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