$(x^2+x+1)^x<1$ How to solve this exponential inequality? $(x^2+x+1)^x<1$
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: $$x<-1$$
?
 A: $(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). ‍  ‍  ‍  ‍  ‍  ‍ 
A: 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 $x < -1$.
A: $$x^2+x+1<1\iff x(x+1)<0\iff -1<x<0$$
But we also need to satisfy $x>0$
Alternatively if $x>0,$
$x+1>1$ and $x(x+1)>0$
A: 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 $x<-1$.
