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Решите задачу:

$log^{2}_{1/2} 4x = log^{2}_{2} 4x = log^{2}_{2}4 + 2log_{2}4 * log_{2}x + log^{2}_{2}x \\ log_{2} \frac{x^2}{8} = log_{2} \frac{1}{8} + 2log_2x \\ log^{2}_{2}4 + 2log_{2}4 * log_{2}x + log^{2}_{2}x + log_{2} \frac{1}{8} + 2log_2x - 8 = 0 \\ 4 + 6log_{2}x + log^{2}_{2}x -11 = 0 \\ t = log_{2}x \\ t^2 + 6t - 7 = 0 \\ D = 36 + 28 = 64 = 8^2 \\ t_{1} = 1 \\ t_{2} = -7 \\ log_{2}x = 1 \\ x = 2 \\ log_{2}x = -7 \\ x=1/128$