Iflog3x+log9–x2+log27x3=9,thenxequals - Iflog3x+log9_x2+log27x3=9,thenxequals - \eqalign{ & {\text{If}}\,{\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9, \cr & {\text{then}}\,\,x\,\,{\text{equals - }} \cr} " /> Iflog3x+log9–x2+log27x3=9,thenxequals - Iflog3x+log9_x2+log27x3=9,thenxequals - \eqalign{ & {\text{If}}\,{\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9, \cr & {\text{then}}\,\,x\,\,{\text{equals - }} \cr} " /> Iflog3x+log9–x2+log27x3=9,thenxequals - Iflog3x+log9_x2+log27x3=9,thenxequals - \eqalign{ & {\text{If}}\,{\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9, \cr & {\text{then}}\,\,x\,\,{\text{equals - }} \cr} " />

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$\begin{aligned} If \log_{3} x + \log_{\underline{9}} x^{2} + \log_{27} x^{3} = 9, \\ then x equals - \end{aligned}$ $\eqalign{ & {\text{If}}\,{\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9, \cr & {\text{then}}\,\,x\,\,{\text{equals - }} \cr}$

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29
327
4None of these

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log_{3} x + \log_{\underline{9}} x^{2} + \log_{27} x^{3} = 9 \\ \Rightarrow \log_{3} x + \log_{32} x^{2} + \log_{33} x^{3} = 9 \\ \Rightarrow \log_{3} x + \frac{2}{2} \log_{3} x + \frac{3}{3} \log_{3} x = 9 \\ \Rightarrow 3 \log_{3} x = 9 \\ \Rightarrow \log_{3} x = 3 \\ \Rightarrow x = 3^{3} = 27 \end{aligned}

$\eqalign{ & {\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + {\log _{32}}{x^2} + {\log _{33}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + \frac{2}{2}{\log _3}x + \frac{3}{3}{\log _3}x = 9 \cr & \Rightarrow 3{\log _3}x = 9 \cr & \Rightarrow {\log _3}x = 3 \cr & \Rightarrow x = {3^3} = 27 \cr}$

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", "text": "

\begin{aligned} If \log_{3} x + \log_{\underline{9}} x^{2} + \log_{27} x^{3} = 9, \\ then x equals - \end{aligned}

$\eqalign{ & {\text{If}}\,{\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9, \cr & {\text{then}}\,\,x\,\,{\text{equals - }} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \log_{3} x + \log_{\underline{9}} x^{2} + \log_{27} x^{3} = 9 \\ \Rightarrow \log_{3} x + \log_{32} x^{2} + \log_{33} x^{3} = 9 \\ \Rightarrow \log_{3} x + \frac{2}{2} \log_{3} x + \frac{3}{3} \log_{3} x = 9 \\ \Rightarrow 3 \log_{3} x = 9 \\ \Rightarrow \log_{3} x = 3 \\ \Rightarrow x = 3^{3} = 27 \end{aligned}

$\eqalign{ & {\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + {\log _{32}}{x^2} + {\log _{33}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + \frac{2}{2}{\log _3}x + \frac{3}{3}{\log _3}x = 9 \cr & \Rightarrow 3{\log _3}x = 9 \cr & \Rightarrow {\log _3}x = 3 \cr & \Rightarrow x = {3^3} = 27 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \log_{3} x + \log_{\underline{9}} x^{2} + \log_{27} x^{3} = 9 \\ \Rightarrow \log_{3} x + \log_{32} x^{2} + \log_{33} x^{3} = 9 \\ \Rightarrow \log_{3} x + \frac{2}{2} \log_{3} x + \frac{3}{3} \log_{3} x = 9 \\ \Rightarrow 3 \log_{3} x = 9 \\ \Rightarrow \log_{3} x = 3 \\ \Rightarrow x = 3^{3} = 27 \end{aligned}

$\eqalign{ & {\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + {\log _{32}}{x^2} + {\log _{33}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + \frac{2}{2}{\log _3}x + \frac{3}{3}{\log _3}x = 9 \cr & \Rightarrow 3{\log _3}x = 9 \cr & \Rightarrow {\log _3}x = 3 \cr & \Rightarrow x = {3^3} = 27 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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