AMC10 2022 A

AMC10 2022 A · Q11

AMC10 2022 A · Q11. It mainly tests Quadratic equations, Exponents & radicals.

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?

Ted 错误地将 $2^m\cdot\sqrt{\frac{1}{4096}}$ 写成了 $2\cdot\sqrt[m]{\frac{1}{4096}}$。对于这两个表达式值相等的全部实数 $m$，它们的和是多少？

(A) 5 5

(B) 6 6

(D) 8 8

(E) 9 9

Answer

Correct choice: (C)

正确答案：（C）

Solution

We are given that \[2^m\cdot\sqrt{\frac{1}{4096}} = 2\cdot\sqrt[m]{\frac{1}{4096}}.\] Converting everything into powers of $2$ and equating exponents, 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.

给定 \[2^m\cdot\sqrt{\frac{1}{4096}} = 2\cdot\sqrt[m]{\frac{1}{4096}}.\] 将一切转换为 $2$ 的幂并令指数相等，有 \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*} 两边乘以 $m$，然后整理为 \[m^2-7m+12=0.\] 由 Vieta 公式，此类 $m$ 值的和为 $\boxed{\textbf{(C) } 7}$。注意上述二次方程得 $m=3$ 或 $m=4$。

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