Question
If $${\log _3}x + {\log _{9}}{x^2} + {\log _{27}}{x^3}$$ $$ = 9,$$ then x equals to -
Answer: Option C
$$\eqalign{
& \Rightarrow {\log _3}x + {\log _9}{x^2} + {\log _{27}}{x^3} = 9 \cr
& \Rightarrow {\log _3}x + {\log _{{3^2}}}{x^2} + {\log _{{3^3}}}{x^3} = 9 \cr
& \Rightarrow {\log _3}x + \frac{2}{2}{\log _3}x + \frac{3}{3}{\log _3}x = 9 \cr
& \Rightarrow 3{\log _3}x = 9 \cr
& \Rightarrow {\log _3}x = 3 \cr
& \Rightarrow x = {3^3} = 27 \cr} $$
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$$\eqalign{
& \Rightarrow {\log _3}x + {\log _9}{x^2} + {\log _{27}}{x^3} = 9 \cr
& \Rightarrow {\log _3}x + {\log _{{3^2}}}{x^2} + {\log _{{3^3}}}{x^3} = 9 \cr
& \Rightarrow {\log _3}x + \frac{2}{2}{\log _3}x + \frac{3}{3}{\log _3}x = 9 \cr
& \Rightarrow 3{\log _3}x = 9 \cr
& \Rightarrow {\log _3}x = 3 \cr
& \Rightarrow x = {3^3} = 27 \cr} $$
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