Proof that $\log_23 +\log_52$ is irrational number 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.
 A: 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.

A: $2^{log_23}.2^{log_52}= 2^{\frac {p}{q}}$ (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.
