A) 64
B) 4
C) 8
D) 32
Correct Answer: D
Solution :
We have, \[{{\log }_{4}}2+{{\log }_{4}}4+{{\log }_{4}}x+{{\log }_{4}}16=6\] \[\Rightarrow \] \[{{\log }_{4}}(2\times 4\times x\times 16)=6\] \[\Rightarrow \] \[128x={{4}^{6}}\] \[\Rightarrow \] \[{{4}^{3}}\times 2x={{4}^{6}}\] \[\Rightarrow \] \[x=\frac{{{4}^{3}}}{2}\] \[\Rightarrow \] \[x=32\]
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