The equation x^{(3/4)(log_2 x)^2 + (log_2 x) | vedclass

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(D) For the given equation to be meaningful,we must have $x > 0$. Taking the logarithm base $2$ on both sides:$((\frac{3}{4})(\log_2 x)^2 + (\log_2 x)

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(D) For the given equation to be meaningful,we must have $x > 0$. Taking the logarithm base $2$ on both sides:$((\frac{3}{4})(\log_2 x)^2 + (\log_2 x) - \frac{5}{4}) \cdot \log_2 x = \log_2(\sqrt{2})$Let $t = \log_2 x$. Then the equation becomes:$(\frac{3}{4}t^2 + t - \frac{5}{4}) \cdot t = \frac{1}{2}$Multiplying by $4$:$(3t^2 + 4t - 5)t = 2$$3t^3 + 4t^2 - 5t - 2 = 0$By testing roots,we find $t = 1$ is a root. Dividing by $(t - 1)$:$(t - 1)(3t^2 + 7t + 2) = 0$$(t - 1)(3t + 1)(t + 2) = 0$So,$t = 1, -2, -1/3$.Since $t = \log_2 x$,we have $x = 2^1 = 2$,$x = 2^{-2} = 1/4$,and $x = 2^{-1/3} = 1/\sqrt[3]{2}$.All three solutions are real,and $1/\sqrt[3]{2}$ is an irrational number. Thus,all statements $(A)$,$(B)$,and $(C)$ are correct.

The equation $x^{(3/4)(\log_2 x)^2 + (\log_2 x) - 5/4} = \sqrt{2}$ has

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