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Erick Clay

Erick Clay

Answered question

2022-05-24

Solve for: 2 log 3 ( x 2 4 ) + 3 log 3 ( x + 2 ) 2 log 3 ( x 2 ) 2 4
My try:
2 log 3 ( x 2 4 ) + 3 log 3 ( x + 2 ) 2 log 3 ( x 2 ) 2 4 log 3 ( x 2 4 ) 2 + 3 log 3 ( x + 2 ) 2 log 3 ( x 2 ) 2 4 log 3 [ ( x 2 ) 2 × ( x + 2 ) 2 ] + 3 log 3 ( x + 2 ) 2 log 3 ( x 2 ) 2 4 log 3 ( x 2 ) 2 + log 3 ( x + 2 ) 2 + 3 log 3 ( x + 2 ) 2 log 3 ( x 2 ) 2 4 log 3 ( x + 2 ) 2 + 3 log 3 ( x + 2 ) 2 4 0 ( )
Put: t = log 3 ( x + 2 ) 2 ( ) t 2 + 3 t 4 0
But I don't know Conditions defined for this math? Could help me?

Answer & Explanation

Boatein

Boatein

Beginner2022-05-25Added 10 answers

The domain of the first inequation is x > 3 or x 3, so the range of your t is t 0. Then you can use t 0 as the domain to solve t 2 + 3 t 4 0, which gets 0 t 1. At last, you just go back to solve 0 t 1

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