A) 1
B) \[lo{{g}_{5}}2\]
C) \[lo{{g}_{2}}5\]
D) \[lo{{g}_{10}}5\]
Correct Answer: C
Solution :
| [c] \[{{\log }_{10}}2,\,{{\log }_{10}}({{2}^{x}}-1)\] and \[{{\log }_{10}}({{2}^{x}}+3)\]are in A.P. |
| Hence, common difference will be same. |
| \[\therefore \,\,{{\log }_{10}}\,({{2}^{x}}-1)-{{\log }_{10}}2\] |
| \[=\log \,({{2}^{x}}+3)-{{\log }_{10}}({{2}^{x}}-1)\] |
| \[\therefore \,\,\,{{\log }_{10}}\left( \frac{{{2}^{x}}-1}{2} \right)={{\log }_{10}}\left( \frac{{{2}^{x}}+3}{{{2}^{x}}-1} \right)\] |
| \[\Rightarrow \,\,\,\frac{{{2}^{x}}-1}{2}=\frac{{{2}^{x}}+3}{{{2}^{x}}-1}\] |
| \[{{({{2}^{x}}-1)}^{2}}=2({{2}^{x}}+3)\] |
| \[{{2}^{2x}}-{{2}^{x+1}}+1={{2}^{x+1}}+6\] |
| \[{{2}^{2x}}-{{2}^{x+2}}=5\] |
| Let \[{{2}^{x}}=y,\]then |
| \[{{y}^{2}}-4y-5=0\] |
| \[{{y}^{2}}-5y+y-5=0\] |
| \[y(y-5)+1(y-5)=0\] |
| \[y=-1,\,y=5\] |
| Therefore, \[{{2}^{x}}=5\] |
| \[x={{\log }_{2}}5.\] |
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