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Difference between revisions of "2019 AMC 12B Problems/Problem 9"

(Improved clarity (with more detail), LaTeX, and formatting)
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Combining our results, <math>x</math> is an integer strictly between <math>4</math> and <math>64</math>, so the number of possible values of <math>x</math> is <math>64 - 4 - 1 = \boxed{\textbf{(B) } 59}</math>.  
 
Combining our results, <math>x</math> is an integer strictly between <math>4</math> and <math>64</math>, so the number of possible values of <math>x</math> is <math>64 - 4 - 1 = \boxed{\textbf{(B) } 59}</math>.  
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==Solution 2 (Somewhat Cheating)==
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Using the triangle inequality, you get <math>\log_2{x}+\log_4{x} > 3</math>. Solving for <math>x</math>, you get <math>x > 4</math>. Now we need an upper-bound for <math>x</math> and since we're dealing with bases of <math>2</math> and <math>4</math>, we're looking for answer choices close to a power of <math>2</math> and <math>4</math>. All the answer choices seem to be around <math>64</math>, and plugging that into the inequality <math>3+\log_4{x} > \log_2{x}</math> we see <math>64</math> is the correct number. Now we have <math>64 > x > 4</math> and the number of integers in between is <math>64-4-1 = \boxed{{59 \textbf{(B)}}}</math>
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==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=8|num-a=10}}
 
{{AMC12 box|year=2019|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:48, 14 January 2020

Problem

For how many integral values of $x$ can a triangle of positive area be formed having side lengths $\log_{2} x, \log_{4} x, 3$?

$\textbf{(A) } 57 \qquad\textbf{(B) } 59 \qquad\textbf{(C) } 61 \qquad\textbf{(D) } 62 \qquad\textbf{(E) } 63$

Solution

For these lengths to form a triangle of positive area, the Triangle Inequality tells us that we need $\log_2{x} + \log_4{x} > 3$, $\log_2{x} + 3 > \log_4{x}$, and $\log_4{x} + 3 > \log_2{x}$. The second inequality is redundant, as it's always less restrictive than the last inequality.

Let's raise the first inequality to the power of $4$. This gives $4^{\log_2{x}} \cdot 4^{\log_4{x}} > 64 \Rightarrow \left(2^2\right)^{\log_2{x}} \cdot x > 64 \Rightarrow x^2 \cdot x > 64$. Thus, $x > 4$.

Doing the same for the second inequality gives $4^{\log_4{x}} \cdot 64 > 4^{\log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64$ (where we are allowed to divide both sides by $x$ since $x$ must be positive in order for the logarithms given in the problem statement to even have real values).

Combining our results, $x$ is an integer strictly between $4$ and $64$, so the number of possible values of $x$ is $64 - 4 - 1 = \boxed{\textbf{(B) } 59}$.

Solution 2 (Somewhat Cheating)

Using the triangle inequality, you get $\log_2{x}+\log_4{x} > 3$. Solving for $x$, you get $x > 4$. Now we need an upper-bound for $x$ and since we're dealing with bases of $2$ and $4$, we're looking for answer choices close to a power of $2$ and $4$. All the answer choices seem to be around $64$, and plugging that into the inequality $3+\log_4{x} > \log_2{x}$ we see $64$ is the correct number. Now we have $64 > x > 4$ and the number of integers in between is $64-4-1 = \boxed{{59 \textbf{(B)}}}$


See Also

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 12 Problems and Solutions

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