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?

write your inequation like that:
$$\log_4(x+1)^2\le \log_4 4+\log_4 x$$ thus $$\log_4 (x+1)^2\le \log_4 4x$$ this is equivalent to $$(x+1)^2\le 4x$$ can you proceed?
