Math Olympiad Problem, which is greater?

Comparison, which one is greater, 3^π or π^3?
This fantastic math problem frightened 200K students! Watch the video and find out the answer!

[@costarich8029]

Really simple rule. If a and b are both greater than e (2.71828...) then the expression with the bigger exponent always wins.

[@jejojoje9521]

And otherwise, the bigger base always wins?

[@costarich8029]

@@jejojoje9521 I tried to include a link, but it seems to have gotten clobbered. If both terms are less than 'e' then yes you are correct. Largest base wins. Otherwise it is complicated as the crossover point follows a fussy curve involving the Lambert W (product log) function. I'll try to post the link to the graph in the next comment to see if it's just my inclusion of the link causing it to get deleted.

[@costarich8029]

Yeah the comment with the link gets auto-deleted. The critical curve where x^y=y^x is solved by using the Lambert W function. x = -yW(-log(y)/y)/log(y) which gives a curve that looks like it asymptotically approaches (1,∞) and (∞,1) while going through the point (e,e).

[@mathwindow]

@@costarich8029

[@Jimmy_Johns]

Es de Costa Rica?

[@Presserp]

I set PI = X + 3 where X is decimal portion of PI. 3^(X+3) vs (X+3)^3. Left side is 3^X*3^3. Since X is < 1, this is (3)*27 which is < 81. The right side expanded is X^33+9X^2+27X+27. Since X

[@ILKERUMAM]

That is cunning solution to prove, using easy math calcs so that anyone understand easily... well done!

[@mekbebtamrat817]

I do not follow your reasoning. In both cases, you managed to establish only upper bounds and you have not actually proved the inequality. In other words, you actually have to show that 3^pi is greater than something and that pi^3 is less than something. In your argument above, you only established that both are less than two numbers. That is not very helpful as 3^pi can still be less than pi^3.

[@unclesmrgol]

It takes seconds to determine that lim(ln(x)/x) as x goes to infinity is zero, and, further, that there's a maximum at e. Since e < 3 < pi, we have everything we need.

[@snarkybuttcrack]

only took me milliseconds, you lose

[@unclesmrgol]

@@snarkybuttcrack Excellent! Congratulations on your most excellent math skills.

[@6310-c5h]

well.. the function f(x) = x^(1/x) is decreasing when x>e. (defferetiate it for proof) hence if e < a < b than a^(1/a) > b^(1/b) hence (a^(1/a))^(ab) > (b^(1/b))^(ab) hence a^b > b^a here 3 > e and pi > e. and 3 < pi. so 3^pi > pi^3 peace to all

[@wostin]

Wait, so for every two numbers when one is greater than the other (a b^a ??? Or am I just confused?

[@6310-c5h]

@@wostin this is true if they both are not less than e (the base of natural lorarithm 2.72..). example e^pi > pi^e and if the both are not greater than e ***upd. AND they both are not less than 1. (a < b) => (a^b < b^a) example 2^2.5 < 2.5^2 and.. if one less than e and another greater than e example 2.5^3 vs. 3^2.5 huh i dunno ¯\_(ツ)_/¯

[@thorinpalladino2826]

@@wostin When b>a, a^b > b^a when a>e 1^10 < 10^1 and 2^3 < 3^2, but 3^4 > 4^3

[@anthonyvalenti9093]

I graphed y=3^x and y=x^3 by hand. They are equal at x=3. And y=3^x is greater after that.

[@patrickpalen9945]

That's how I did it too! Then watching the video I must say making them both into the same function and differentiating was a brilliant trick!

[@mekbebtamrat817]

Nice one

[@betrand.F]

Pi is between 3 and 4 3 cubed is the same as 3 cubed 3 to the 4th is greater than 4 cubed Therefore 3 raised to the pi is greater than pi cubed. This is a more simple approach, but it not necessarily proves it*😅

[@swedishpsychopath8795]

Or just use induction combined with intuition. Substitute PI with 2. You then have 3 to the power of 2 (which is 9) on the left side. And you have 2 to the power of 3 (which is 8) on the right side. So is 9 bigger than 8? YES! And this is true for all numbers N. If you have PI instead of 2 you know the left hand side will always "win".

[@allanflippin2453]

I considered this approach, but wasn't able to completely prove it. Suppose we start the sequence with 1^2 and 2^1, the left side is less than. Going to 2^3 and 3^2, the result is greater than as you showed. Working the next few (3^4 4^3. etc) they lead to the left side greater than. Intuitively, if this continues to infinity, the two sides seem to approach equality asymptotically. But how does one prove it? If you're at point N and the left side is greater, how to prove that point N+1 also has left side greater? I couldn't figure a method to prove it that doesn't involve calculus (and likely log as well).

