Algebra-2

1 [by instinct :P]

hey are u guys goin to ask each option and when i say no , then the one remaining will be the answer

arey see .. i think 2y=3z=x then all the powers wud be zero...

so. it will be 1+2+3=6

no its not six

so options remaining are b and c

then its (c) for sure [3]

applying A.M≥G.M to equation

≥3³√x^{(log2y-log3z)}.(2y)^(log3z-logx).(3z)^(logx-log2y)

≥ 3³√e^{logx(log2y-log3z)}.e^{log2y(log3z-logx)}.e^{log3z(logx-log2y}

≥3³√e⁰
=3

nishant bhaiya is this method valid and is this correct

hey ek Q.
is it (2y) power

or 2 (y) power ??[7]

i guess it is correct...

@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)

@asish...

I'm sorry....i didn't see post #13

but 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)^....