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Algebra

Blazing goIITian

Joined:	26 Mar 2009
Post:	1428

13 Jul 2009 17:00:23 IST

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Logarithms

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2x^logx + 3x ^-logx = 5

Solve

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Comments (3)

suguna

Cool goIITian

Joined: 23 Nov 2008

Posts: 77

13 Jul 2009 17:52:27 IST

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x = 1.eqn may be true if

x^{log x}=1.rate me if correct

BALGANESH

Blazing goIITian

Joined: 21 Dec 2007

Posts: 1025

13 Jul 2009 18:26:31 IST

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2x^logx +1/ 3x^logx = 5x^logx= y2y+3/y = 52y^2 -5y + 3 =0solving y=1 or y=3/2 case 1-x^logx=1 log x= 1for this x=1this is the required soltuion x=1

Sushobhan Sen

Blazing goIITian

Joined: 1 Jul 2009

Posts: 1416

14 Jul 2009 21:12:59 IST

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2x^logx + 3x ^-logx = 5

Let x^logx= t

The eq. becomes

2t + (3/t) = 5

2t² - 5t + 3 = 0

2t² -2t-3t+3=0

(t-1)(2t-3)=0

t = 1, 3/2

If x^logx=1, then (logx)²=0 [taking log on both sides] or x =1

If x^logx=3/2, then (logx)²= log (3/2) or x = antilog[sq.rt. log(3/2)]

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