# Is base conversion in $\mathsf{TC^0}$?

$$\mbox{Problem}_{j,i,q}(x_1,\dots,x_n)$$: Given an integer $$\sum_{i''=0}^{n-1}2^{i''}x_{i''+1}$$ in binary output the $$j$$th binary digit of the $$i$$th base-$$q$$ digit (where $$q$$ is not necessarily a prime).

$$\mbox{Problem}_{j,i,q}(x_1,\dots,x_n)=j\mbox{ th bit of }y_i\mbox{ in }\sum_{i'=0}^{m-1}q^{i'}y_{i'+1}=\sum_{i''=0}^{n-1}2^{i''}x_{i''+1}$$ where $$y_{i'}\in\{0,1,\dots,q-1\}$$ at every $$i'\in\{1,\dots,m\}$$ and $$m=\lceil\log_q\sum_{i''=0}^{n-1}2^{i''}x_{i''+1}\rceil$$.

Is $$\mbox{Problem}_{j,i,q}(x_1,\dots,x_n)\in\mathsf{uniformTC^0}$$ or at least $$\mathsf{non-uniformTC^0}$$?