Exponential Comparison

Level 2

How many integer values of x x satisfy the inequality 32 243 < ( 2 3 ) x 2 < 9 4 ( 8 27 ) x ? \frac{32}{243} < \left( \frac{2}{3} \right)^{x^2} < \frac{9}{4} \cdot \left( \frac{8}{27} \right)^x ?

4 4 1 1 2 2 3 3

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1 solution

Kenneth Choo
Feb 20, 2016

32 243 < ( 2 3 ) x 2 < 9 4 ( 8 27 ) x ( 2 3 ) 5 < ( 2 3 ) x 2 < ( 2 3 ) 2 ( 2 3 ) 3 x ( 2 3 ) 5 < ( 2 3 ) x 2 < ( 2 3 ) 2 + 3 x 5 > x 2 > 2 + 3 x \frac { 32 }{ 243 } <{ (\frac { 2 }{ 3 } ) }^{ { x }^{ 2 } }<\frac { 9 }{ 4 } \cdot { (\frac { 8 }{ 27 } ) }^{ x }\\ \Rightarrow { (\frac { 2 }{ 3 } ) }^{ 5 }<{ (\frac { 2 }{ 3 } ) }^{ { x }^{ 2 } }<{ (\frac { 2 }{ 3 } ) }^{ -2 }\cdot { (\frac { 2 }{ 3 } ) }^{ 3x }\\ \Rightarrow { (\frac { 2 }{ 3 } ) }^{ 5 }<{ (\frac { 2 }{ 3 } ) }^{ { x }^{ 2 } }<{ (\frac { 2 }{ 3 } ) }^{ -2+3x }\\ \Rightarrow 5>{ x }^{ 2 }>-2+3x

Solving x x in inequalities,

x 2 5 < 0 5 < x < 5 { x }^{ 2 }-5<0\\ \Rightarrow -\sqrt { 5 } <x<\sqrt { 5 }

x 2 3 x + 2 > 0 ( x 2 ) ( x 1 ) > 0 x > 2 , x < 1 { x }^{ 2 }-3x+2>0\\ \Rightarrow (x-2)(x-1)>0\\ \Rightarrow x>2,x<1

We then have the solutions

5 < x < 1 2 < x < 5 -\sqrt { 5 } <x<1\\ 2<x<\sqrt { 5 }

The integer solutions of x x are 2 , 1 , 0 -2,-1,0 .

Great! Thanks for the solution!

Calvin Lin Staff - 5 years, 3 months ago

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I found that there's no solution here so I thought "why not provide one?"

Kenneth Choo - 5 years, 3 months ago

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That's really helpful :)

I love that Brilliant's community is made up of members like you who are helping each other improve, and learning from the experience.

Calvin Lin Staff - 5 years, 3 months ago

Only 2 where the zero comes

Hemanth N - 1 year, 2 months ago

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