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(This is from a proof by contradiction, so that's why the equality does not actually hold. Edited for brevity; I don't think I've omitted anything pertinent to my questions.)

[...] The monotonicity of $f$ implies that $2^u<3^v$ if and only if $3^u<6^v$, $u$, $v$ being positive integers. Taking logarithms this means that $\frac{v}{u}>\log_2 3$ if and only if $\frac{v}{u}>\log_3 6$. Since rationals are dense, it follows that $\log_2 3 = \log_3 6$. This can be written as $\log_2 3 = \frac{1}{\log_2 3}+1$, and so $\log_2 3$ is the positive solution of the quadratic equation $x^2−x −1=0$ [...]

  • Question 1

When he concludes that:

Since rationals are dense, it follows that $\log_2 3 = \log_3 6$.

What is the reason for this? Wouldn't this only be true if $\frac{v}{u}$ goes to $0$?

  • Question 2

[$\log_2 3 = \log_3 6$] can be written as $\log_2 3 = \frac{1}{\log_2 3}+1$.

This one is the most confusing. How did he jump from $x=y$ to $x = \frac{1}{x}+1$?

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    @joriki ooh, ok. I have to learn to pay better attention to details. thanks!2011-06-27

2 Answers 2

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Here is help for Question 2:
First note that by the change of base formula, we have that $\log_{3}6=\frac{\log_{2}6}{\log_{2}3}.$

Also, $\log_{2}6=\log_{2}3+\log_{2}2.$

Combining the above results should resolve all the issues.

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Nana's answer is fine. Alternatively, ${1\over\log_23}+1=\log_32+1=\log_32+\log_33=\log_36$