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Algebra

New kid on the Block

 Joined: 7 Feb 2007 Post: 7
7 Feb 2007 11:58:57 IST
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a problem
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

log x/4log 2=log y/6log 2=log z/3klog 2 and x.x.x.y.y.z=1 find k

Hot goIITian

Joined: 18 Dec 2006
Posts: 146
7 Feb 2007 16:22:59 IST
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logx/log16=logy/log64=logz/klog8=m(say)

taking log both sides in second given equation and substituting from above equation
3mlog16 +2mlog64 +mklog8=0
8^k=16*16*16*64*64=8^8
hence,k=8

Blazing goIITian

Joined: 15 Dec 2006
Posts: 321
7 Feb 2007 17:12:14 IST
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Hi malay,
Ur ans had a small flaw in it.........
It's (8)-k not (8)k  ;
So, k = - 8 .......ans.

Blazing goIITian

Joined: 4 Jan 2007
Posts: 486
7 Feb 2007 17:25:32 IST
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logx/4log2=logy/6log2
3logx=2logy
x3=y2..................(1)
logx/4log2=logz/3k log2
logx/4=log(x-3y-2)/3k

logx/4=log(x-6)/3k
1/4= -6/3k
k=  - 8  Ans

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