A) \[{{\log }_{3}}4\]
B) \[1-{{\log }_{3}}4\]
C) \[1-{{\log }_{4}}3\]
D) \[{{\log }_{4}}3\]
Correct Answer: B
Solution :
[b]: The given numbers are in A.P. \[\therefore \]\[2{{\log }_{9}}({{3}^{1-x}}+2)=lo{{g}_{3}}({{4.3}^{x}}-1)+1\] \[\Rightarrow \]\[2{{\log }_{{{3}^{2}}}}({{3}^{1-x}}+2)=lo{{g}_{3}}({{4.3}^{x}}-1)+{{\log }_{3}}3\] \[\Rightarrow \]\[\frac{2}{2}{{\log }_{3}}({{3}^{1-x}}+2)=lo{{g}_{3}}[3({{4.3}^{x}}-1)]\] \[\Rightarrow \]\[{{3}^{1-x}}+2=3({{4.3}^{x}}-1)\] \[\Rightarrow \]\[\frac{3}{y}+2=12y-3,\]where \[y={{3}^{x}}\] \[\Rightarrow \]\[12{{y}^{2}}-5y-3=0\] \[y=\frac{-1}{3}\]or\[\frac{3}{4}\Rightarrow {{3}^{x}}=\frac{-1}{3}\]or\[{{3}^{x}}=\frac{3}{4}\] \[x={{\log }_{3}}(3/4)\Rightarrow x=1-lo{{g}_{3}}4.\]
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