“Coffin” Problem: which is larger?

In this video we determine whether log_2 3 or log_3 5 is larger.

This is one of the 'Coffin' problems, used by the Maths Department of the Moscow State University, as their oral interview questions. 'Coffin' problems were designed to have short and simple problem statements, elegant solutions, but very hard to find under an oral exam environment. You may find more information of such problems from Ms. Tanya Khovanova here:

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[linecheung]

log base 2 of 3:log base 3 of 5
=log3/log2:log5/log3
=log3*log3:log2*log5
=6log3*log3:6log2*log5
=2log3*3log3:3log2*2log5
=log9*log27:log8:log25
>1
So, log base 2 of 3>log base 3 of 5

[MelkorNoir]

You can double each number from the start.
2 log_2 (3) = log_2 (9) which is between 3 and 4. Also 2 log_3 (5) = log_3 (25) which is between 2 and 3. So 2 log_2 (3) > 2 log_3 (5) which implies log_2 (3) > log_3 (5)

[lalitmarwaha]

I just watched Edward Frenkel's podcast called Coffin Problems
Thanks for the video
We really appreciate them

[michaldugas]

How did you find 3/2 ? Is tgere any method to find rational number between two logs?

[richardfredlund]

let log base 2 of 3 = y and log base 3 of 5 = z then
2^y=3 and 3^z=5 so multiplying these equations by 5 and 3 gives 5*2^y=3^{z+1}=15 substituting 2^y in for 3 we have:
5*2^y=(2^y)^{z+1}=2^{y(z+1)}
taking logs base 2 of both sides gives
y+log_2 (5) =y(z+1). Now 5>2^2 so a:=log_2(5)>2 also y>1 so
(y+a)/y = z+1 implies y>z.

[Jop_pop]

Alternative: use base 3^9 for your logarithm and compare log(3)^2=1/81 to log(2)*log(5). Multiplying isn't that hard to show log(2) is just below 1/14 and log(5) is just below 1/6. So log(2)*log(5)< 1/84 < 1/81 = log(3)^2. Divide by log(3)*log(2) and use change of base to get the desired result.

[leventkocalan]

Maybe the same way but we can say like this:
log(base 8 of 9)>1
(2/3)log (base 2 of 3)>1
log (base 2 of 3)>3/2... (×)

log (base 27 of 25)<1
(2/3)log (base 3 of 5)<1
log (base 3 of 5)<3/2... (××)

From (×) and (××)
log(base 3 of 5) < 3/2 < log(base 2 of 3).

[alanx]

book suggestion to learn to solve this?

why compare with 3/2?

[jongyonp]

If I was given this problem in an oral exam, I would've failed...

[edwardjcoad]

3 is the mid point between 2^1 and 2^2 while 5 is that it is less than the mid point of 3 and 9 and therefore log2 (3) > log3(5)? too hand wavy?