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 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

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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