A) \[{{\log }_{3}}4\]
B) \[1-{{\log }_{3}}4\]
C) \[1-{{\log }_{4}}3\]
D) \[{{\log }_{4}}3\]
Correct Answer: B
Solution :
Since, \[1,\,{{\log }_{3}}\sqrt{{{3}^{1-x}}+2},{{\log }_{3}}({{4.3}^{x}}-1)\]are in AP. \[\therefore 2{{\log }_{3}}{{({{3}^{1-x}}+2)}^{1/2}}={{\log }_{3}}3+{{\log }_{3}}({{4.3}^{x}}-1)\] \[\Rightarrow \] \[{{\log }_{3}}({{3}^{1-x}}+2)={{\log }_{3}}3({{4.3}^{x}}-1)\] \[\Rightarrow \] \[{{3}^{1-x}}+2={{12.3}^{x}}-3\] Let \[{{3}^{x}}=t\] \[\therefore \frac{3}{t}+2=12t-3\Rightarrow 12{{t}^{2}}-5t-3=0\] \[\Rightarrow \] \[(3t+1)\,(4t-3)=0\] \[\Rightarrow \] \[t=-\frac{1}{3},\frac{3}{4}\] \[\Rightarrow \] \[{{3}^{x}}=\frac{3}{4}\] \[\left( \because {{3}^{x}}\ne -\frac{1}{3} \right)\] \[\Rightarrow \] \[{{\log }_{3}}\left( \frac{3}{4} \right)=x\] \[\Rightarrow \] \[x={{\log }_{3}}3-{{\log }_{3}}4\] \[\Rightarrow \] \[x=1-{{\log }_{3}}4\]
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