## Wednesday, April 8, 2015

### $\scriptsize \log_{24}(48)+\log_{12} (54)>4$

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

For any logarithms problem (be it solving a logarithmic equation or proving a logarithmic inequality), one would usually use the change of base formula to rewrite the two logarithms terms so that they have the same base. Also, one would be sorely tempted to rewrite the LHS of the inequality by breaking down the figures 12, 24, 48 and 52 into the product of prime numbers:

[MATH]\color{yellow}\bbox[5px,green]{12=2^2\cdot 3}[/MATH], [MATH]\color{yellow}\bbox[5px,purple]{24=2^3\cdot 3}[/MATH], [MATH]\color{yellow}\bbox[5px,blue]{48=2^4\cdot 3}[/MATH] and [MATH]\color{yellow}\bbox[5px,orange]{54=2\cdot 3 ^3}[/MATH]

>Thus,

$\left(\log_{24}(48) \right)^2+ \left(\log_{12}(54) \right)^2$

$=\left(\log_{2^3\cdot 3}(2^4\cdot 3) \right)^2+ \left(\log_{2^2\cdot 3}(2\cdot 3 ^3) \right)^2$

In this case, we naturally want to change the base for both logarithmic terms to a base of 2:

$\left(\log_{24}(48) \right)^2+ \left(\log_{12}(54) \right)^2$

$=\left(\log_{2^3\cdot 3}(2^4\cdot 3) \right)^2+ \left(\log_{2^2\cdot 3}(2\cdot 3 ^3) \right)^2$

$=\left(\dfrac{\log_2 (2^4\cdot 3)}{\log_2 (2^3\cdot 3)}\right)^2+ \left(\dfrac{\log_2 (2^2\cdot 3)}{\log_2(2\cdot 3 ^3)} \right)^2$

$=\left(\dfrac{4+\log_2 (3)}{3+\log_2 (3)}\right)^2+ \left(\dfrac{2+\log_2 (3)}{1+3\log_2 (3)} \right)^2$

Okay, now, what would be the next best move to conclude that the above expression is greater than 4?

We can, of course, try to minimize the numerator and at the same time maximize the denominator so we could determine the $x$, where

$\left(\log_{24}(48) \right)^2+ \left(\log_{12}(54) \right)^2=\left(\dfrac{4+\log_2 (3)}{3+\log_2 (3)}\right)^2+ \left(\dfrac{2+\log_2 (3)}{1+3\log_2 (3)} \right)^2 >x$

It is obvious that

$\log_2 3>\log_2 2$

and it requires some thought to arrive at $\log_2 3<1.6$ where:

$256>243$

$2^8>3^5$

$2^{\dfrac{8}{5}}_{\phantom{i}}>3$

$\log_2 3<\dfrac{8}{5}$

$\log_2 3<1.6$

Hence, our best bet now is:

$\left(\log_{24}(48) \right)^2+ \left(\log_{12}(54) \right)^2$

$=\left(\dfrac{4+\log_2 (3)}{3+\log_2 (3)}\right)^2+ \left(\dfrac{2+\log_2 (3)}{1+3\log_2 (3)} \right)^2$

$>\left(\dfrac{4+\log_2 (2)}{3+\log_2 (3)}\right)^2+ \left(\dfrac{2+\log_2 (2)}{1+3\log_2 (3)} \right)^2$

$>\left(\dfrac{5}{3+\log_2 (3)}\right)^2+ \left(\dfrac{3}{1+3\log_2 (3)} \right)^2$

$>\left(\dfrac{5}{3+1.6)}\right)^2+ \left(\dfrac{3}{1+3(1.6)} \right)^2$

$>\left(\dfrac{5}{3+1.6}\right)^2+ \left(\dfrac{3}{1+3(1.6)} \right)^2$

$>1.449013125$

Argh!!! That doesn't help one bit at all! What we wanted to prove is that $\left(\log_{24}(48) \right)^2+ \left(\log_{12}(54) \right)^2>4$ and it would be useless to prove it greater than any value that is less than 4.

It appears like we have reached an impasse, when we thought we could improve our work by recognizing $\log_2 3>\log_2 2$ and $\log_2 3<1.6$, but the eventual result leaves too much to be desired.

Please don't feel aggravated or stupid, I would say all teachers and professional educators would agree that to have some idea to nibble at is far better than we are blanking ourselves out.

At this juncture, we would temporarily admit defeat since the above approach did not work and we HAVE to think of an alternative, the credible one to prove that $\left(\log_{24}(48) \right)^2+ \left(\log_{12}(54) \right)^2 >4$.

Would you want to try the problem out by picking up a pencil and start to scribble? If you failed after attempting, please stay tuned as we will reveal the solution to it soon, and if you solved it, would you pretty please to share your solution with us and at the same time check the shared solution of us so you are one step closer to become a real problem solver and 21st century critical thinker?