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Problem is to prove that $$\log_23 +\log_52$$ is irrational number.
My attempt:
I try to write number like $$\log_23 +\frac{1}{\log_25}$$ but I didn't get anything(proof by contradiction). I also try to find polynomial such that given number is zero point but also without success.

Martin Sleziak
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josf
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    Where does this problem come from ? It appears quite difficult to prove the irrationality of this number. – Peter Jan 25 '20 at 10:34
  • This smells like one can show transcendence from some Gelfond-Schneider-like theorem, which may however not available to the OP? – Hagen von Eitzen Jan 25 '20 at 10:45
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    @josf Do you have some reason for thinking this has an elementary solution? – almagest Jan 25 '20 at 11:02
  • You may have to prove/disprove that $\log_2 3$ and $\log_5 2$ are linearly independent over the rational numbers. –  Jan 25 '20 at 11:50
  • If $\exists p, q \in \mathbb{N}$ such that $\log_23 +\log_52 = \frac{p}{q}$, then $2^{\log_23 +\log_52} = 2^{\frac{p}{q}}$ which is algebraic (an integer raised to a rational power..). Wolfram alpha says $2^{\log_23 +\log_52}$ is transcendental, which gives you the contradiction needed - but I don't know how to prove that myself (: – Blueyedaisy Jan 30 '20 at 13:14
  • Lindman Weistress Might be useful over here. – maths student Feb 02 '20 at 13:17
  • @Daphna If you write the number as $3 \cdot 2^{\log_5 2}$, W|A correctly says that it's unknown whether or not the number is transcendental. – Maxim Feb 03 '20 at 10:51

2 Answers2

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Refer to Qiaochu Yuan's answer from (Question 986227), which states that, contingent on the currently unproven Schanuel's conjecture, the logarithms of the primes, $\ln 2,\, \ln3\, \,\ln 5\,\ldots$, are algebraically (not just linearly) independent over $\mathbb{Q}$. The stronger statement would follow that $\log_2 3+\log_5 2$ is transcendental and consequentially irrational.

Proof. By unique prime factorization, the logarithms of the primes are linearly independent over $\mathbb{Q}$. If $p_1, p_2, \dots$ is an enumeration of the primes, then by Schanuel's conjecture it follows that $\mathbb{Q}(\log p_1, \log p_2, \dots \log p_k)$ has transcendence degree at least $k$, hence exactly $k$, for all $k$. $\Box$

Your problem will likely remain open until Schanuel's conjecture is resolved. To quote GH from MO's answer to (Math Overflow: Question 185540),

...[that $\log_35+\log_25$ is irrational] is probably true, but proving it might be out of reach at the moment.

Jam
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$2^{log_23}.2^{log_52}= 2^{\frac {p}{q}}$ (where p and q are natural numbers as the number is graeter than 0).

(and $a^{log_bc}=c^{log_ba}$=>$2^{log_23}=3$)

therefore $3^q . 2^{qlog_52}=2^p$ (we know that $3^q$ and $2^p$ are natural number so we now just need to show $2^{qlog_52}$ is always irrational for any natural 'q').

$2^{qlog_52}=\frac{m}{n}$ (we know $\frac{m}{n}$ is not pf the form $2^k$ where $k$ is a natural number becaise $qlog_52$ cannot be natural)

therefore $q=log_25.log_2(\frac{m}{n})$=c (where q is a natural number,and if $log_2\frac{m}{n}$ becomes rational then $log_25$ being irrational $q$ becomes irrational which is not possible.)

$log_25=a,log_2\frac{m}{n}=b$ ,where a and b are both irrational.I dont know how to proceed further , irrational $\times$ irrational can be rational.

aryan bansal
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