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2019 AMC 12B Problems/Problem 9: Difference between revisions

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-DrJoyo
-DrJoyo
==Solution 2==
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.
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>.
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>.
Thus, x is an integer strictly between <math>64</math> and <math>4</math>: <math>64 - 4 - 1 = 59</math>.
- Robin's solution


==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 14:56, 14 February 2019

Problem

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

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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