2022 AMC 10A Problem 11

Every number here is a power of $2$, so write everything as a power of $2$ and compare exponents.

1. Rewrite $4096$. $2^{10} = 1024$, so $2^{12} = 4096$. That means $\frac{1}{4096} = 2^{-12}$.
2. Left side. A square root halves the exponent: $\sqrt{2^{-12}} = 2^{-6}$. So
$$2^m \cdot 2^{-6} = 2^{m-6}.$$
3. Right side. An $m$-th root divides the exponent by $m$: $\sqrt[m]{2^{-12}} = 2^{-12/m}$. So
$$2 \cdot 2^{-12/m} = 2^{1 - 12/m}.$$
4. Compare. Two powers of $2$ are equal only when their exponents are equal:
$$m - 6 = 1 - \frac{12}{m}.$$
5. Solve. Multiply both sides by $m$ (which cannot be $0$): $m^2 - 6m = m - 12$, so
$$m^2 - 7m + 12 = 0, \qquad (m - 3)(m - 4) = 0.$$
6. Check both. For $m = 3$: the left side is $8 \cdot \frac{1}{64} = \frac18$, and the right side is $2 \cdot \frac{1}{16} = \frac18$. For $m = 4$: the left side is $16 \cdot \frac{1}{64} = \frac14$, and the right side is $2 \cdot \frac18 = \frac14$.
7. The sum is $3 + 4 = 7$.

The answer is $\boxed{\textbf{(C)}\ 7}$.

When every number in sight is a power of the same base, rewrite in that base: roots turn into division of exponents, and the problem becomes an ordinary equation about the exponents. Here that equation had $m$ on the bottom of a fraction, so clearing it gave a quadratic.

Since $4096 = 2^{12}$, we have $\frac{1}{4096} = 2^{-12}$. Convert each radical to a power of $2$:
$$2^m \cdot \sqrt{2^{-12}} = 2^m \cdot 2^{-6} = 2^{m-6}, \qquad 2 \cdot \sqrt[m]{2^{-12}} = 2 \cdot 2^{-12/m} = 2^{1 - 12/m}.$$
Powers of $2$ are equal exactly when their exponents are equal:
$$m - 6 = 1 - \frac{12}{m} \quad\Longrightarrow\quad m^2 - 7m + 12 = 0 \quad\Longrightarrow\quad (m-3)(m-4) = 0.$$
Both $m = 3$ and $m = 4$ are legitimate (neither is $0$), and by Vieta their sum is $7$. Check $m = 3$: $8 \cdot \frac{1}{64} = \frac18$ and $2 \cdot \frac{1}{16} = \frac18$.

The answer is $\boxed{\textbf{(C)}\ 7}$.