<s>[INST] System of logarithmic and exponential equations $\log_{7}(x^2-x+1)=\log_{2}(y^2-1)-\log_{2}(x+1)$ Solve the following system in $\mathbb{R}$.
$$\log_{7}(x^2-x+1)=\log_{2}(y^2-1)-\log_{2}(x+1)$$
$$\log_{7}(y^2-y+1)=\log_{2}(z^2-1)-\log_{2}(y+1)$$
$$\log_{7}(z^2-z+1)=\log_{2}(x^2-1)-\log_{2}(z+1)$$
I considered the functions $f,g:[1,\infty]\longrightarrow \mathbb{R}$
$f(x)=\log_{7}(x^2-x+1)+\log_{2}(x+1) $ and $g(x)=\log_{2}(x^2-1)$ both being increasing on $[1,\infty]$
And so the system is equivalent with:
$f(x)=g(y) \\ f(y)=g(z)\\f(z)=g(x)$ 
Suppose that $x \geq y$ then $f(x) \geq f(y) \Leftrightarrow $ $g(y) \geq g(z) \Leftrightarrow y \geq z $ then $f(y) \geq f(z) \Leftrightarrow $ $g(z) \geq g(x) \Leftrightarrow z \geq x $ this implies $x \geq y \geq z \geq x$ so $x=y=z$
Therefore the system can be rewritten as
$\log_{7}(x^2-x+1)=\log_{2}(x-1)=a$
So I have a new system
$x^2-x+1=7^a \\ x-1=2^a$
But i don't know how to prove that the only solution of this system is $x=3$ , any ideas?
 [/INST] Substitute $x=2^a+1$ in the first equation to get:
$$4^a+2^a+1=7^a\implies \left(\frac{4}{7}\right)^a+\left(\frac{2}{7}\right)^a+\left(\frac{1}{7}\right)^a=1$$
Clearly, the left hand side is decreasing, so we can have at most one solution, and that is $a=1$, which gives $x=3$.
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