<s>[INST] Solve the equation $2^{x^2 + x} + \log_2x=2^{x+1}$ 
Solve the equation $2^{x^2 + x} + \log_2x=2^{x+1}$ where $x$ is real.

I tried to use derivatives, without success.
It's obvious that $x=1$ is solution. 
Also, if $x \gt 1$ then $2^{x^2 + x} \gt 2^{x+1}$ and $\log_2x \gt 0$ therefore the equation has no solutions $x \gt 1$.
 [/INST] It's clear that $x=1$ is a solution. Let's look at two cases: $x>1$ and $x<1$.
If $x>1$, then $\log_2x>0$ and $x^2+x>x+1$, which implies that $2^{x^2+x}>2^{x+1}$. It's impossible.
If $0<x<1$, then $\log_2x<0$ and $x^2+x<x+1$, which implies that $2^{x^2+x}>2^{x+1}$. It's impossible.
In summary, we only have one solution $x=1$.
</s>