2019 AMC 12B Problems/Problem 9

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

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2019 AMC 12B Problems/Problem 9

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