If {x^{frac{3}{4}(log_3 x)^2 + log_3 x - frac | vedclass

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(A) Given equation: ${x^{\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}} = 3^{1/2}}$.Taking $\log_3$ on both sides:$(\frac{3}{4}(\log_3 x)^2 + \log_

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a single positive integer value

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(A) Given equation: ${x^{\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}} = 3^{1/2}}$.Taking $\log_3$ on both sides:$(\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}) \cdot \log_3 x = \log_3(3^{1/2})$.Let $y = \log_3 x$. Then:$(\frac{3}{4}y^2 + y - \frac{5}{4})y = \frac{1}{2}$.Multiply by $4$:$(3y^2 + 4y - 5)y = 2$.$3y^3 + 4y^2 - 5y - 2 = 0$.By testing values,$y = 1$ is a root: $3(1)^3 + 4(1)^2 - 5(1) - 2 = 3 + 4 - 5 - 2 = 0$.Dividing by $(y-1)$,we get $(y-1)(3y^2 + 7y + 2) = 0$.$(y-1)(3y+1)(y+2) = 0$.So,$y = 1, y = -1/3, y = -2$.Since $y = \log_3 x$,we have $x = 3^1 = 3$,$x = 3^{-1/3} = \frac{1}{\sqrt[3]{3}}$,and $x = 3^{-2} = \frac{1}{9}$.All three values are positive,but only $x = 3$ is an integer. Thus,there is only one positive integer value.

If ${x^{\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}} = \sqrt{3}}$,then $x$ is:

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