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Joined:	7 Feb 2007
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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

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malay

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Joined: 18 Dec 2006

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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

vinu

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7 Feb 2007 17:12:14 IST

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Hi malay,

Ur ans had a small flaw in it.........

It's (8)^-knot (8)^k ;

So, k = - 8 .......ans.

yahiyafirdous

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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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