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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Jun 2007 22:34:01 IST
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Hey ! I solved this question , but was getting a slightly different answer . Pls check - Solve for all values of x : logx(2x-3/4) > 2 . Dont forget to take all cases !!!!!
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Umang |
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19 Jun 2007 22:54:40 IST
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1/2 and 3/2?
and what cases r u saying.. maybe i left loads out..
constraints: x is greater than zero, greater than 3/8,x cannot = 1, 0.
what else?
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19 Jun 2007 22:57:02 IST
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x has to be positive...( 3/8 , infinity ) - { 1 }...base cant be 1...
so...x^2 - 2x + 3/4 < 0
implies... x belongs to ( 1/3 , 1/2 )...
therefore....the reqd values of x are .....( 3/8 , 1/2 )....ANS
n dont 4gt 2 rate me...
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............tseb eht ma i |
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19 Jun 2007 22:58:33 IST
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just combine all ur results and form one result The answer is x belongs to ( 3/8 , 1/2 ) union ( 1, 3/2)
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Umang |
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19 Jun 2007 22:58:56 IST
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case 1) x>1 2x - 3/4 >x2 (2x-3)(2x-1)<0 x  (1/2,3/2) but acc to case ans for case is x (1,3/2) case 2) 0<x<1 (2x-3)(2x-1) >0 acc. to case ans is (0,1/2) net result is x (0,1/2)  (1,3/2) |
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19 Jun 2007 23:00:48 IST
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Hey him26.89 ,
i hav already posted the answer . All r getting it different
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Umang |
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19 Jun 2007 23:05:37 IST
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Hey him ! u left one case , and thats 2x-3/4>0 . here u get x>3/8 . Just combine and u get the correct answer . Thank u all for contributing !!!!!
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Umang |
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19 Jun 2007 23:09:49 IST
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hey you are sharp him2689!! good job
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