A) \[{{\log }_{3}}4\]
B) \[1-{{\log }_{3}}4\]
C) \[1-{{\log }_{4}}3\]
D) \[{{\log }_{4}}3\]
Correct Answer: B
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 \[\frac{3}{4}\] \[\Rightarrow \] \[x={{\log }_{3}}\left( \frac{3}{4} \right)\]\[\left( \because \,\,\,\,{{3}^{x}}\ne \frac{1}{3} \right)\] \[\Rightarrow \] \[x=1-{{\log }_{3}}4\]
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