If sqrt{log_3 x^{16}} + 9 log_{27} sqrt[3]{fr | vedclass

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(A) Given equation: $\sqrt{\log_3 x^{16}} + 9\log_{27}\sqrt[3]{\frac{3}{x}} = 5$.Simplify the first term:$\sqrt{\log_3 x^{16}} = \sqrt{16\log_3 x} = 4

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(A) Given equation: $\sqrt{\log_3 x^{16}} + 9\log_{27}\sqrt[3]{\frac{3}{x}} = 5$.Simplify the first term:$\sqrt{\log_3 x^{16}} = \sqrt{16\log_3 x} = 4\sqrt{\log_3 x}$.Simplify the second term:$9\log_{27}\sqrt[3]{\frac{3}{x}} = 9 \cdot \log_{3^3} (\frac{3}{x})^{1/3} = 9 \cdot \frac{1}{3} \cdot \frac{1}{3} \log_3(\frac{3}{x}) = 1 \cdot (\log_3 3 - \log_3 x) = 1 - \log_3 x$.Substitute these into the equation:$4\sqrt{\log_3 x} + 1 - \log_3 x = 5$.Rearrange the equation:$4\sqrt{\log_3 x} - \log_3 x = 4$.Let $t = \sqrt{\log_3 x}$,then $t^2 = \log_3 x$:$4t - t^2 = 4 \Rightarrow t^2 - 4t + 4 = 0$.Factor the quadratic equation:$(t - 2)^2 = 0 \Rightarrow t = 2$.Since $t = \sqrt{\log_3 x} = 2$,we have $\log_3 x = 4$.Therefore,$x = 3^4 = 81$.

If $\sqrt{\log_3 x^{16}} + 9 \log_{27} \sqrt[3]{\frac{3}{x}} = 5$,then $x = \dots$.

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