A) \[{{\log }_{3}}4\]
B) \[1-{{\log }_{3}}4\]
C) \[1-{{\log }_{4}}3\]
D) \[{{\log }_{4}}3\]
Correct Answer: B
Solution :
\[1,\,\,{{\log }_{3}}\sqrt{{{3}^{1-x}}+2},\,\,{{\log }_{3}}(4\cdot {{3}^{x}}-1)\]are in\[AP\] \[\therefore \]\[2{{\log }_{3}}{{({{3}^{1-x}}+2)}^{1/2}}=1+{{\log }_{3}}(4\cdot {{3}^{x}}-1)\] \[{{\log }_{3}}({{3}^{1-x}}+2)={{\log }_{3}}3+{{\log }_{3}}(4\cdot {{3}^{x}}-1)\] \[{{\log }_{3}}({{3}^{1-x}}+2)={{\log }_{3}}[3(4\cdot {{3}^{x}}-1)]\] \[{{3}^{1-x}}+2=3(4\cdot {{3}^{x}}-1)\] \[3\cdot {{3}^{-x}}+2=12\cdot {{3}^{x}}-3\] Let\[{{3}^{x}}=t\] \[\therefore \] \[\frac{3}{t}+2=12t-3\] \[3+2t=12{{t}^{2}}-3t\] \[12{{t}^{2}}-5t-3=0\] \[(12{{t}^{2}}-9t+4t-3)=0\] \[3t(4t-3)+1(4t-3)=0\] \[\therefore \] \[t=-\frac{1}{3},\,\,\frac{3}{4}\] \[{{3}^{x}}=\frac{3}{4}\] \[\Rightarrow \] \[x={{\log }_{3}}\left( \frac{3}{4} \right)={{\log }_{3}}3-{{\log }_{3}}4\] \[\Rightarrow \] \[x=1-{{\log }_{3}}4\]
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