Ted mistakenly wrote $2^m\cdot\sqrt{\frac{1}{4096}}$ as $2\cdot\sqrt[m]{\frac{1}{4096}}.$ What is the sum of all real numbers $m$ for which these two expressions have the same value?
$\textbf{(A) } 5 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9$
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{align*} 2^m\cdot(2^{-12})^{\frac12} &= 2\cdot (2^{-12})^{\frac1m} \\ 2^{m-6} &= 2^{1-\frac{12}{m}} \\ m-6 &= 1-\frac{12}{m}. \end{align*}
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 $\boxed{\textbf{(C) } 7}.$
Note that $m=3$ or $m=4$ from the quadratic equation above.
~MRENTHUSIASM
~KingRavi