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

Revision as of 01:18, 8 September 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 $2,$ 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 surd 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

@@ Line 23: / Line 23: @@
 ~KingRavi
-==Solution 2==
+==Solution 2 (Logarithms)==
-Note that <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>.
+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 then use the additive properties of logarithms to split them up: <cmath>\log_2{(2^{m})} + \log_2{(4096^{-1/2})} = \log_2{2} + \log_2{(4096^{-1/m})}.</cmath>
+Using the power rule, the fact that <math>4096 = 2^{12},</math> and bringing the exponents down, we get
+<cmath>\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*}</cmath>
+from which <math>m = 3</math> or <math>m = 4</math>. Therefore, the answer is <math>3+4 = \boxed{\textbf{(C) } 7}.</math>
+- youtube.com/indianmathguy
+==Solution 3==
+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
-==Video Solution 1==
+==Video Solution 1 ==
 https://youtu.be/UmaCmhwbZMU
 ~Education, the Study of Everything
-== Video Solution (Easy) ==
+== 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

Page

Toolbox

Search