Let $$u = ({\log_2 x})^2 - 6 {\log_2 x} + 12$$ where x is a real number. Then the equation $$x^u = 256$$, has
$$x^u = 256$$
Taking log to the base 2 on both the sides,
$$u * \log_{2}{x} = \log_{2}{256}$$
=>$$[({\log_2 x})^2 - 6 {\log_2 x} + 12] * \log_{2}{x} = 8$$
$$(log_2 x)^3 - 6(log_2 x)^2 + 12log_2 x = 8$$
Let $$log_2 x = t$$
$$t^3 - 6t^2 +12t - 8 = 0$$
$$(t-2)^3 = 0$$
Therefore, $$log_2 x = 2$$
=> $$x = 4$$ is the only solution
Hence, option B is the correct answer.
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