A) 61
B) 17
C) 68
D) 34
Correct Answer: C
Solution :
[C]\[5\,\,({{\log }_{y}}x+{{\log }_{x}}y)=26,\] \[xy=64\] |
\[\Rightarrow 5\left( {{\log }_{y}}x+\frac{1}{{{\log }_{y}}x} \right)=26\] |
Put \[{{\log }_{y}}x=t\] |
\[\therefore 5\,\,({{t}^{2}}+1)=26t\]\[\Rightarrow 5{{t}^{2}}-26t+5=0\]\[\Rightarrow t=5,\,\,\frac{1}{5}\] |
\[\therefore {{\log }_{y}}x=5,\,\,\frac{1}{5}\] |
\[x={{y}^{5}},\] \[{{y}^{1/5}}\] |
But \[xy=64\] \[\Rightarrow {{y}^{5}}.y=64\] or \[{{y}^{1/5}}.y=64\] |
\[\therefore \,\,y=2\] or 32 |
\[\therefore \] Solution set will be (32, 2) & (2, 32) |
\[\therefore \,\,a+b+c+d=32+2+2+32=68\] |
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