Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3 (Logarithms))
m (→‎Solution 3)
 
(7 intermediate revisions by 4 users not shown)
Line 8: Line 8:
  
 
We are given that <cmath>2^m\cdot\sqrt{\frac{1}{4096}} = 2\cdot\sqrt[m]{\frac{1}{4096}}.</cmath>
 
We are given that <cmath>2^m\cdot\sqrt{\frac{1}{4096}} = 2\cdot\sqrt[m]{\frac{1}{4096}}.</cmath>
−
Converting everything into powers of <math>2,</math> we have
+
Converting everything into powers of <math>2</math> and equating exponents, we have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
2^m\cdot(2^{-12})^{\frac12} &= 2\cdot (2^{-12})^{\frac1m} \\
 
2^m\cdot(2^{-12})^{\frac12} &= 2\cdot (2^{-12})^{\frac1m} \\
Line 14: Line 14:
 
m-6 &= 1-\frac{12}{m}.
 
m-6 &= 1-\frac{12}{m}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
We multiply both sides by <math>m</math>, then rearrange as <cmath>m^2-7m+12=0.</cmath>
+
We multiply both sides by <math>m,</math> then rearrange as <cmath>m^2-7m+12=0.</cmath>
 
By Vieta's Formulas, the sum of such values of <math>m</math> is <math>\boxed{\textbf{(C) } 7}.</math>
 
By Vieta's Formulas, the sum of such values of <math>m</math> is <math>\boxed{\textbf{(C) } 7}.</math>
  
 
Note that <math>m=3</math> or <math>m=4</math> from the quadratic equation above.
 
Note that <math>m=3</math> or <math>m=4</math> from the quadratic equation above.
  
−
~MRENTHUSIASM
+
~MRENTHUSIASM ~KingRavi
  
−
~KingRavi
+
==Solution 2 (Logarithms)==
−
 
 
−
==Solution 2==
 
−
 
 
−
Since surd roots are conventionally positive integers, assume <math>m</math> is an integer, so <math>m</math> can only be <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>6</math>, and <math>12</math>. <math>\sqrt{\frac{1}{4096}}=\frac{1}{64}</math>. Testing out <math>m</math>, we see that only <math>3</math> and <math>4</math> work. Hence, <math>3+4=\boxed{\textbf{(C) }7}</math>.
 
−
 
 
−
~MrThinker
 
−
 
 
−
==Solution 3 (Logarithms)==
 
  
 
We can rewrite the equation using fractional exponents and take logarithms of both sides: <cmath>\log_2{(2^{m}\cdot4096^{-1/2}}) = \log_2{(2\cdot4096^{-1/m})}.</cmath>  
 
We can rewrite the equation using fractional exponents and take logarithms of both sides: <cmath>\log_2{(2^{m}\cdot4096^{-1/2}}) = \log_2{(2\cdot4096^{-1/m})}.</cmath>  
Line 43: Line 35:
 
from which <math>m = 3</math> or <math>m = 4</math>. Therefore, the answer is <math>3+4 = \boxed{\textbf{(C) } 7}.</math>
 
from which <math>m = 3</math> or <math>m = 4</math>. Therefore, the answer is <math>3+4 = \boxed{\textbf{(C) } 7}.</math>
  
−
- abed_nadir
+
- youtube.com/indianmathguy
 +
 
 +
==Solution 3==
 +
 
 +
Since third roots are conventionally positive integers, assume <math>m</math> is an integer, so <math>m</math> can only be <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>6</math>, and <math>12</math>. <math>\sqrt{\frac{1}{4096}}=\frac{1}{64}</math>. Testing out <math>m</math>, we see that only <math>3</math> and <math>4</math> work. Hence, <math>3+4=\boxed{\textbf{(C) }7}</math>.
 +
 
 +
~MrThinker
  
 
==Video Solution 1 ==
 
==Video Solution 1 ==
Line 50: Line 48:
 
~Education, the Study of Everything
 
~Education, the Study of Everything
  
−
== Video Solution (Easy) ==
+
== Video Solution 2 ==
 +
 
 +
https://youtu.be/x716XmDDY9w
 +
 
 +
== Video Solution 3 ==
  
 
https://youtu.be/r-27UOzrL00
 
https://youtu.be/r-27UOzrL00
  
 
~Whiz
 
~Whiz
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/0kkc4-y8TkU
 +
 +
~IceMatrix
  
 
== See Also ==
 
== See Also ==

Latest revision as of 22:35, 10 November 2024

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 1

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.

~MRENTHUSIASM ~KingRavi

Solution 2 (Logarithms)

We can rewrite the equation using fractional exponents and take logarithms of both sides: \[\log_2{(2^{m}\cdot4096^{-1/2}}) = \log_2{(2\cdot4096^{-1/m})}.\] We can then use the additive properties of logarithms to split them up: \[\log_2{(2^{m})} + \log_2{(4096^{-1/2})} = \log_2{2} + \log_2{(4096^{-1/m})}.\] Using the power rule, the fact that $4096 = 2^{12},$ and bringing the exponents down, we get \begin{align*} m - 6 &= 1 - \frac{12}{m} \\ m + \frac{12}{m} &= 7 \\ m^{2} + 12 &= 7m \\ m^{2} - 7m + 12 &= 0 \\ (m-3)(m-4) &= 0, \end{align*} from which $m = 3$ or $m = 4$. Therefore, the answer is $3+4 = \boxed{\textbf{(C) } 7}.$

- youtube.com/indianmathguy

Solution 3

Since third roots are conventionally positive integers, assume $m$ is an integer, so $m$ can only be $1$, $2$, $3$, $4$, $6$, and $12$. $\sqrt{\frac{1}{4096}}=\frac{1}{64}$. Testing out $m$, we see that only $3$ and $4$ work. Hence, $3+4=\boxed{\textbf{(C) }7}$.

~MrThinker

Video Solution 1

https://youtu.be/UmaCmhwbZMU

~Education, the Study of Everything

Video Solution 2

https://youtu.be/x716XmDDY9w

Video Solution 3

https://youtu.be/r-27UOzrL00

~Whiz

Video Solution by TheBeautyofMath

https://youtu.be/0kkc4-y8TkU

~IceMatrix

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_11&oldid=233181"