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 .

log 5 8 2 x 2 - x - 3 8 ≥ 1

Given that : log 5 8 2 x 2 - x - 3 8 ≥ 1 As we know, log a b ≥ c ⇒ b ≥ a c and a > 0 , a ≠ 0 , b > 0 Since the base of the log is between 0 & 1 So, 2 x 2 - x - 3 8 ≤ 5 8 ⇒ 16 x 2 - 8 x - 3 ≤ 5 ⇒ 16 x 2 - 8 x - 3 - 5 ≤ 0 ⇒ 16 x 2 - 8 x - 8 ≤ 0 ⇒ 2 x 2 - x - 1 ≤ 0 ⇒ 2 x 2 - 2 x + x - 1 ≤ 0 ⇒ 2 x ( x - 1 ) + ( x - 1 ) ≤ 0 ⇒ ( x - 1 ) ( 2 x + 1 ) ≤ 0 ⇒ x ∈ [ - 1 2 , 1 ] . . . ( 1 ) and Again, 2 x 2 - x - 3 8 > 0 ⇒ 16 x 2 - 8 x - 3 > 0 ⇒ 16 x 2 - 12 x + 4 x - 3 > 0 ⇒ 4 x ( 4 x - 3 ) + ( 4 x - 3 ) > 0 ⇒ ( 4 x - 3 ) ( 4 x + 1 ) > 0 ⇒ x ∈ ( ∞ , - 1 4 ) ∪ ( 3 4 , ∞ ) . . . ( 2 ) Now, from ( 1 ) & ( 2 ) x ∈ [ - 1 2 , 1 ] & x ∈ ( ∞ , - 1 4 ) ∪ ( 3 4 , ∞ ) ⇒ x ∈ [ - 1 2 , - 1 4 ) ∪ ( 3 4 , 1 ]

3 4 6 x + 10 - x 2 < 27 64

( 3 4 ) 6 x + 10 - x 2 < ( 3 4 ) 3 taking log both sides with base 3 4 ⇒ 6 x + 10 - x 2 > 3 ⇒ - x 2 + 6 x + 10 - 3 > 0 ⇒ - x 2 + 6 x + 7 > 0 ⇒ x 2 - 6 x - 7 < 0 ⇒ x 2 - 7 x + x - 7 < 0 ⇒ x ( x - 7 ) + ( x - 7 ) < 0 ⇒ ( x - 7 ) ( x + 1 ) < 0 ⇒ x ∈ ( - 1 , 7 )

Solve: log x 2 · log 2 x 2 . log 2 4 x > 1

Given: log x 2 . log 2 x 2 . log 2 4 x > 1 ∴ x > 0 , x ≠ 1 , 2 x ≠ 1 i . e . x ≠ 1 2 Now, 1 log 2 x . 1 log 2 2 x . log 2 4 + log 2 x > 1 2 + log 2 x log 2 x . log 2 2 + log 2 x > 1 ⇒ 2 + log 2 x log 2 x 1 + log 2 x > 1 Let log 2 x = t ⇒ 2 + t t . 1 + t - 1 > 0 ⇒ t + 2 - t 2 - t t t + 1 > 0 ⇒ t 2 - 2 t t + 1 < 0 ⇒ t - 2 t + 2 t t + 1 < 0 By sign-scheme method, t ∈ - 2 , - 1 ∪ 0 , 2 ∴ log 2 x ∈ - 2 , - 1 ∪ 0 , 2 ∴ - 2 < log 2 x < - 1 or 0 < log 2 x < 2 ∴ 2 - 2 < x < 2 - 1 or 2 0 < x < 2 2 ∴ 2 - 2 < x < 2 - 1 or 1 < x < 2 2