2022 AMC 10A Problems/Problem 11: Difference between revisions

Revision as of 14:15, 14 October 2023

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 $3+4 = \boxed{\textbf{(C) } 7}.$

- abed_nadir

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 (Easy)

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

~Whiz

@@ Line 23: / Line 23: @@
 ~KingRavi
-==Solution 2==
+==Solution 2 (Logarithms)==
-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>
@@ Line 44: / Line 38: @@
 - abed_nadir
+==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 ==

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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