Question 410 of 519MediumMCQ
If 1, log9(31−x+2), log3(4⋅3x−1) are in A.P., then x equals:
Solutions · 1
Official solution
\[1,\,{\log _9}\left( {{3^{1 - x}} + 2} \right),{\log _3}\left( {{{4.3}^x} - 1} \right)\] are in A.P.
\[ \Rightarrow 2{\log _9}\left( {{3^{1 - x}} + 2} \right)\]
\[\,\,\,\,\,\,\,\,\,\] \[ = 1 + {\log _3}\left( {{{4.3}^x} - 1} \right)\]
\[ \Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)\]
\[\,\,\,\,\,\,\,\,\,\] \[ = {\log _3}3 + {\log _3}\left( {{{4.3}^x} - 1} \right)\]
\[ \Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)\]
\[\,\,\,\,\,\,\,\,\,\] \[ = {\log _3}\left[ {3\left( {{{4.3}^x} - 1} \right)} \right]\]
\[ \Rightarrow {3^{1 - x}} + 2 = 3\,\left( {{{4.3}^x} - 1} \right)\]
\[ \Rightarrow {3.3^{ - x}} + 2 = {12.3^x} - 3.\]
Put 3x=t
\[ \Rightarrow {3 \over t} + 2 = 12t - 3\]
or 12t2−5t−3=0;
Hence \[t = - {1 \over 3},{3 \over 4} \Rightarrow {3^x} = {3 \over 4}\]
(as \[{3^x}\,\, \ne \,\, - ve\] )
\[ \Rightarrow x = {\log _3}\left( {{3 \over 4}} \right)\]
or \[x = {\log _3}3 - {\log _3}4\]
\[ \Rightarrow x = 1 - {\log _3}4\]
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