About irrational logarithms

Question

Could someone provide, please, a proof of the theorem below?

"Being $x$ and $b$ integers greater than $1$, which can not be represented as powers of the same basis (positive integer) and integer exponent, then the logarithm of $x$, in base $b$, is an irrational number."

Well... Assuming that the logarithm is a rational number $p/q$ ($p$ and $q$ are relatively prime integers), I know I can write $x^q = b^p$, but I can not conclude from this fact that $b$ and $x$ are powers of a same integer (and with integer exponent).

Answer 1

Hint:

If $\log_b\;x$ were a rational number $\frac{p}{q}$, then

$$x^q=b^p$$

Now, go back to your initial assumption...

Edited to add: This might be of interest.