Solve for $x$ $2\log_4(x+1) \le 1+\log_4x$ $2\log_4(x+1) \le 1+\log_4x$
I did:
$$2\log_4(x+1) \le 1+\log_4x \Leftrightarrow \log_4(x^2+1) \le 1+\log_4(x) \Leftrightarrow \log_4(\frac{x^2+1}{x}) \le 1 \Leftrightarrow 4 \ge \frac{x^2+1}{x} \Leftrightarrow 4x \ge n^2 +1 \Leftrightarrow 0\ge x^2 -4x +1$$
Using the quadratic formula this becomes $x \le 2 + \sqrt{3}$ $\lor$ $x\le 2 - \sqrt{3}$
But my book says that the solution is $[1,1] = [1]$
What did I do wrong?

$(x+1)^2 \ne x^2+1$, in general. If you fix that error the equation is now $x^2-2x+1 \le 0 \iff (x-1)^2 \le 0 \iff x=1$
