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 $\log_{2} x, \log_{4} x, 3$ ?

Solution

The lower bound for x would be x=4, where the sides of the triangle would be (2,1,3). The upper bound for x would be x=4, where the sides of the triangle would be (6,3,3). The number of integers strictly between 4 and 64 is 64 - 4 - 1 = 59

-DrJoyo

Solution 2

Note that $log_2{x} + log_4{x} > 3$ , $log_2{x} + 3 > log_4{x}$ , and $log_4{x} + 3 > log_2{x}$ . The second one is redundant, as it's less restrictive in all cases than the last.

Let's raise the first to the power of $4$ . $4^{log_2{x}} \cdot 4^{log_4{x}} > 64 \Rightarrow x^2 \cdot x > 64$ . Thus, $x > 4$ .

Doing the same for the second nets us: $4^{log_4{x}} \cdot 64 > 4^{log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64$ .

Thus, x is an integer strictly between $64$ and $4$ : $64 - 4 - 1 = 59$ .

- Robin's solution

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

Contents

Problem

Solution

Solution 2

See Also