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I know that $\ln n$ is transcendental for all integer $n>1$. But does this still hold for non-integer rational values of $n>1$? For example, is $\ln 1.5$ transcendental?

EDIT: Somehow managed to overlook the fact that cases like $\ln e=1$ are not transcendental :P The question has been revised to only include rational numbers $n$.

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    $\begingroup$ $\ln e$, for example, is not transcendental. $\endgroup$ – Qiaochu Yuan Mar 14 '15 at 22:39
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No, what about: $$ \ln(e^2)=2 $$ And $2$ is certanly not trancendental.


It is however possible to restrict the values of $n$ further than integers. By the Lindemann–Weierstrass theorem, any algebraic number $\alpha\ne1$ has $\ln\alpha$ as trancendental.

Since any rational number is algebraic, and $1.5$ is a rational number, $\ln1.5$ is trancendental.

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  • $\begingroup$ >.< Thanks for pointing it out, and the question's been edited to be clearer. $\endgroup$ – user3932000 Mar 14 '15 at 22:43
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    $\begingroup$ @user3932000 I have updated the answer to restrict further which values are trancendental. $\endgroup$ – Alice Ryhl Mar 14 '15 at 22:45
  • $\begingroup$ So yes, $\ln 1.5$ is transcendental? $\endgroup$ – user3932000 Mar 14 '15 at 22:54
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    $\begingroup$ @user3932000 Yes $\endgroup$ – Alice Ryhl Mar 14 '15 at 22:55
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    $\begingroup$ @user3932000 It means that wolfram alpha acts weird when given decimal numbers, as a fraction however: wolframalpha.com/input/?i=is+ln%283%2F2%29+transcendental $\endgroup$ – Alice Ryhl Mar 14 '15 at 23:00
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$\qquad$ This is probably the equivalent of nuking a mosquito, but if $\ln x$ would be transcendental for every value of x, then the logarithmic function would be discontinuous in every point, since the transcendental numbers form a dense, but not continuous, subset of the reals. However, we know that the natural logarithm is continuous, hence contradiction. $($As for the amended version of the initial post, the answer is yes$)$.

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