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