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 $n$.
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.
.< Thanks for pointing it out, and the question's been edited to be clearer.
– user3932000 Mar 14 '15 at 22:43$\qquad$ 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$)$.
