Which is larger?

Geometry Level 3

A = max ⁡ x ∈ R ( log ⁡ 2 3 ) sin ⁡ x , B = max ⁡ x ∈ R ( log ⁡ 3 2 ) sin ⁡ x A = \max_{x \in \mathbb{R}} \left( \log_2 3 \right)^{\sin x }, \qquad B = \max_{x \in \mathbb{R}} \left( \log_3 2 \right)^{\sin x }

Which is larger, A A or B ? B?

A B Both are equal

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Calvin Lin by Calvin Lin

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3 solutions

Sandeep Bhardwaj
Jun 17, 2015

Given : A = max ⁡ x ( log ⁡ 2 3 ) sin ⁡ x and B = max ⁡ x ( log ⁡ 3 2 ) sin ⁡ x = max ⁡ x ( log ⁡ 2 3 ) − sin ⁡ x \text{ Given : } A = \max_x \left( \log_2 3 \right)^{\sin x } \text{ and } \\ B = \max_x \left( \log_3 2 \right)^{\sin x }=\max_x \left( \log_2 3 \right)^{- \sin x }

A = log ⁡ 2 3 ( when sin ⁡ x = 1 ) A=\log_23 ( \text{when } \sin x=1 )

B = ( log ⁡ 3 2 ) − 1 = l o g 2 3 ( when sin ⁡ x = − 1 ) B=\left( \log_32 \right)^{-1}=log_23 ( \text{when } \sin x=-1)

Hence both A and B are equal ! \text{ Hence both A and B are equal ! }

enjoy ! \text{enjoy !}

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Moderator note:

Most people think that just because log ⁡ 2 3 > 1 > log ⁡ 3 2 \log_2 3 > 1 > \log_3 2 , then that A A must be greater than B B . The usage of sin ⁡ x \sin x is a hidden way to make use of the interval [ − 1 , 1 ] [-1, 1] .

Nice problem , becomes more interesting when base is sinx

sandeep Rathod - 5 years, 12 months ago
Etaash Katiyar
Jun 18, 2015

l o g 2 3 = l o g 3 l o g 2 log_2 3 = \frac {log3}{log2}

l o g 3 2 = l o g 2 l o g 3 log_3 2 = \frac {log2}{log3}

since − 1 ≤ s i n x ≤ 1 -1 \leq sinx \leq 1 the largest values can be given when sin x = -1 or 1

l o g 3 l o g 2 1 = l o g 2 l o g 3 − 1 \frac {log3}{log2} ^1 = \frac {log2}{log3} ^{-1}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Your inequality signs are the wrong way around. You're saying − 1 ≥ 1 -1\ge 1

Isaac Buckley - 5 years, 12 months ago
Humberto Bento
Jun 17, 2015

Both expressions are a function of (sin x). And we want the max value of x, not the maximum value of the expression.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you explain in more detail? How does that show that they are equal?

Calvin Lin Staff - 5 years, 12 months ago

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Unless I'm reading it incorrectly, A is the value of x that maximizes the expression. Therefore, as sin is periodic, there are infinit values of x that maximize the expression. This is valid for both expressions. I'm not reading it as the maximum value of the the expression!

Humberto Bento - 5 years, 12 months ago

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No, A A is the maximum value, over all real values of x x , that ( log ⁡ 2 3 ) sin ⁡ x \left(\log_2 3 \right)^{\sin x } can take. So, even though the maximum can occur at many different values, the value of A A is uniquely determined (assuming that the maximum of the function exists).

Calvin Lin Staff - 5 years, 12 months ago

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@Calvin Lin – I understand you. But that's not the way I learned it! It would be easy to prove your way too!

Humberto Bento - 5 years, 12 months ago

You're reading it incorrectly. max ⁡ x ∈ R sin ⁡ x \displaystyle\max_{x \in \mathbb{R}} \sin x (for simplification, instead of ( log ⁡ 2 3 ) sin ⁡ x (\log_2 3)^{\sin x} ) is the maximum value of sin ⁡ x \sin x over all x ∈ R x \in \mathbb{R} ; in other words, max ⁡ { sin ⁡ x ∣ x ∈ R } \max \{\sin x | x \in \mathbb{R}\} . You're reading it as { x | x ∈ R ∧ sin ⁡ x = max ⁡ y ∈ R sin ⁡ y } \displaystyle \left\{x \middle| x \in \mathbb{R} \wedge \sin x = \max_{y \in \mathbb{R}} \sin y \right\} (which is a set, not a number).

Ivan Koswara - 5 years, 12 months ago

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