Math Olympiad Problem, which is larger?

[@allanflippin2453]

I think it is easier to prove the right side is greater. Suppose we take the 50th root of both sides. Then the comparison is between 70 and 60 to 6/5 power. The right side can be restated as 60 times the fifth root of 60. It's fairly easy to show that the fifth root of 60 is greater than 2 (2^5 = 32). So the right side is > 120. 70 < 120.

[@robertveith6383]

Yes.

[@patricksheldon5859]

Take the tenth root of each, factor the bases to (7•10) and (6•10), distribute the exponents, divide each by 10 to the 5, and then you factor the left side to 7 squared times 7 squared times 7 and the right side to 6 cubed times 6 cubed times 10. Right side is greater.

[@iirekm]

Simpler but approximate way: use formula x^y = e^(y*log x). Now after taking log of both sides we have 50*log(70) vs 60*log(60), both logs are almost the same, we have 50 vs 60 -> right side is bigger.
Formula x^y = e^(y*log x) is also useful in other cases (e.g. differentiation, integration, computer algorithm analysis), so it's always good to remember it.

[@xyz.ijk.]

@@iirekm this is how I approached it as well and found it to be a much simpler solution

[@michallesz2]

Just look at numbers and powers. A power of 50 means that the number will consist of 50 digits and a power of 60 means that the number will consist of 60 digits, so 60>50
10^1 = 10 (1 )
10^2 = 100 (2 )
10^3 = 1000 ( 3 )
10^50= 10000000 (50 )
10^60= 10000000 (60 )

[@09stir]

That's pretty sick 👍

[@GulibleKarma20]

If 70^50 > 60^60, (70^50) / (60^60) > 1
(70^50) / (60^60) = (70^50) / (60^50 * 60^10)
= ((70^50) / (60^50)) * (1/(60^10))
= (7/6)^50 * (1/(60^10))
as 7/6 is only a bit bigger than 1, it is clear this ratio far less than 1
thus 60^60 > 70^50

[@SidneiMV]

K = 70⁵⁰/60⁶⁰
K = (70⁵⁰70¹⁰)/(60⁶⁰7¹⁰)
K = (70⁶⁰)/(60⁶⁰7¹⁰)
K = (70/60)⁶⁰/7¹⁰
K = (1 + 10/60)⁶⁰/7¹⁰
(1 + 10/60)⁶⁰ < e¹⁰ < 7¹⁰ => K < 1
70⁵⁰/60⁶⁰ < 1
*70⁵⁰ < 60⁶⁰*

[@natashamiljevic1220]

Excellent and more elegant solution!

[@Gideon_Judges6]

@@SidneiMV I don't get the 7^10 terms in the denominator. As best as can tell they should have been 70^10 like was multiplied on the numerator. I thought it was a typo but it's repeated on multiple lines.

[@siddharthchabra9022]

@@Gideon_Judges6 this is incorrect it should be 70^10 in denominator
limit n-> infinity (1+1/n)^n = e is an identity so your step of lhc

[@ParalyticAngel]

Off course 60^60 is greater. Exponential growth is killing it.^^

[@mehulsingla6565]

Can we do it by using log on both sides and then checking which side is greater

[@dejankaranovic]

Intresting problem to solve and find evidence to prove even there is math roules to conclude very fast wich is bigger.(check exp)

[@patricklaenen3468]

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[@romank.6813]

Which is larger: e^(π^(π^e)) or π^(e^(e^π))?

[@dickottel]

@@romank.6813 yo mama 😸

[@Киноварь]

Мгновенно решил по теореме "степень пизже основания"

[@petersusilo9588]

70^50 = (70^5)^10 = 16807000000^10
60^60 = (60^6)^10 = 46656000000^10
So, 60^60 is larger than 70^50

[@yuusufliibaan1380]

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