# How to prove this logarithmic inequality?

#### anemone

##### MHB POTW Director
Staff member
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
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

#### anemone

##### MHB POTW Director
Staff member
Hi CB, a big thank for your help in making it so straightforward and simple for me!

Thanks.