When is $\log_{b}(a)$ irrational, for natural numbers $a,b$?

Question

I've seen this result usually given as "if $a$ and $b$ are coprime, then $\log_{b}(a)$ is irrational." Wolfram|Alpha even says it is "unknown" whether or not $\log_{12}(6)$ is irrational, but I think it is easy to show this is irrational (unless I'm missing something obvious):

By way of contradiction, suppose $\dfrac{\ln(6)}{\ln(12)} = \dfrac{n}{m}$ with integers $n,m$. Without loss of generality, take $n,m$ to be natural numbers.

Then $m\ln(6) = n\ln(12)$. Hence, $6^{m} = 12^{n}$, so we have $2^{m}3^{m} = 2^{2n}3^{n}$, and so:

$2^{m-2n}3^{m} = 3^{n}$

Since $3^{n}$ is odd, so is $2^{m-2n}3^{m}$, therefore $m-2n = 0$, and $m = 2n$. Hence $\dfrac{\ln(6)}{\ln(12)} = \dfrac{1}{2}$. This is a contradiction, so $\dfrac{\ln(6)}{\ln(12)}$ is irrational.

This leads me to believe that $a,b$ being coprime is a sufficient condition to say that $\log_{b}(a)$ is irrational, but not a necessary one. My question is, can we give a stronger result regarding whether or not $\log_{b}(a)$ is irrational? Can we give a necessary and sufficient condition? Does anyone know of a specific example of $a,b$ where the irrationality of $\log_{b}(a)$ really is unknown?

Answer 1

The number $\log_b{a}$ is irrational iff no integer power of $a$ is an integer power of $b$. Indeed, if $a^m=b^n$ then $m\log_b a=n$, so $\log_b a=n/m$. Conversely, if $\log_b a=n/m$, then $\log_b a^m=n,$ so $b^n=a^m$.

For example for $a=6, b=12$, $a^m=b^n$ means $2^m3^m=2^{2n}3^n$ which is impossible.

Remark Note that if $a=p_1^{k_1}...p_s^{k_s}, b=p_1^{l_1}...p_s^{l_s}$ then $a^m=b^n$ means $p_1^{mk_1}...p_s^{mk_s}=p_1^{nl_1}...p_s^{nl_s}$ or $mk_1=nl_1,...,mk_s=nl_s$. This implies that $k_i\ne 0$ iff $l_i\ne 0$ and all numbers $l_i/k_i$ are the same for every $i$ such that $k_i\ne 0$.

Answer 2

The Gelfond-Schneider theorem answers this directly, and for a much broader class of numbers. If $a, b > 0$ are algebraic numbers, and $b \neq 1$, then $\log_b a$ is either rational or transcendental. The first case holds exactly when (as the other answer states) $a$ and $b$ have a common integer power.

Answer 3

Note that when $a,b$ are both positive integers and $b \neq 1$, there is a relatively efficient algorithm to determine when $\log_b(a)$ is rational without having to consider prime factorization or possibly dealing with large powers:

(1) If $b \leq a$, then check whether $\frac{a}{b}$ is an integer:

• If so, then replace $a$ with $\frac{a}{b}$ and repeat step (1).
• If not, then stop, it's irrational.

(2) If $a=1$, then stop, it's rational.

(3) Swap $a$ and $b$ and go to step (1).

Additionally, if you keep track of the steps, you can easily get the continued fraction representation of the logarithm assuming it's rational.

The proof this algorithm is correct still uses the fundamental theorem of arithmetic as in your question and some of the other answers, but in application, all we need is to determine divisibility.

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For instance, $\log_{12}(6)$ sets $b=12, a=6$, it goes something like (skipping the steps that don't do anything):

• Swap $a$ and $b$, so $a=12$ and $b=6$
• Set $a = 12/6 = 2$
• Swap $a$ and $b$, so $a=6$ and $b=2$
• Set $a=6/2 = 3$
• Since $b =2 \leq 3 = a$ but $3/2$ isn't an integer, $\log_{12}(6)$ is irrational.