|^{x^2-x-2}<|x|^0$$ It is equivalent to: $$1) \begin{cases} |x|>1 \\ x^2-x-2<0 \end{cases} \ \ \text{OR} \ \ 2) \begin{cases} 0<|x|<1 \\ x^2-x-2>0 \end{cases}.$$ Can you solve the two systems and unite their solutions?