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Differential Calculus

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31 Mar 2009 10:37:55 IST

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log{(log)²-5logx+6}...domain

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pranav agrawal

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31 Mar 2009 10:42:58 IST

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log{(log)²-5logx+6}

now {(log)²-5logx+6} > 0 ( basic requirement for log to be defined)

now put log x = t

so t² - 5t + 6 >0

(t-3)(t-2) > 0

so t>3 or t<2

hence x> 1000 or 0<x< 100

rate if correct~~~~~

sneha

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31 Mar 2009 10:43:29 IST

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x > 0

and log²x -5 logx + 6 > 0

let log x = t

then t² -5t +6 >0

this gives (t-2) (t-3) >0

this means t >3 or t<2

this means logx >3 or logx < 2

sneha

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31 Mar 2009 10:44:51 IST

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this gives logx > log 1000 or logx < log100

hence, x<100 or x>1000

hence, 0< x < 100 or x > 1000

saharsha kumar keshkar

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31 Mar 2009 10:45:25 IST

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acc to ur ques,...it will be possible only if.

(logx)²-5logx+6>0

and log x is defined when x>0...

hence...

(logx - 2)(log x -3)>0

logx<2 & logx>3

x<100...which gives x belongs to (0,100)...as we know that x cannot be -ve..

x>1000..

hence answer is=> x (0 , 100) U (1000 , )

correct me if wrong.......

saharsha kumar keshkar

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31 Mar 2009 10:47:01 IST

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hello! it is a log function..and x cannot be -ve

sneha

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31 Mar 2009 10:48:31 IST

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if you are saying about my answer, it is the same as yours

sagar panda

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Joined: 5 Mar 2009

Posts: 25

31 Mar 2009 10:53:48 IST

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thanx!!!!!!!!!!!!!!

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