1y>o
each of x and xy and x^2y >0 , not equal to 1.
that means x>0 and xnotequal to 1
so x ε R^+ -{1}
If logxy*logxyy*logx2yy =1/6 the express y as a real function of x and indicate the domain of definition .
i have tried it and got x+xy+x2y=y6 now what next?
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6 Answers
1
62\frac{\log y}{\log x}\times\frac{\log y}{\log xy}\times\frac{\log y}{\log x^2y} =1/6
\frac{1}{\log_y x}\times\frac{1}{\log_y x+1}\times\frac{1}{2\log_y x+1} =\frac{1}{6}
{\log_y x}\times(\log_y x+1)\times(2\log_y x+1)={6}
t\times(t+1)\times(2t+1)={6}
t=1 is oen of the roots..
Can you find the others?
1i suppose we are asked to find the domain of definition and not the solutions .
11st see#3 in nishant bhaiya's forum he has taken
yt=x;
since v r talkin bout real value 4 y,it means y>0;so no matter wats the value of t......x is always positive. OR more appropritely solve the equation in #3 & find the solution 4 t,if itz negative den 0<x<1....or if +ve den x>1........hope u got it[1]
1i solved it the other two roots are non real so clearly y=x is the only solution for the question