11 [by instinct :P]
if x,y,z are positive then minimum value of
xlog2y-log3z+2ylog3z-logx+3zlogx-log2y
a)1 b)12 c)3 d)6
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16 Answers
1hey are u guys goin to ask each option and when i say no , then the one remaining will be the answer
106arey see .. i think 2y=3z=x then all the powers wud be zero...
so. it will be 1+2+3=6
1asish give the solution, i have got only the answers , but i dont have solution, so plz give the solution, even i ll show u my half solution
1applying A.M≥G.M to equation
≥33√x(log2y-log3z).(2y)(log3z-logx).(3z)(logx-log2y)
≥ 33√elogx(log2y-log3z).elog2y(log3z-logx).elog3z(logx-log2y
≥33√e0
=3
nishant bhaiya is this method valid and is this correct
11@amit...
you are absolutely correct (post #10)
@asish...
even what you were thinking was rite (post #6)
but...you made 1 mistake.
x=2y=3z : true
but...i suppose,in the next step you put x=y=z=1.So you got 1+2+3=6.Instead if you had put x=1,y=1/2,z=1/3 i.e. x=2y=3z=1 you wud have got the correct ans !!! (1+1+1=3)
106but i dint put x=y=1 then how can
x=2y=3z ... i was just mentioning that the powers wud cancel... then it wud be x0 + 2*y0 + 3*z0 = 6..
in that ur making the mistake of (2y).... and i thot 2*(y)....