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  4. If log 2(5.2x+1), log 4(21-x+1) and 1 are in AP, then x equals

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Q. If $ {{\log }_{2}}({{5.2}^{x}}+1),{{\log }_{4}}({{2}^{1-x}}+1) $ and 1 are in AP, then $ x $ equals

JamiaJamia 2009

Solution:

The given numbers are in AP. $ \therefore $ $ 2{{\log }_{4}}({{2}^{1-x}}+1)={{\log }_{2}}({{5.2}^{x}}+1)+1 $ $ \Rightarrow $ $ 2{{\log }_{2}}\left( \frac{2}{{{2}^{x}}}+1 \right)={{\log }_{2}}({{5.2}^{x}}+1)+{{\log }_{2}}2 $ $ \Rightarrow $ $ \frac{2}{2}{{\log }_{2}}\left( \frac{2}{{{2}^{x}}}+1 \right)={{\log }_{2}}({{5.2}^{x}}+1)2 $ $ \Rightarrow $ $ {{\log }_{2}}\left( \frac{2}{{{2}^{x}}}+1 \right)={{\log }_{2}}({{10.2}^{x}}+2) $ $ \Rightarrow $ $ \frac{2}{{{2}^{x}}}+1={{10.2}^{x}}+2 $ $ \Rightarrow $ $ \frac{2}{y}+1=10y+2, $ where $ {{2}^{x}}=y $ $ \Rightarrow $ $ 10{{y}^{2}}+y-2=0\Rightarrow (5y-2)(2y+1)=0 $ $ \Rightarrow $ $ y=\frac{2}{5} $ or $ y=\frac{-1}{2}\Rightarrow {{2}^{x}}=\frac{2}{5} $ or $ {{2}^{x}}=\frac{-1}{2} $ $ \Rightarrow $ $ x={{\log }_{2}}\left( \frac{2}{5} \right) $ $ \Rightarrow $ $ x={{\log }_{2}}2-{{\log }_{2}}5 $ $ \Rightarrow $ $ x=1-{{\log }_{2}}5 $