How to prove if log is rational/irrational I'm an English major, now doubling in computer science. The first course I'm taking is Discrete Mathematics for Computer Science, using the MIT 6.042 textbook.
Within the first chapter of the book's practice problems, they ask us multiple times to prove that some log function is either rational or irrational.
Specific cases make more sense than others, but I would really appreciate any advice on how to approach these problems. Not how to carry them out algebraically, but what thought constructs are necessary to consider a log being (ir)rational.
For example, in the case of $\sqrt{2}^{2\log_2 3}$, proving that $2\log_23$ is irrational (and therefore $a^b$, when $a=\sqrt{2}$ and $b=2\log_23$, is rational) is not an easily solvable problem. I understand the methods of proofs, but the rules of logs are not intuitive to me. 
A section from my TF's solution is not something I would know myself to construct:
Since $2 < 3$, we know that $\log_23$ is positive
(specifically it is greater than $1$), and hence so is $2\log_23$. Therefore, we can assume that $a$ and $b$ are two positive integers.
Now $2\log_2 3=a/b$ implies $2^{2\log_2 3}=2^{a/b}$.
Thus $$2^{a/b}=2^{2\log_2 3} = 2^{\log_2 3^2} =3^2 =9\text{,}$$
and hence $2^a = 9^b$.
Any advice on approaching thought construct to logs would be greatly appreciated!
 A: To say that $\log_2 3$ is rational would mean that there are integers $m$, $n$ such that $\log_2 3=\dfrac mn$.  That would imply that $2^{m/n}=3$, so that $2^m=3^n$.  But that says an even number equals an odd number, which is impossible, so $\log_2 3$ cannot be rational.
But that doesn't help much in figuring out whether $2^{2\log_2 3}$ is rational.  One can write
$$
2^{2\log_2 3} = \left(2^{\log_2 3}\right)^2 = 3^2 = 9,\tag1
$$
so that is rational.  But in doing that you don't need to know anything at all about rational or irrational numbers until that final step where you observe that $9$ is rational.  Line $(1)$ above is just application of standard rules governing exponents and logarithms.
A: I came up with what I believe to be an original formulation of proving that $\log_ab $ is irrational.
If $\log_ab$ is rational, it is equal to some $a/b$, where $ a,b \in \mathbb{Z} $, where $ \mathbb{Z} $ is the set of integers.
If $\log_2 3=a/b$, then 
Thus $$2^{a} =3^b = N\text{,}$$
Where $N$ is some number $ \in \mathbb{Z} $
Now, since 2 and 3 are both prime, if this N exists, it has a prime factorization of $2 \cdotp 2 \cdotp 2 \cdotp ... \cdotp 2 $ and also of $3\cdotp3\cdotp3\cdotp3\cdotp...\cdotp3 $
( $a$ number of 2's and $b$ number of 3's)
But any number $ N $ can only have one unique prime factorization, so no such $ a,b $ can exist.
This shows that any $\log_x y$ where $a, b \in \mathbb{Z}$ and $a, b $ are mutually prime is irrational, since $x^{a} =y^b $ cannot be true, and all prime numbers (including 2 and 3) are mutually prime by definition.
