Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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. – Quixotic 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
up vote 3 down vote accepted

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: $$\frac{a^{1/\log_a(b/a)}}{b^{1/\log_b(b/a)}} = \frac{1}{ab}.$$

share|cite|improve this answer
Thanks, I realized that in general $$\frac{a^{ 1/\log_a(\frac{b}{a})}}{b^{1/\log_b(\frac{b}{a})}}=\frac{1}{a\times b}$$ – Quixotic 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. – Quixotic Jun 22 '12 at 16:31

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

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

\begin{equation} \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}} \end{equation}

Using the identity we have that

\begin{equation} 3^{\rm{log}_{\frac{3}{4}}3}=3\cdot 4^{\rm{log}_{\frac{3}{4}}3} \end{equation}

replacing we get

\begin{equation} \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} \end{equation}

share|cite|improve this answer

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

Taking the log base $4$, I get


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

share|cite|improve this answer

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.

share|cite|improve this answer
How $4^{\frac{\log_4 3}{\log_4 3-1}}=3^{\frac{1}{\log_4 3-1}}$? – Quixotic 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. – Quixotic Jun 22 '12 at 17:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.