# Difference between revisions of "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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 