Lesson goal: Radical Exponents Puzzle (AMC 10A Problem 11)

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In the 2022 AMC 10A competition, Problem #11 explores what happens when exponents and radicals are misplaced:

Contest Problem Statement:
"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?"

Converting Radicals to Powers of 2

First, note that $4096 = 2^{12}$, which means:
$$\frac{1}{4096} = 2^{-12}$$
Now simplify each expression as a power of $2$:
  1. Left-Hand Side (LHS): The square root is the $\frac{1}{2}$ power:
    $$\sqrt{\frac{1}{4096}} = \left(2^{-12}\right)^{1/2} = 2^{-6} = \frac{1}{64}$$
    Multiplying by $2^m$:
    $$\text{LHS} = 2^m \cdot 2^{-6} = 2^{m - 6}$$
  2. Right-Hand Side (RHS): The $m$-th root is the $\frac{1}{m}$ power:
    $$\sqrt[m]{\frac{1}{4096}} = \left(2^{-12}\right)^{1/m} = 2^{-12/m}$$
    Multiplying by $2 = 2^1$:
    $$\text{RHS} = 2^1 \cdot 2^{-12/m} = 2^{1 - 12/m}$$

Equating Exponents

Because the base $2$ is the same on both sides, the expressions are equal if and only if their exponents are equal:
$$m - 6 = 1 - \frac{12}{m}$$
Add $6$ to both sides:
$$m = 7 - \frac{12}{m}$$
Multiplying by $m$ (with $m \ne 0$):
$$m^2 = 7m - 12 \implies m^2 - 7m + 12 = 0$$
Factoring with CodeByMath's symbolic factor() function:
$$(m - 3)(m - 4) = 0$$
The solutions are $m = 3$ and $m = 4$. Their sum is $3 + 4 = 7$!
factor("m^2 - 7*m + 12")
lhs = function(m) return 2^m * math.sqrt(1/4096) end
rhs = function(m) return 2 * (1/4096)^(1/m) end
sum_m = 3 + 4
Move the mouse over a dotted box for more information.

By Vieta's formulas, for any quadratic $m^2 - bm + c = 0$, the sum of the roots is simply the linear coefficient $b = 7$, matching choice (C) 7.

Now you try. Replace ???? with 3 and click Run. Then try evaluating lhs(5) and rhs(5) to confirm that other values of $m$ do not make the expressions equal!

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