[@Patrik6920]

@@allanflippin2453 ..did graph this a^x=x^a , at e thers one intersection point, under and over thers two under ae, a^x is always greater after the second intersection point for very lage values of a the first intersection point goes toward 1 and the second is x=a^a tested a between 0 < a < 1 000 000 ..and its always true... edit: actually 0's not tested as 0^0 is undefined, just close to zero values tip: use a log axis scale, makes life easier ;D (a^x becomes a traight line)

[@allasar]

​@@allanflippin2453I think his point is, we inuitively know the answer without proof, so if the question is just "which is larger?" we'd get the points. But I agree, we need proof since intuition can be wrong, and an answer (even a correct one) without proof is dissatisfying.

[@allanflippin2453]

@@allasar Yup, I am dissatisfied that I can't prove it :D

[@samueldeandrade8535]

​@@allanflippin2453 so, you are trying to prove N^{N+1} > (N+1)^N for N≥3? First, we can prove 2^k < (k+1)! for k≥2. Indeed, for k=2, 2² = 4 < 6 = 3! = (2+1)! ✓ If it is valid for k≥2, then 2 2×2^k < (k+2)×(k+1)! => 2^{k+1} < (k+2)! Done. We proved 2^k < (k+1)! for k≥2, which implies 1/(k+1)! < 1/2^k (*1*) for k≥2. Now we prove sum 1/k! < 3 (*2*) for any k. Indeed, 1/0! = 1 1/1! = 1 1/2! = 1/2 1/3! = 1/6 < 1/2², by (*1*) 1/4! = 1/24 < 1/2³, by (*1*) ... Adding everything, sum 1/k! < 1+1+1/2+1/2²+... < 1+2 = 3 So we proved (*2*). Now we prove (1+1/N)^N ≤ sum_k 1/k! (*3*) for any N, with k = 0,...,N. Indeed, expanding (1+1/N)^N using the binomial formula (1+1/N)^N = sum_k C(N,k)(1/N)^k with k=0,...,N and C(N,k) = N!/(k!(N-k)!) So, fixing k, 1≤k≤N, we have the term C(N,k)(1/N)^k = (N!/(k!(N-k)!))(1/N)^k = (N(N-1)...(N-k+1)/k!)(1/N)^k = (N/N)((N-1)/N)...((N-k+1)/N)(1/k!) = 1(1-1/N)...(1-(k-1)/N)(1/k!) ≤ 1/k! because for each factor 1-j/N, j=0,...,k-1, 1-j/N ≤ 1 So we proved (*3*). This means we have, for any N, (1+1/N)^N ≤ sum_k 1/k! , by (*3*) < 3 , by (*2*) proving (1+1/N)^N < 3 for any N, which implies (N+1)^N < 3N^N (!!) Finally ... for N≥3, we obtain the inequality, (N+1)^N < 3N^N , by (!!) ≤ N×N^N , by 3≤N = N^{N+1} Done. That's a way.

[@komunist431]

π³ = 3 ^ (3 * ln(π)/ln(3)) . 3 * ln(π)/ln(3) < 3,126 < 3,14 < π . 3^π > π³ .

[@betabenja]

7:12 sounds like you're threatening me on the street

[@SubbharaoApplasamy]

Oh My Dear. Those Advance Mathematics still give me nightmares 😢

[@tullfan2560]

You could use an induction proof. If π was equal to 3, then 3^3 = 3^3. if π was equal to 4, then 3^4 > 4^3 (81>64) if π was equal to 5, then 3^5 > 5^3 (243>125) But π = 3.142... and 3 > π > 4 > 5 Hence 3^π > π^3

[@robertveith6383]

That is not a correct proof. Pi is not equal to 3.142... That is nor an induction proof.

[@robertveith6383]

That is not an induction proof.

[@tullfan2560]

@@robertveith6383 What do you mean pi is not equal to 3.142... That is actually rounded. It should be 3.14159... But, what I'm really saying is that it is between 3 and 4. If it's not induction, what do you call it?

[@robertveith6383]

​​@@tullfan2560 Pi, of course, is *not* equal to 3.142... The three dots (an elliipsis), means the digits continue after the 2, which they do not. Pi = 3.141... instead. Induction will involve a base case, and using a variable, for instance.

