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The title is a little nonsensical (so feel free to edit as you see fit)

Logb(n) where n is a power of b produces a rational number; for example; Log2(8) = 3

But, Log2(3) = 1.5849625007211561814537389439478165087598144076924810604557526545...

Thus, Log2(3) produces an irrational number.

Is the same true for any Logb(n) where n is not a power of b, or are there cases where this produces a rational number?

Matthew Layton
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  • You need to clarify what constraints make this nontrivial. For example, are you asking when $\log_bn$ is a non-integral rational number for integers $b,,n>2$? If so, hint: writing such a logarithm as $p/q$ with $p,,q$ co-prime, consider the prime factorizations of each side of $b^p=n^q$ to prove the most general solution is $b=a^q,,n=a^p$, such as in $\log_48=\frac32$. – J.G. Mar 14 '23 at 15:12
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    There are some non-integer examples like $\log_2 (2 \sqrt{2} ) = \frac{3}{2}$. – Matthew Leingang Mar 14 '23 at 15:12
  • @J.G. For context, I've been investigating under which circumstances a decimal expansion terminates, repeats periodically, or continues infinitely with no repetition, so I guess my question is, does Logb(n) where n is not a power of b always produces the latter? – Matthew Layton Mar 14 '23 at 15:16
  • Do you consider $8$ "a power of $4$", or merely consider $4,,8$ each powers of $2$? (By the way, looking at a few digits of $\log_23$ doesn't prove they never repeat, but my first comment tells you how to verify $\log_23$ is irrational.) – J.G. Mar 14 '23 at 15:18
  • @J.G. So there is a perfect example of what I was asking. Log4(8) = 1.5 and 4^1.5 = 8, thus, the answer is no, right? (because 1.5 has a terminating decimal expansion)? – Matthew Layton Mar 14 '23 at 15:22
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    "but log2(3)=1.584962500... thus log2(3) produces an irrational number" This alone is not sufficient proof. It may be evidence that leads us to believe that it may be irrational, but a number beginning with 1.584962500 is not necessarily going to be irrational. For example 1.584962500 itself is rational. – JMoravitz Mar 14 '23 at 15:24
  • For positive integers $b$ and $n$, the Fundamental Theorem of Arithmetic can be helpful in showing that $\log_b n$ is irrational (when it is in fact irrational). For instance such an argument can be made to show that $\log_2 3$ is irrational. – paw88789 Mar 14 '23 at 15:36
  • With regards to "Is it true that any $\log_b(n)$ where $n$ is not a rational power of $b$ will be irrational" The answer is... of course. That is precisely what that means. If $n$ is not a rational power of $b$ then it follows that $n$ is an irrational power of $b$ (barring nonsensical situations like log base 1 or the logarithm of zero, etc...). That is no more interesting than saying "Is it true that any real number which is not rational is irrational?" – JMoravitz Mar 14 '23 at 15:39
  • If your question was specifically about integers and integer powers, consider $\log_4(8)=1.5$. Here $8$ is not an integer power of $4$, though it is a rational power of $4$. – JMoravitz Mar 14 '23 at 15:40

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The logarithm function is the inverse of the exponential function. So $\log_b x = a \iff x = b^a$.

Thus, if $\log_b x = a$ is rational, then we can write $a = \frac{m}{n}$ where $m$ and $n$ are integers, and so $x = b^\frac{m}{n}$. But by applying power rules we can see that this is equivalent to $x = (b^m)^\frac{1}{n} = \sqrt[n]{b^m}$, i.e. $x$ is the $n$th root of $b^m$.

All of these steps are reversible, meaning that if $x$ is a rational power of $b$, then $\log_b x$ is a rational number and vice versa.

ConMan
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