Solve inequality
$\log_x\left[\log_2\left(4^x-6\right)\right]\le1$
$\log_x\left[\log_2\left(4^x-6\right)\right]\le\log_xx$
$4^x-6>0\Rightarrow x>\log_46 $
For $x\in(0;1)$
$\log_x\left[\log_2\left(4^x-6\right)\right]\le\log_xx\\\log_2(4^x-6)\ge x\\4^x-6\ge2^x\Rightarrow4^x-2^x-6\ge0\Rightarrow(2^x-3)(2^x+2)\ge0 \Rightarrow 2^x\ge3\Rightarrow x\ge\log_23\not\in(0;1)$
For $x\in(1;\infty) \wedge x>\log_46 \Rightarrow x\in(\log_46;\infty)$
$\log_x\left[\log_2\left(4^x-6\right)\right]\le\log_xx\\\log_2(4^x-6)\le x\\4^x-6\le2^x\Rightarrow 4^x-2^x-6\le0\Rightarrow (2^x-3)(2^x+2)\le0 \Rightarrow 2^x\le3 \Rightarrow x\le\log_23\\\log_46...\log_23,\frac{\log_26}{\log_24}...\frac{\log_23}{\log_22},\frac{\log_26}{2}...\frac{\log_23}{1},\log_2\sqrt{6}..<.\log_23\\\Rightarrow x\in(\log_46;\log_23\rangle$
But wolfram say: $x\in\langle\log_2\sqrt{7};\log_23\rangle$
What I do wrong?
You correctly used $4^x-6>0\Rightarrow x>\log_46$.
However $4^x-6$ must be greater than this because we need $\log_2(4^x-6)> 0.$
Therefore we must have $4^x-6>1$ i.e. $4^x>7$ which gives Wolfram's answer.