A) \[lo{{g}_{3}}4\]
B) \[1-lo{{g}_{3}}4\]
C) \[1-lo{{g}_{4}}3\]
D) \[lo{{g}_{4}}3\]
Correct Answer: B
Solution :
\[1,{{\log }_{3}}\sqrt{{{3}^{1-x}}+2},{{\log }_{3}}({{4.3}^{x}}-1)\]are in A.P. \[2{{\log }_{3}}{{({{3}^{1-x}}+2)}^{\frac{1}{2}}}=1+{{\log }_{3}}({{4.3}^{x}}-1)\] \[{{\log }_{3}}({{3}^{1-x}}+2)={{\log }_{3}}3+{{\log }_{3}}({{4.3}^{x}}-1)\] \[{{\log }_{3}}({{3}^{1-x}}+2)={{\log }_{3}}[3({{4.3}^{x}}-1)]\] \[{{3}^{1-x}}+2=3({{4.3}^{x}}-1)\] \[{{3.3}^{-x}}+2={{12.3}^{x}}-3\] Let \[{{3}^{x}}=t\] \[\frac{3}{t}+2=12t-3\] \[3+2t=12{{t}^{2}}-3t\] \[12{{t}^{2}}-5t-3=0\] \[(12{{t}^{2}}-9t+4t-3)=0\] \[3t(4t-2)+1(4t-3)=0\] \[t=-\frac{1}{3},\frac{3}{4}\] \[{{3}^{x}}=\frac{3}{4}\] \[\Rightarrow \] \[x={{\log }_{3}}\left( \frac{3}{4} \right)={{\log }_{3}}3-{{\log }_{3}}4\] \[\Rightarrow \] \[x=1-{{\log }_{3}}4\]
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