Is there any way to prove that the ratio of two natural logarithms is rational or irrational? Take the natural logarithms of $a = 25$ and $b = 6$, for example. Can you prove $\ln(a)/\ln(b)$ rational or irrational?
Asked
Active
Viewed 800 times
3
-
2Not sure this helps, but recall that $\frac{\ln(a)}{\ln(b)} = \log_b(a)$. – Viktor Vaughn Jan 22 '16 at 19:56
-
2$\frac{\ln(a)}{\ln(b)}=\log_b(a)$, which can be rational or irrational, depending on the values of $a$ and $b$. – barak manos Jan 22 '16 at 19:57
-
1http://math.stackexchange.com/questions/166441/what-is-the-simplest-way-to-prove-that-the-logarithm-of-any-prime-is-irrational – Dan Brumleve Jan 22 '16 at 20:27
2 Answers
5
$\ln(a)/\ln(b) = n/d$ (where $n,d$ are integers) iff $a^d = b^n$. If $a$ and $b$ are positive integers, this can be decided using the prime factorizations of $a$ and $b$: $a$ and $b$ have the same prime factors, and the ratios of the exponents are the same. In your example, $25$ has prime factor $5$ but $6$ does not, so the answer is no.
Robert Israel
- 470,583
2
$ln(a)/ln(b)=p/q$ implies $qln(a) =pln(b)$ so $e^{pln(b)}=e^{qln(a)}$ which means $b^p=a^q$
Tsemo Aristide
- 89,587