Does the set of all logarithms with a radical base and argument belong to the set of all radicals? A simple yes, no answer will suffice, an explanation would be wonderful.
EDIT 1 Can a logarithm with a base and argument expressible as "radical" be a root to a polynomial with integer coefficients when the root cannot be expressed as otherwise "radical"? Note that for this logarithm to be a root of a polynomial with integer coefficients, it cannot be transcendental.
So, let's say by "radical" you mean an element of the field $$\large F=\mathbb{Q}(\{a^{1/r}:a,r\in\mathbb{N}\}),$$ or if you want to allow "radicals" of negative integers, an element of the larger field $$\large L=F(\{e^{2\pi i/n}:n\in\mathbb{N}\}).$$ Then no, there are many logarithms with "radical" base and argument that are not themselves "radicals". First, observe that any element of $L$ is an algebraic number (there are algebraic numbers that are not elements of $L$, but that is irrelevant to this question). Now apply the Gelfond-Schneider theorem to the equation $$2^{\log_2(3)}=3$$ for example. $2$ and $3$ are algebraic numbers, so that $\log_2(3)$ must either be rational or transcendental. It cannot be rational; otherwise we get $$\begin{align*} \log_2(3)&=\frac{n}{m}\\\\ 3&=2^{n/m}\\\\ 3^m&=2^n \end{align*}$$ which is impossible by the unique factorization of integers (or put more simply, because the left side is odd and the right side is even). Therefore $$\log_2(3)$$ is a logarithm with a "radical" base and exponent that is transcendental, and therefore certainly cannot be "radical" itself.