# How to prove $\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$

How could we prove that
$$\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$$ I have reduced it the form $$\frac{4^{\ln(4)/\ln(3/4)}}{3^{\ln(3)/\ln(3/4)}}$$

I am not sure what to do next to get snappy solution. Any ideas?

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What does the subindex represent? Oh, it's not a subindex, just the result of lousy typesetting (because of the over-use of \left and \right. I'll fix it. – Arturo Magidin Jun 22 '12 at 15:49
Oops sorry Arturo. – VelvetThunder Jun 22 '12 at 15:52
Please, your typesetting is pretty bad. Now you've undone the fix I did. – Arturo Magidin Jun 22 '12 at 15:52
Well, you almost undid yet again my fixes. But you got your edit in before mine, so I overrode it. – Arturo Magidin Jun 22 '12 at 15:55
@Fool: I'm not late, I'm just wasting my time because you can't write the question you actually mean. Thanks for that. – Arturo Magidin Jun 22 '12 at 16:10

This refers to the original question, which had the left hand side equal to $\frac{1}{2}$ instead of $\frac{1}{12}$.

They are not equal.

Your simplification is correct. Then we can rewrite the left hand side as $$\left(\frac{4^{\ln(4)}}{3^{\ln(3)}}\right)^{1/\ln(3/4)}$$ so raising both sides of the equation to the $\ln(3/4)$ power, we get that the equation would be equivalent to $$\frac{4^{\ln(4)}}{3^{\ln(3)}} \stackrel{?}{=} \frac{1}{2}^{\ln(3/4)}.$$ Rewriting $4^{\ln(4)}$ as $e^{(\ln 4)^2}$, $3^{\ln(3)}$ as $e^{(\ln(3))^2}$, and $\left(\frac{1}{2}\right)^{\ln(3/4)}$ as $e^{-\ln(2)\ln(3/4)}$, the equality would be equivalent to $$\left(\ln 4\right)^2 - \left(\ln 3\right)^2 \stackrel{?}{=} -\ln(2)\ln\frac{3}{4}.$$

Now, $\ln(4) = 2\ln(2)$, and $\ln\frac{3}{4} = \ln 3 - 2\ln 2$. So the left hand side is equal to $$4(\ln 2)^2 - (\ln 3)^2$$ while the right hand side is equal to $$-\ln(2)(\ln 3 - 2\ln 2) = 2(\ln 2)^2 - (\ln 2)(\ln 3).$$

But $$4(\ln 2)^2 - (\ln 3)^2 \approx 0.714863$$ and $$2(\ln 2)^2 - (\ln 2)(\ln 3) \approx 0.199406$$

As corrected, the right hand side now be, after the simplification $$\left(\frac{1}{12}\right)^{\ln(3/4)} = \exp\left(-\ln(12)\ln(3/4)\right).$$ The exponent can be simplified: \begin{align*} -\ln(12)\ln(3/4) &= -\left(\ln(3)+2\ln(2)\right)\left(\ln(3)-2\ln(2)\right)\\ &= \left(\ln(3)+2\ln(2)\right)\left(2\ln(2)-\ln(3)\right)\\ &= \left(2\ln(2)\right)^2 - \left(\ln 3\right)^2\\ &= 4(\ln 2)^2 - (\ln 3)^2. \end{align*} Since this is the same as the exponent of $e$ on the left hand side, we do indeed have $$\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}.$$

There's nothing special about $3$ and $4$. Replacing them with arbitrary positive numbers $a$ and $b$ will lead to $$\exp\left((\ln(a))^2 - (\ln(b))^2\right) \stackrel{?}{=} \exp\left(-\ln(ab)(\ln(b/a)\right)$$ which of course holds, since $$-\ln(ab)\ln(b/a) = (\ln a + \ln b)(\ln a - \ln b)$$ giving the equality you have in the comment: $$\frac{a^{1/\log_a(b/a)}}{b^{1/\log_b(b/a)}} = \frac{1}{ab}.$$

