## Thursday, April 9, 2015

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

Have you tried out any other approaches that you could think of to solve the previous "$\log_{24}(48)+\log_{12} (54)>4$"?

At any rate, that is one very tough problem and it could cause you lose sleep over it if you're persistent enough and have the absolute determination to prove the statement. But, we know as it have been our experience that nothing would be more frustrating and aggravating that when we put up the titanic fight to solve for the problem, the attempts at the solution were never correct. After the complete failures, you began to wonder if the extra effort would hardly be worth it and if you should just ditch the problem in the first place once you realized there is nothing that you could have thought to tackle the problem successfully.

Yes, this seems like it would be a task of immense intelligence effort to prove for this logarithmic inequality where $\left(\log_{24}(48) \right)^2+ \left(\log_{12}(54) \right)^2 >4$.

We will lead you to the solution at this post and read on, as the solution will open your eyes:

First, note that

\begin{align*}\log_{24}(48)&=\log_{24}(24\cdot 2)\\&=1+\log_{24}(2)\end{align*}

Now, our effort should focus on to determine what value exactly is $x$, when [MATH]\color{yellow}\bbox[5px,blue]{\log_{24}(2)>x}[/MATH]?

This is equivalent in checking out what possible values $x$ could be if we have $2^{\dfrac{1}{x}}_{\phantom{i}}>24$?

When we put it that way, isn't it obvious that $2^{\dfrac{1}{x}}_{\phantom{i}}$ can take on any value starts from 25?

We of course want the smallest and "sweet" value from the list. The thing that we also need to bear in mind is we need to prove the LHS of the inequality is greater than 4.

We all know $2^5=32>24$, as $2^{\dfrac{1}{x}}_{\phantom{i}}>24$, so we have $\dfrac{1}{x}=5$ or simply $x=\dfrac{1}{5}$.

From this we obtain [MATH]\color{yellow}\bbox[5px,blue]{\log_{24}(2)>\dfrac{1}{5}}[/MATH]  and hence

[MATH]\color{yellow}\bbox[5px,blue]{\begin{align*}(\log_{24}(48))^2&=(\log_{24}(24\cdot 2))^2\\&=\left(1+\log_{24}(2)\right)^2>\left(1+\dfrac{1}{5}\right)^2>1.44\end{align*}}[/MATH]

Work thing out in the similar fashion for the second term, i.e. $$\log_{12}(54)$$, we see that

\begin{align*}\log_{12}(54)&=\log_{12}\left(12\cdot \dfrac{54}{12}\right)\\&=1+\log_{12}\left(\dfrac{54}{12}\right)\\&=1+\log_{12}(4.5)\end{align*},

Now, what is $y$ if [MATH]\color{yellow}\bbox[5px,purple]{\log_{12}(4.5)>y}[/MATH]?

In other words, we want to figure out the $y$ value as in $4.5^{\dfrac{1}{y}}_{\phantom{i}}>12$. You might want to tell me to proceed with $4.5^2=20.25>12$, yes, of course that is correct, but would this lead to the answer? We will see:

If $\dfrac{1}{y}=2,\,\,y=\dfrac{1}{2}$, then [MATH]\color{yellow}\bbox[5px,purple]{\log_{12}(4.5)>\dfrac{1}{2}}[/MATH]  and hence

[MATH]\color{yellow}\bbox[5px,purple]{\begin{align*}(\log_{12}(54))^2&=\left(\log_{12}\left(12\cdot \dfrac{54}{12}\right)\right)^2\\&=\left(1+\log_{12}\left(\dfrac{54}{12}\right)\right)^2>\left(1+\dfrac{1}{2}\right)^2=\dfrac{9}{4}=2.25\end{align*}}[/MATH]

But note that $\log_{24}(48))^2+\log_{12}(54))^2>1.44+2.25=3.69$ and $3.69<4$. That means our effort in determining the $y$ value has gone ALL the way wrong...we need a better value of $\dfrac{1}{y}$ so that when it sums up with 1.44 will give us a 4...

Should this be another headache problem, because it seems like it is not exactly clear to us what value of $y$ should we use, such that we obtained the sum of $1.44+(1+y)^2=4$.

No, to find out y from the equation above is purely arithmetic:

$1.44+(1+y)^2=4$

$y=+\sqrt{4-1.44}-1=0.6=\dfrac{3}{5}$

Back substituting this to see if $y=\dfrac{3}{5}$ works, we have:
[MATH]\color{yellow}\bbox[5px,green]{4.5^{\dfrac{1}{y}}_{\phantom{i}}=4.5^{\dfrac{5}{3}}_{\phantom{i}}>12}[/MATH]

And we need to determine if the above inequality holds. But how? By using your calculator! You must now check if

$4.5^{\dfrac{5}{3}}_{\phantom{i}}>12$

$4.5^5_{\phantom{i}}>12^3\,\,\,\rightarrow\,1845.28125>1728$ Yes, that inequality holds, hurray!

You could also check it the following way:

$\left(\dfrac{9}{2}\right)^5_{\phantom{i}}>12^3$

$9^5_{\phantom{i}}>2^5\cdot 12^3\,\,\,\rightarrow\,59049>55296$ This suggests that the inequality [MATH]\color{yellow}\bbox[5px,green]{4.5^{\dfrac{1}{y}}_{\phantom{i}}=4.5^{\dfrac{5}{3}}_{\phantom{i}}>12}[/MATH] is indeed true.

To sum up, we could write:

Since $2^5=32>24$, this gives us $\log_{24}(2) > \dfrac{1}{5}$ and $\left(\dfrac{9}{2}\right)^5_{\phantom{i}}>12^3$, this suggests $\log_{12}(4.5) > \dfrac{3}{5}$

And therefore

\begin{align*}(\log_{24}(48))^2+(\log_{12}(54))^2&=\left(1+\log_{24}(2)\right)^2+\left(1+\log_{12}(4.5)\right)^2\\&>\left(1+\dfrac{1}{5}\right)^2+\left(1+\dfrac{3}{5}\right)^2=4\,\,\,\text{and we're done.}\end{align*}