A) \[x{{\log }_{2}}(1-{{2}^{x}})\]
B) \[x+{{\log }_{2}}(1-{{2}^{x}})\]
C) \[{{\log }_{2}}(1-{{2}^{x}})\]
D) \[x{{\log }_{2}}({{2}^{x}}+1)\]
Correct Answer: B
Solution :
[b] \[{{2}^{2y}}-{{2}^{y}}+{{2}^{x}}(1-{{2}^{x}})=0\] put \[{{2}^{y}}=t\] \[{{t}^{2}}-t+{{2}^{x}}(1-{{2}^{x}})=0\] where \[{{t}_{1}}={{2}^{{{y}_{1}}}}\] and \[{{t}_{2}}={{2}^{{{y}_{2}}}}\] \[{{t}_{1}}{{t}_{2}}={{2}^{x}}(1-{{2}^{x}})\] \[{{2}^{{{y}_{1}}+{{y}_{2}}}}={{2}^{x}}(1-{{2}^{x}})\] \[{{y}_{1}}+{{y}_{2}}=x+{{\log }_{2}}(1-{{2}^{x}})\]
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