Solving the logarithimic inequality log_2(x)/(2) + (log_2x^2)/(log_2 (2/x)) <= 1 several times but keeping getting wrong answers.

kadejoset

kadejoset

Answered question

2022-07-15

Solving the logarithimic inequality log 2 x 2 + log 2 x 2 log 2 2 x 1
I tried solving the logarithmic inequality:
log 2 x 2 + log 2 x 2 log 2 2 x 1
several times but keeping getting wrong answers.

Answer & Explanation

Selden1f

Selden1f

Beginner2022-07-16Added 14 answers

Let log 2 x = A, then log 2 x 2 = 2 log 2 x = 2 A and log 2 2 x = log 2 2 log 2 x = 1 A. So the given inequality becomes:
( A 1 ) + 2 A 1 A 1.
Consequently we get
4 A A 2 1 1 A 1.
Furthermore you get
5 A A 2 2 1 A 0.
Hopefully you can solve from here.
kominis3q

kominis3q

Beginner2022-07-17Added 2 answers

Put u = log 2 ( x 2 ) = log 2 x 1
Note that log 2 ( x 2 ) = 2 ( u + 1 ), and log 2 ( 2 x ) = u
Hence inequality becomes
u 2 ( u + 1 ) u 1 u 2 3 u 2 u 0 ( u α ) ( u β ) u 0
where α = 3 + 17 2 ,   β = 3 17 2  
u β ,   0 u α ,   log 2 ( x 2 ) 3 17 2 ,   0 log 2 ( x 2 ) 3 + 17 2 log 2 x 5 17 2 ,   1 log 2 x 5 + 17 2
As x > 0,
0 < x 2 5 17 2 ,   2 x 2 5 + 17 2

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