Find the product of the positive roots of the equation ( 2008 ) ( x ) log 2008 x = x 2 .

Given that 2008 x log 2008 x = x 2 Apply the logarithm both sides, we get log 2008 2008 + log 2008 x log 2008 x = log 2008 x 2 ⇒ 1 2 log 2008 ( 2008 ) + log 2008 x log 2008 x = 2 log 2008 x ⇒ 1 2 + ( log 2008 x ) 2 = 2 ( log 2008 x ) ⇒ 2 ( log 2008 x ) 2 - 4 log 2008 x + 1 = 0 ⇒ log 2008 x = 4 ± 16 - 4 × 2 4 = 4 ± 2 2 4 = 1 ± 2 2 Let x 1 and x 2 are the roots of equation , then log 2008 x 1 = 1 + 2 2 , log 2008 x 2 = 1 - 2 2 ⇒ log 2008 x 1 + log 2008 x 2 = 2 ⇒ log 2008 x 1 x 2 = 2 ⇒ x 1 x 2 = ( 2008 ) 2

Solve : log 3 ( x + | x - 1 | ) = log 9 ( 4 x - 3 + 4 | x - 1 | )

Given that log 3 ( x + | x - 1 | ) = log 9 ( 4 x - 3 + 4 | x - 1 | ) Case : 1 When 0 ≤ x < 1 , ⇒ log 3 ( x - ( x - 1 ) ) = log 9 ( 4 x - 3 - 4 ( x - 1 ) ) ⇒ log 3 ( 1 ) = log 9 ( 1 ) = 0 Case : 2 When x ≥ 1 , ⇒ log 3 ( x + x - 1 ) = log 9 ( 4 x - 3 + 4 ( x - 1 ) ) ⇒ log 3 ( 2 x - 1 ) = log 9 ( 8 x - 7 ) = 1 2 log 3 8 x - 7 ⇒ log 3 ( 2 x - 1 ) 2 = log 3 ( 8 x - 7 ) ⇒ 4 x + 1 - 4 x = 8 x - 7 ⇒ 4 x + 8 = 12 x ⇒ x - 3 x + 2 = 0 ⇒ x - 2 x - 1 = 0 ⇒ x = 4 , x = 1 ∴ x ∈ 0 , 1 ∪ 4 .

Find the value of x satisfying the equation, log 3 3 x 3 + log x 3 x 3 · log 3 x 3 + log 3 x / 3 3 + log x 3 / x 3 · log 3 x 3 = 2

Given log 3 3 x 3 + log x 3 x 3 · log 3 x 3 + log 3 x / 3 3 + log x 3 / x 3 · log 3 x 3 = 2 ⇒ log 3 3 x 1 / 3 + log x 3 x 1 / 3 · log 3 x 3 + log 3 x / 3 1 / 3 + log x 3 / x 1 / 3 · log 3 x 3 = 2 Using log a m n = n log a m , ⇒ 1 3 log 3 3 x + 1 3 log x 3 x · 3 log 3 x + 1 3 log 3 x / 3 + 1 3 log x 3 / x · 3 log 3 x = 2 Using log a m n = log a m + log a n , log a m n = log a m - log a n , and log a a = 1 , ⇒ 1 + log 3 x + 1 + log x 3 · log 3 x + log 3 x - 1 + log x 3 - 1 · log 3 x = 2 Using log a x · log x a = 1 ⇒ log 3 x 2 + 2 log 3 x + 1 + log 3 x 2 - 2 log 3 x + 1 = 2 ⇒ log 3 x + 1 2 + log 3 x - 1 2 = 2 ⇒ log 3 x + 1 + log 3 x - 1 = 2 . . . i We know that x = x , x ≥ 0 - x , x < 0 and if log a x = b , x > 0 , a > 0 , a ≠ 1 then x = a b . Case I : log 3 x > 1 , ⇒ x > 3 Then from the equation i , we have ⇒ log 3 x + 1 + log 3 x - 1 = 2 ⇒ log 3 x + log 3 x = 2 ⇒ log 3 x = 1 ⇒ x = 3 , but we have x > 3 , hence there is no solution for log 3 x > 1 or x > 3 . Case I I : - 1 ≤ log 3 x ≤ 1 , ⇒ 1 3 ≤ x ≤ 3 . ⇒ log 3 x + 1 + 1 - log 3 x = 2 ⇒ 2 = 2 , hence the equation is satisfied for all values of x in 1 3 , 3 but for x = 1 , the base of the logarithm is not defined, hence the solution is x ∈ 1 3 , 3 - { 1 } . Case I I I : log 3 x < - 1 ⇒ x < 1 3 ⇒ - log 3 x - 1 - log 3 x + 1 = 2 ⇒ - 2 log 3 x = 2 ⇒ log 3 x = - 1 , ⇒ x = 1 3 . Hence, the solution is x ∈ 1 3 , 3 - { 1 } since base can not be 1 .