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If $x>0$ and $\log _{3} x+\log _{3}(\sqrt{x})+\log _{3}(\sqrt[4]{x})+$ $\log _{3}(\sqrt[8]{x})+\log _{3}(\sqrt[16]{x})+\ldots=4$, then $x$ equals
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9
Given: $\log _{3} x+\log _{3}(\sqrt{x})+\log _{3}(\sqrt[4]{x})$
$\quad+\log _{3} \sqrt[8]{x}+\log _{3}(\sqrt[16]{x})+-=4$
$\Rightarrow \log _{3} x^{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+--\infty}=4$
$\Rightarrow \log _{3} x^{\frac{1}{1-\frac{1}{2}}}=4 \quad\left[\because \mathrm{S}_{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}}\right]$
$\Rightarrow \log _{3} \mathrm{x}^{2}=4 \Rightarrow \mathrm{x}^{2}=3^{4} \Rightarrow \mathrm{x}=9$
$\quad+\log _{3} \sqrt[8]{x}+\log _{3}(\sqrt[16]{x})+-=4$
$\Rightarrow \log _{3} x^{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+--\infty}=4$
$\Rightarrow \log _{3} x^{\frac{1}{1-\frac{1}{2}}}=4 \quad\left[\because \mathrm{S}_{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}}\right]$
$\Rightarrow \log _{3} \mathrm{x}^{2}=4 \Rightarrow \mathrm{x}^{2}=3^{4} \Rightarrow \mathrm{x}=9$
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