Prove the logarithmic inequality Prove that: (log_(24)(48))^2+(log_(12)(54))^2>4 I tried to put t=log_2 3 and get the equation 6t^4+32t^3+22t^2-84t-74>0 . But I can't do anything with it...

racmanovcf

racmanovcf

Answered question

2022-10-28

Prove the logarithmic inequality
Prove that: ( log 24 48 ) 2 + ( log 12 54 ) 2 > 4
I tried to put t = log 2 3 and get the equation
6 t 4 + 32 t 3 + 22 t 2 84 t 74 > 0. But I can't do anything with it...

Answer & Explanation

wlanauee

wlanauee

Beginner2022-10-29Added 17 answers

Let l = log 2 3, so log 3 2 = 1 l and l < 8 5 since 2 8 > 3 5 , and l > 11 7 since 2 11 < 3 7
Then log 24 48 = log 2 48 log 2 24 = 4 + l 3 + l > 4 + 8 / 5 3 + 8 / 5 = 28 23 and log 12 54 = log 3 54 log 3 12 = 3 + log 3 2 1 + 2 log 3 2 = 3 + 1 / l 1 + 2 / l = 3 l + 1 l + 2 > 3 ( 11 7 ) + 1 11 7 + 2 = 8 5
so ( log 24 48 ) 2 + ( log 12 54 ) 2 > ( 28 23 ) 2 + ( 8 5 ) 2 > 4
racmanovcf

racmanovcf

Beginner2022-10-30Added 2 answers

I would suggest moving the four over to the left, and the log with the 54 to the right. Then try changing the 4 into a 2 2 , which you will want to change to a logarithm in base 24. From there you can turn all of your logarithms into ratios of natural logs. You'll notice you have a difference of two squares on the left, which you can start to factor. It should clear up around here.

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