I know that $\log_e b$ is transcendental for some algebraic $b$.
Is there any results about the irrationality/transcedence of numbers of the form $\log_a b$ where $a$ is integer and $b$ is integer/rational/algebraic etc.
Or anything close to this.
Edit: If we choose an even $a$ and an odd $b$ we need to show that exists integers $p,q$ such that $a^p=b^q$ which is not possible, the same with both even numbers. And odd numbers eg. $\log_3 5 = p/q \implies 3^p = 5^q$ and we need to show that there is no integers $p,q$ that satifies this equality.
From this post we can apply the same argument to show that $\log$ of even base and odd argument (and vice-versa) are transcedental.
As I mentioned in my comment, you have enough information in the edits you've made to answer this.
Specifically, for integers $a,b \geq 2,$ the assumption that $\log_ab$ is rational implies there are positive integers $p,q$ such that $a^p = b^q,$ so the prime factorizations of $a$ and $b$ must have the same primes, e.g. $$ a = p_1^{k_1} p_2^{k_2}\cdots p_n^{k_n} \\ b = p_1^{m_1} p_2^{m_2} \cdots p_n^{m_n}$$ where $k_i, m_i > 0$ for all $i$ and also $$\frac{m_1}{k_1} = \frac{m_2}{k_2} = \cdots = \frac{m_n}{k_n}$$ which provides a test for whether $\log_ab$ is rational or irrational. If it fails this test, then $\log_ab$ is necessarily irrational and then [by Gelfond-Schneider] it must be transcendental.
[As a sidenote: Note that your comment about $a$ being even and $b > 1$ odd (or vice versa) is just an example in which $2$ appears in the prime factorization of one of the numbers and not the other, so it fails in the first step]
Similarly, if we keep $a \geq 2$ an integer and let $b$ be an algebraic number which is not $a^r$ for any rational number (of which there are many), then $\log_ab$ is once again irrational and therefore transcendental by Gelfond-Schneider.