A) \[\frac{5}{2}\]
B) \[log_{3}^{2}\]
C) \[\frac{7}{2}\]
D) \[log_{2}^{5}\]
Correct Answer: D
Solution :
[d] \[\because log2,log\left( {{2}^{x}}-1 \right),log\left( {{2}^{x}}+3 \right)\] be in A.P. \[\Rightarrow 2log{{\left( {{2}^{x}}-1 \right)}^{2}}=log2+log\left( {{2}^{x}}+3 \right)\] \[\Rightarrow log{{({{2}^{x}}-1)}^{2}}=log2({{2}^{x}}+3)\] \[\Rightarrow {{\left( {{2}^{x}}-1 \right)}^{2}}=2\left( {{2}^{x}}+3 \right)\] \[\Rightarrow {{(y-1)}^{2}}=2(y+3)\,\,\,\,\,\,\,[Let{{2}^{x}}=y]\] \[{{y}^{2}}-2y+1=2y+6\] \[{{y}^{2}}-4y-5=0\] \[\Rightarrow \left( x-5 \right)\left( x+1 \right)=0\] \[x=5,-1\] \[y=-1\]is considered \[\therefore y=5\] \[\Rightarrow {{2}^{x}}=5\] Taking log on base 2 on both sides. We have \[x=lo{{g}_{2}}=5\] Hence, option [d] is correct.
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