A) 0
B) \[\frac{1}{5}\]
C) 1
D) \[\frac{2}{5}\]
Correct Answer: C
Solution :
(c): \[\frac{1}{2}lo{{g}_{10}}25-2lo{{g}_{10}}4+lo{{g}_{10}}32+lo{{g}_{10}}1\] \[=\frac{1}{2}{{\log }_{10}}{{(5)}^{2}}-2{{\log }_{10}}{{(2)}^{2}}+{{\log }_{10}}{{(2)}^{5}}+0\]
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