A) 3
B) 1
C) 2
D) 0
Correct Answer: B
Solution :
\[{{\log }_{4}}(x-1)={{\log }_{2}}(x-3)={{\log }_{{{4}^{1/2}}}}(x-3)\] \[\Rightarrow \] \[{{\log }_{4}}(x-1)=2{{\log }_{4}}(x-3)\] \[\Rightarrow \] \[{{\log }_{4}}(x-1)={{\log }_{4}}{{(x-3)}^{2}}\] \[\Rightarrow \] \[(x-1)={{(x-3)}^{2}}\] \[{{x}^{2}}+9-6x=x-1\] \[\Rightarrow \] \[{{x}^{2}}-7x+10=0\] \[\Rightarrow \] \[(x-2)(x-5)=0\] \[\Rightarrow \] \[x=2\] or \[x=5\] Hence,\[x=5\] [\[\because \,\,x=2\]makes\[\log (x-3)\]undefined.]
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