If y = $$(\log_2 x)^2 - 6 \log_2 x + 12$$ is a positive number, then the equation $$x^y = 256$$ has-

Explanation

A solution for x $$x^y = 256$$ ∴ $$x^y = 4^4$$ or $$(16)^2$$ Therefore, x = 4 then y = 4 x = 16 then y = 2 According to the question y = $$(\log_2 x)^2 - 6 \log_2 x + 12$$ x = 4 then y = $$(\log_2 4)^2 - 6 \log_2 4 + 12$$ y = $$[\log_2 (2)^2]^2 - 6 \log_2 (2)^2 + 12$$ y = $$[2 \log_2 2]^2 - 6 \times 2 \log_2 2 + 12$$ y = $$2^2 - 12 + 12 = 4$$ ∴ loga a = $$\frac{\log a}{\log a} = 1$$ x = 16 then y = $$[\log_2 2^4]^2 - 6 [\log_2 2^4] + 12$$ y = 16 – 24 + 12 = 4 If, x = 16 then y = 2 Therefore, x = 4 satisfies the equation. Hence, there is only one possible solution to the equation.