\(\cfrac{1}{log_{x}xy}\) + \(\cfrac{1}{log_{y}xy}\)1/logx xy + 1/log

1.	\(\cfrac{1}{log_{x}xy}\) + \(\cfrac{1}{log_{y}xy}\)1/logx xy + 1/logy xy A) 0 B) 1 C) -1 D) 2
Answer» Correct option is (B) 1 \(\frac{1}{log_x\,xy} + \frac{1}{log_y\,xy} \) \(=\frac{1}{log_x\,x+log_x\,y} + \frac{1}{log_y\,x+log_y\,y} \) \((\because\) log AB = log A + log B) \(=\frac{1}{1+log_x\,y} + \frac{1}{1+log_y\,x} \) \((\because log_aa=1)\) \(=\frac{1}{1+log_x\,y} + \frac{1}{1+\frac1{log_x\,y}} \) \((\because log_a\,b=\frac1{log_b\,a})\) \(=\frac{1}{1+log_x\,y} + \frac{log_x\,y}{1+log_x\,y} \) \(=\frac{1+log_x\,y}{1+log_x\,y}=1\) Correct option is B) 1

\(\cfrac{1}{log_{x}xy}\) + \(\cfrac{1}{log_{y}xy}\)1/logx xy + 1/logy xy A) 0 B) 1 C) -1 D) 2

Answer»

Correct option is (B) 1

\(\frac{1}{log_x\,xy} + \frac{1}{log_y\,xy} \) \(=\frac{1}{log_x\,x+log_x\,y} + \frac{1}{log_y\,x+log_y\,y} \) \((\because\) log AB = log A + log B)

\(=\frac{1}{1+log_x\,y} + \frac{1}{1+log_y\,x} \) \((\because log_aa=1)\)

\(=\frac{1}{1+log_x\,y} + \frac{1}{1+\frac1{log_x\,y}} \) \((\because log_a\,b=\frac1{log_b\,a})\)

\(=\frac{1}{1+log_x\,y} + \frac{log_x\,y}{1+log_x\,y} \)

\(=\frac{1+log_x\,y}{1+log_x\,y}=1\)

Correct option is B) 1