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

(Solution 3 (Logarithms))
(Solution 3 (Logarithms): Fixed errors and LaTeX.)
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==Solution 3 (Logarithms)==
 
==Solution 3 (Logarithms)==
  
We can rewrite the equation using fractional exponents and take logarithms of both sides:
+
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>. 
  
<math>log_2{(2^{m})(4096^{-1/2}}) = log_2{(2)(4096^{-1/m})}</math>
+
Therefore, the answer is <math>3+4 = \boxed{\textbf{(C) } 7}.</math>
 
 
We can then use the additive properties of logarithms to split them up:
 
 
 
<math>log_2{2^{m}} + log_2{4096^{-1/2}} = log_2{2} + log_2{4096^{-1/m}}</math>
 
 
 
Using the power rule, the fact that <math>4096 = 2^{12}</math>, and bringing the exponents down, we get:
 
 
 
<math>m - 6 = 1 - \frac{12}{m}</math>
 
 
 
<math>m + \frac{12}{m} = 7</math>
 
 
 
<math>m^{2} + 12 = 7m</math>
 
 
 
<math>m^{2} - 7m + 12 = 0</math>
 
 
 
<math>(m-3)(m-4) = 0</math>
 
 
 
<math>m = 3</math> and <math>m = 4</math> 
 
 
 
Since our two values for m are <math>3</math> and <math>4</math>, our final answer is <math>3+4 = \boxed{\textbf{(C) } 7}</math>
 
  
 
- abed_nadir
 
- abed_nadir

Revision as of 14:14, 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

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

Solution 3 (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

Video Solution 1

https://youtu.be/UmaCmhwbZMU

~Education, the Study of Everything

Video Solution (Easy)

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

~Whiz

See Also

2022 AMC 10A (ProblemsAnswer KeyResources)
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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