2022 AMC 10A Problems/Problem 11

Problem

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?

$\textbf{(A) } 5 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9$

Solution

We are given that $2^m\cdot\sqrt{\frac{1}{4096}} = 2\cdot\sqrt[m]{\frac{1}{4096}}.$ Converting everything into powers of $2,$ 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 and factor as $(m-3)(m-4)=0.$ Therefore, we have $m=3$ or $m=4.$ The sum of such values of $m$ is $3+4=\boxed{\textbf{(C) } 7}.$

~MRENTHUSIASM

Alternatively, once we reach $m-6 = 1-\frac{12}{m}, we rearrange to get$ m-7+\frac{12}{m}=0 $. Multiplying both sides by$ m $, we have$ m^2-7m+12=0 $. Since were asked to find the sum of all possible values of$ m $, we use vieta’s formula to get the sum of the roots is$ 7=\boxed{C}$ ~KingRavi

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2022 AMC 10A Problems/Problem 11

Problem

Solution

See Also