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  4. If 1, log 9(31-x+2), log 3(4.3x-1) are in AP, then x equals

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Q. If $ 1,\,{{\log }_{9}}({{3}^{1-x}}+2),{{\log }_{3}}({{4.3}^{x}}-1) $ are in AP, then x equals

J & K CETJ & K CET 2004

Solution:

Since, given numbers are in AP.
$ \therefore $ $ 2{{\log }_{9}}({{3}^{1-x}}+2)={{\log }_{3}}({{4.3}^{x}}-1)+1 $
$ \Rightarrow $ $ 2{{\log }_{{{3}^{2}}}}\,({{3}^{1-x}}+2)={{\log }_{3}}(({{4.3}^{x}}-1)+{{\log }_{3}}3 $
$ \Rightarrow $ $ \frac{2}{2}{{\log }_{3}}({{3}^{1-x}}+2)={{\log }_{3}}[3({{4.3}^{x}}-1)] $
$ \Rightarrow $ $ {{3}^{1-x}}+2=3({{4.3}^{x}}-1) $
$ \Rightarrow $ $ \frac{3}{y}+2=12y-3, $ where $ y={{3}^{x}} $
$ \Rightarrow $ $ 12{{y}^{2}}-5y-3=0 $
$ \Rightarrow $ $ y=-\frac{1}{3} $ or $ \frac{3}{4} $
$ \Rightarrow $ $ {{3}^{x}}=-\frac{1}{3} $ or $ \frac{3}{4} $
$ \Rightarrow $ $ x={{\log }_{3}}\left( \frac{3}{4} \right) $ $ \left( \because \,\,\,\,{{3}^{x}}\ne \frac{1}{3} \right) $
$ \Rightarrow $ $ x=1-{{\log }_{3}}4 $