A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
\[{{\operatorname{logl}}_{x}}4+{{\log }_{x}}16+{{\log }_{x}}64=12\] \[{{\log }_{x}}{{2}^{2}}+{{\log }_{x}}{{2}^{4}}+{{\log }_{x}}{{2}^{6}}=12\] \[2{{\log }_{x}}2+4{{\log }_{x}}2+6{{\log }_{x}}2=12\] \[\Rightarrow \] \[12{{\log }_{x}}2=12\] \[\Rightarrow \] \[{{\log }_{x}}2=1\]\[\Rightarrow \]\[x=2\]
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