Transcendentality of the $\log$ of the golden mean

We know that $\phi$, the golden ratio, is algebraic. Is it known whether $\log(\phi)$ is algebraic?

Thank you!

PS. I am not in number theory, so I apologize in advance if this is obvious.

-
You mean we know that $\phi$ is algebraic? – Qiaochu Yuan May 27 '11 at 8:20
Certainly. Typo. Thanks! Fixed it. – William May 27 '11 at 8:28

$\log (\phi)$ is transcendental. The Lindemann–Weierstrass theorem implies that if $\alpha$ is a nonzero algebraic number, then $e^\alpha$ is transcendental. So since $\phi$ is algebraic, $\log (\phi)$ is transcendental.

-
Whoops. The statement of Lindemann-Weierstrass is slightly stronger than I remember. – Qiaochu Yuan May 27 '11 at 9:04
The trick is to have forgotten the statement entirely, so you have to look it up. – Chris Eagle May 27 '11 at 9:09
I see. Thank you. – William May 27 '11 at 9:22