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:$(\frac{3}{4}(\log_2 x)^2 + \log_2 x - \frac{5}{4}) \log_2 x = \log_2 \sqrt{2} = \frac{1}{2}$.Let $t = \log_2 x$. Then the equation becomes:$(\frac{3}{4}t^2 + t - \frac{5}{4}) t = \frac{1}{2}$.Multiplying by $4$:$(3t^2 + 4t - 5) t = 2 \Rightarrow 3t^3 + 4t^2 - 5t - 2 = 0$.By testing roots,we find $(t - 1)$ is a factor:$(t - 1)(3t^2 + 7t + 2) = 0 \Rightarrow (t - 1)(3t + 1)(t + 2) = 0$.Thus,$t = 1, -2, -1/3$.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 irrational. Therefore,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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