$(x^2+x+1)^x<1$ How to solve this exponential inequality?

Question

$(x^2+x+1)^x<1$

How to solve it? because in the answers I get this

\begin{cases} (x^2+x+1)>1\\ x<0 \end{cases}

And I understand why this one above works and the one below here doesn't: \begin{cases} (x^2+x+1)<1\\ x>0 \end{cases}

I don't understand how the case that has $x<0$ in it satisfies the answer that that the inequality is true for: $$x<-1$$ ?

Answer 1

$(x^2+x+1)^x<1 \iff $ $x\ln (x^2+x+1)<0$

Now if the product of 2 numbers is $0$ then one must be positive and the other one must be negative, so either $x<0$ and $\ln (x^2+x+1)>0$ (i.e. $x^2+x+1>1$)

Or $x>0$ and $\ln (x^2+x+1)<0$ (i.e. $x^2+x+1<1$)

Now you said you know why the second possibility is impossible to happen, so for the first one, we want $x^2+x+1>1 \implies x(x+1)>0$ and now by the same thinking, the product of 2 numbers is positive if they are both negative or if they are both positive, i.e. either $x>0$ and $x+1>0$ (which is not possible since we know from the begining that we are studying the case where $x<0$) or $x<0$ and $x+1<0 \implies x<-1$

Finally if you want those 2 conditions to be true at the same time $x<0$ and $x<-1$, you take all the numbers such that $x<-1$ because they satisfy both conditions (the intersection of the 2 intervals). ‍ ‍ ‍ ‍ ‍ ‍

Answer 2

$$x^2+x+1<1\iff x(x+1)<0\iff -1<x<0$$

But we also need to satisfy $x>0$

Alternatively if $x>0,$

$x+1>1$ and $x(x+1)>0$

Answer 3

We have that $$(x^2+x+1)^x<1 \iff x\log(x^2+x+1) < \log 1$$

that is

$$x\log(x^2+x+1) <0$$

and since

• $\log(x^2+x+1)>0 \iff x^2+x+1>1 \iff x(x+1)>0 \iff x<-1 \lor x>0$
• $\log(x^2+x+1)<0 \iff x^2+x+1<1 \iff x(x+1)<0 \iff -1<x<0$

the solution is given by $x<-1$.

Answer 4

The inequality is equivalent to $$ x \log(x^2 + x + 1) < 0. $$ For that to happen the factors must have opposite sign.

If $x>0$ then $x^2 + x + 1 > 1$ so its logarithm is positive too. So that case produces no solutions.

If $x < 0$ then we want to know when the second factor is positive for negative $x$. That happens when $$ x^2 + x + 1 > 1 $$ so $$ x^2 + x = x(x+1) > 0. $$ For negative $x$ that means $x + 1 < 0$ so $x < -1$.