imc

We are given that $$2^m\cdot \sqrt{\frac{1}{4096}}=2\cdot \sqrt[m]{\frac{1}{4096}}.$$ Converting everything into powers of $2,$ we have $$\begin{array}{rl}{2}^m\cdot (2^{-12}{)}^{\frac{1}{2}}& =2\cdot (2^{-12}{)}^{\frac{1}{m}}\\ 2^{m-6}& =2^{1-\frac{12}{m}}\\ m-6& =1-\frac{12}{m}.\end{array}$$ We multiply both sides by $m$, then rearrange as $$m^2-7m+12=0.$$ By Vieta's Formulas, the sum of such values of $m$ is $\boxed{7}.$ Note that $m=3$ or $m=4$ from the quadratic equation above.