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Unformatted text preview: ( x )(3 x2) = 1 3 x 22 x1 = 0 ( x1)(3 x + 1) = 0 So x = 1 ,1 3 . lg ( u ) = 1 u = 10 1 = 10 lg ( u ) =1 3 u = 101 3 = 1 3 √ 10 So u = 10 , 1 3 √ 10 . 6. Find x : x 11 2 (lg x ) 21 2 √ 10 = 0 First of all, x > 0. x 11 2 (lg x ) 21 2 √ 10 = 0 x 11 2 (lg x ) 2 = 1 2 √ 10 log 10 [ x 11 2 (lg x ) 2 ] = log 10 1 2 √ 10 [11 2 (log 10 x ) 2 ] log 10 ( x ) =1 2 Set y = log 10 ( x ). y1 2 y 3 =1 2 y 32 y1 = 0 By remainder theorem, 1 is a root. y 32 y1 = ( y + 1)( y 2y1) By quadratic formula, the other two roots are: y = 1 ± p (1) 24 * 1 * (1) 2 y = 1 ± √ 5 2 Therefore: log 10 ( x ) =1 x = 1 10 OR log 10 ( x ) = 1 ± √ 5 2 x = 10 1 ± √ 5 2 x = √ 10 * 10 ± √ 5 2 So we get three possible x values....
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 Fall '10
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 Linear Algebra, Algebra, Quadratic equation, Harshad number, LG, 48 digits

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