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Thanks, I realized that in general $$\frac{a^{ 1/\log_a(\frac{b}{a})}}{b^{1/\log_b(\frac{b}{a})}}=\frac{1}{a\times b}$$ – VelvetThunder Jun 22 '12 at 16:29
@Arturo: I just saw that. For some reason the system isn’t automatically updating anything for me today. – Brian M. Scott Jun 22 '12 at 16:29
@Brian: No problem! – Arturo Magidin Jun 22 '12 at 16:29
When I was posting the question (the first time) there was no preview for me. Hence, I posted that version quickly to know how the latex look like. – VelvetThunder Jun 22 '12 at 16:31

This is another way, using the basics about logarithms. You can use just the identity $$b^{\rm{log}_{\frac{b}{a}}b}=b\cdot a^{\rm{log}_{\frac{b}{a}}b}$$

Note that $\frac{1}{\rm{log}_x{\frac{3}{4}}} =$ log$_\frac{3}{4} \ x$, you can use that to get

$$\frac{4^{\frac{1}{\rm{log}_4(3/4)}}}{3^{\frac{1}{\rm{log}_3(3/4)}}} = \frac{4^{\rm{log}_{3/4}4}}{3^{\rm{log}_{3/4}3}}$$

Using the identity we have that

$$3^{\rm{log}_{\frac{3}{4}}3}=3\cdot 4^{\rm{log}_{\frac{3}{4}}3}$$

replacing we get

$$\frac{4^{\rm{log}_{3/4}4}}{3^{\rm{log}_{3/4}3}}=\frac{4^{\rm{log}_{3/4}4}}{3\cdot 4^{\rm{log}_{\frac{3}{4}}3}}=\frac{4^{\rm{log}_{3/4}4-\rm{log}_{3/4}3}}{3}=\frac{4^{\rm{log}_{3/4}(4/3)}}{3}=\frac{4^{-1}}{3}=\frac{1}{12}$$

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$$\frac{4^{1/\log_4 3/4}}{3^{1/\log_3(3/4)}}=\frac{4^{\frac1{\log_43-1}}}{3^{\frac1{1-\log_34}}}=\frac{4^{\frac1{\log_43-1}}}{3^{\frac1{1-(\log_43)^{-1}}}}=\frac{4^{\frac1{\log_43-1}}}{3^{\frac{\log_43}{\log_43-1}}}=\left(\frac4{3^{\log_43}}\right)^{\frac1{\log_43-1}}$$

Taking the log base $4$, I get

$$\frac1{\log_43-1}\Big(1-(\log_43)^2\Big)=-(1+\log_43)\;.$$

Clearly $\log_4\frac1{12}=-\log_412=-(1+\log_43)$.

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Equivalently, we want to prove that $$4^{1+\frac{1}{\log_4(3/4)}}=3^{\frac{1}{\log_3(3/4)}-1}.\tag{1}$$ We play a little with the left-hand side of $(1)$. We have $$4^{1+\frac{1}{\log_4(3/4)}}=4^{1+\frac{1}{\log_4 3-1}}=4^{\frac{\log_4 3}{\log_4 3-1}}=3^{\frac{1}{\log_4 3-1}}.$$ The right-hand side of $(1)$ can be written as $$3^{\frac{\log_3 4}{1-\log_3 4}}.$$ We have expressed the left-hand side and the right-hand side as a power of $3$, and need to show that the exponents match. This is an easy consequence of the fact that $(\log_s t)(\log_t s)=1$.

Remark: Of course there is nothing special about $3$ and $4$. Also, it would be more attractive to symmetrize, and write $4/3$ in some places, and $3/4$ in others.

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How $4^{\frac{\log_4 3}{\log_4 3-1}}=3^{\frac{1}{\log_4 3-1}}$? – VelvetThunder Jun 22 '12 at 17:00
$4^{a/b}=(4^a)^{1/b}$. Here $a=\log_4 3$, so $4^a=3$. – André Nicolas Jun 22 '12 at 17:06
That explains it. Thanks. – VelvetThunder Jun 22 '12 at 17:08