A) \[\frac{2ac+1}{2a+abc+1}\]
B) \[\frac{2ac+1}{2a+c+a}\]
C) \[\frac{2ac+1}{2c+ab+a}\]
D) None of these
Correct Answer: D
Solution :
Now, \[{{\log }_{140}}=63={{\log }_{{{2}^{2}}\times 5\times 7}}(3\times 3\times 7)\] \[=\frac{{{\log }_{2}}(3\times 3\times 7)}{{{\log }_{2}}({{2}^{2}}\times 5\times 7)}=\frac{{{\log }_{2}}3+{{\log }_{2}}3+{{\log }_{2}}7}{2{{\log }_{2}}2+{{\log }_{2}}5+{{\log }_{2}}7}\] \[=\frac{2a+\frac{1}{c}}{2+b+\frac{1}{c}}=\frac{2ac+1}{2c+bc+1}\]
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