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How to prove this logarithmic inequality?

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
Hi all, I've been having a hard time trying to solve the following inequality:

Prove that $\displaystyle \left(\log_{24}(48) \right)^2+\displaystyle \left(\log_{12}(54) \right)^2 >4$

I've tried to change the bases to base-10 log and relating all the figures (12, 24, 48, and 54) in terms of 2 and 3 but only to make the problem to be more confounded.

Could I get some hints on how to tackle this problem?

Any help would be deeply appreciated.

Thanks!

P.S. This question was originally asked here (Logarithm) at MMF.
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
Hi all, I've been having a hard time trying to solve the following inequality:

Prove that $\displaystyle \left(\log_{24}(48) \right)^2+\displaystyle \left(\log_{12}(54) \right)^2 >4$

I've tried to change the bases to base-10 log and relating all the figures (12, 24, 48, and 54) in terms of 2 and 3 but only to make the problem to be more confounded.

Could I get some hints on how to tackle this problem?

Any help would be deeply appreciated.

Thanks!

P.S. This question was originally asked here (Logarithm) at MMF.
\(\log_{24}(48)=1+\log_{24}(2)\)

But \(2^5 \gt 24\) so \(\log_{24}(2) \gt 1/5\)

Also: \(\log_{12}(54)=1+\log_{12}(4.5)\), and \(4.5^5>12^3\) so \(\log_{12}(4.5)>3/5\)

Hence:
\[ (\log_{24}(48))^2 + (\log_{12}(54))^2 \gt 1.2^2+1.6^2 =4 \]

CB
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
Hi CB, a big thank for your help in making it so straightforward and simple for me!

Thanks.