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Difference between revisions of "2019 AMC 12B Problems/Problem 9"
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==Solution==
 
==Solution==
  
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Note that <math>\log_2{x} + \log_4{x} > 3</math>, <math>\log_2{x} + 3 > \log_4{x}</math>, and <math>\log_4{x} + 3 > \log_2{x}</math>. The second one is redundant, as it's less restrictive in all cases than the last.  
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For these lengths to form a triangle of positive area, the Triangle Inequality tells us that we need <math>\log_2{x} + \log_4{x} > 3</math>, <math>\log_2{x} + 3 > \log_4{x}</math>, and <math>\log_4{x} + 3 > \log_2{x}</math>. The second inequality is redundant, as it's always less restrictive than the last inequality.  
  
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Let's raise the first to the power of <math>4</math>. <math>4^{\log_2{x}} \cdot 4^{\log_4{x}} > 64 \Rightarrow x^2 \cdot x > 64</math>. Thus, <math>x > 4</math>.  
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Let's raise the first inequality to the power of <math>4</math>. This gives <math>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</math>. Thus, <math>x > 4</math>.  
  
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Doing the same for the second nets us: <math>4^{\log_4{x}} \cdot 64 > 4^{\log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64</math>.  
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Doing the same for the second inequality gives <math>4^{\log_4{x}} \cdot 64 > 4^{\log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64</math> (where we are allowed to divide both sides by <math>x</math> since <math>x</math> must be positive in order for the logarithms given in the problem statement to even have real values).
  
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Thus, x is an integer strictly between <math>64</math> and <math>4</math>: <math>64 - 4 - 1 = 59</math>.  
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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>.  
  
 
==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 18:56, 18 February 2019

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 $64 - 4 - 1 = \boxed{\textbf{(B) } 59}$.

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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