How to solve log_(x^2)−3(x^2+6x)<logx(x+2)?

kadejoset 2022-07-17 Answered
How to solve log x 2 3 ( x 2 + 6 x ) < log x ( x + 2 )?
How to solve the following inequality
log x 2 3 ( x 2 + 6 x ) < log x ( x + 2 )   ??
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Answers (1)

escobamesmo
Answered 2022-07-18 Author has 18 answers
Note first that x A B where A = ( 3 , 2 ) , B = ( 2 , ), for the question to make sense.
Now if x A, note that 0 < x 2 3 < 1 < x < x + 2 < x 2 + 6 x, so LHS < 0, while R H S > 0 so this is a solution.
For x B, we have 1 < x < x + 2 and 1 < x 2 3 < x 2 + 6 x, so both LHS and RHS are positive. Here we have the equivalent inequality:
log x ( x + 2 ) > log x 2 3 ( x 2 + 6 x ) log ( x 2 3 ) log ( x + 2 ) > log x log ( x 2 + 6 x ) log x ( x 2 3 ) > log x + 2 ( x 2 + 6 x )
Now note that x 2 3 < x 2 while x 2 + 6 x > ( x + 2 ) 2 for x B

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