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Difference between revisions of "2022 AMC 10A Problems/Problem 11"

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

Revision as of 01:10, 12 November 2022

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

See Also

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

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