2022 AMC 10A Problem 11

Problem 11 of 25IntermediateAlgebra

Ted mistakenly wrote

2^m\cdot\sqrt{\dfrac{1}{4096}}

2\cdot\sqrt[m]{\dfrac{1}{4096}}.

What is the sum of all real numbers

m

for which these two expressions have the same value?

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Solution

We can rewrite

4096

2^{12},

\dfrac{1}{4096} = 2^{-12}.

Then if we equate the given expressions, we get

2^m \cdot 2^{-6} = 2 \cdot 2^{\frac{-12}{m}}.

Equating the exponents, we get

m - 6 = 1 + \dfrac{-12}{m}.

Multiplying by

m,

we get

m^2 - 6m = m - 12

and so

m^2 - 7m + 12 = 0

(m-4)(m-3)

m=4,~m=3

Therefore, we can see that the sum of the solutions is

7.

Thus, C is the correct answer.