Log 2(7x-1)-log2x^2=<3

Решите задачу:

$log_2(7x-1)-log_2x^2 \leq 3\; ,\; \; ODZ:\; 7x-1\ \textgreater \ 0,\; x\ \textgreater \ \frac{1}{7}\\\\log_2\frac{7x-1}{x^2} \leq log_22^3\\\\\frac{7x-1}{x^2}\leq 2^3\\\\\frac{7x-1}{x^2}-8 \leq 0\\\\\frac{7x-1-8x^2}{x^2} \leq 0\\\\x^2 \geq 0\; pri\; x\in R\; \; \to \; \; -8x^2+7x-1 \leq 0\\\\8x^2-7x+1 \geq 0\\\\D=49-32=17\\\\x_1=\frac{7-\sqrt{17}}{2},\; x_2=\frac{7+\sqrt{17}}{2}\\\\Znaki:\; \; ++(\frac{1}{7})+++(x_1)---(x_2)+++\\\\x\in (\frac{1}{7},\frac{7-\sqrt{17}}{2})U(\frac{7+\sqrt{17}}{2},+\infty )$