[@tullfan2560]

@@robertveith6383 You can see that when the difference between the number and its exponent increases, the value of 3^N - N^3 increases. The difference between the two terms is zero when N=3 and monotonically increases as N gets bigger. As π is greater than 3, 3^π will be greater than π^3.

[@zionfultz8495]

Simple to solve really. 3^x = x^3 when x = 3. 3^4 = 81, and 4^3 = 64 therefore 3^4 > 4^3. Therefore 3^pi > pi^3. The only hand wavy thing was 3^x = x^3 when x = 3. You should technically prove this is the only real solution. But since cube root's alternative solutions and logarithms alternative solutions only matter in the complex I do believe it is easy to see there is only one.

[@jamesharmon4994]

I love your substitution of x for pi. This makes the problem obvious. You can graph 3^x and x^3.

[@jamesharmon4994]

I love your substitution of x for pi. If you graph 3^x and x^3, it becomes clear that for any value of x greater than 3, 3^x is greater than x^3.

[@yurenchu]

Sorry, but your logic appears to be flawed. According to your logic, we can also say: 3^x = x^3 when x = 3 . 3^2 = 9 , and 2^3 = 8 , therefore 3^2 > 2^3 . Therefore (you'd conclude that) 3^e > e^3 . Which is obviously incorrect, because 3^e < e^3 .

[@thedeathofbirth0763]

Thank you. Clear explanation without going off topic. I am glad I checked out your video. I am looking forward to more postings and going through what you already have made.

[@mathwindow]

@@thedeathofbirth0763 Thanks!

[@xoo2009]

Compare 3^x & x^3. When X=2, 9>8; When X=4, 81>64. 2

[@xanderlopez3458]

Thank you for the consistent posts. I get a little smarter every time. Your explanations always seem to come together, even if it takes me half an hour or so to get some concepts. 😊

[@mathwindow]

@@xanderlopez3458 Thanks!

[@Akenfelds1]

There's a much simpler way to prove the same outcome. Alas, writing in a KZbin comment isn't the most efficient way to explain it.

[@sevhenry]

It is much simpler to take the log base 3 on each side. Log 3 = 1 in that base and log PI il a little greater than 1.

[@mmarques2736]

@@sevhenry log_3(\pi) > log_3(3), but that doesn't tell us if 3 log_3(\pi) is greater of lower than \pi log_3(3), so it is not that simple

[@jonathanbirchley]

Yes, I feel it's the logical first step to take. It makes the result easy to see from the (well-known) fact that log(x) increases slower than x if log>1, and faster if log

[@212ntruesdale]

Really enjoyed the analysis! Interesting that you take the derivative with the product rule, not the quotient rule. I used to do that, but then decided it was better to just memorize the quotient rule.

[@jonmetaphorist1327]

So it's always easily comparable when both >e or both

[@Potemkin2000]

I just imagined 3^x vs x^3 as a graph and "felt" the answer - easy peasy

[@micke_mango]

Since this, and similar integer problems, are so common on math channels, 5 years ago I chose to remember this rule of thumb: if the exponents are larger than e, the larger exponent wins over the larger base. I think it holds true, at least for the problems I have come across

[@icesurf5815]

Sooo, everyone’s got some complex solution process but has no one literally just plugged it into a calculator, I got the answer in seconds

[@kadardur]

Is saying 3^pi>3^3.14>3.15^3>pi^3 is enough?

[@sandeepagarwal7387]

Excellent. Very well explained... with a simple yet concise flow...

[@markcairns9574]

I just come to the comments for the answer and realised why i had to resit my maths to get a pass.

[@kennethgee2004]

yawn 3^pi is large as it is closer to e. this is already forever answered. in any a^b versus b^a if a,b > e and given a

[@robertveith6383]

Write sentences! What you have written are groups of words that do not have capitalized first words or needed punctuation. Do not expect readers to look at your sloppy, lazy, ignorant post.

[@El_Carrito_del_Helao]

Nice, but unnecessarily detailed. Essentially ln(3) ~ ln(pi), and then pi*ln(3) > 3*ln(pi)

[@maxmustermann3938]

This one I can get behind. I'd say for a more rigorous proof, we'd use the property that (d ln(x)/dx) < 1 for x > 1 and ln(1)=0 which means that ln(x) grows strictly slower in that interval and is guaranteed to be a smaller value than the constant function f(x)=x

[@HoD999x]

"we cannot calculate 3^pi so let's calculate ln(pi) in our heads"

[@yiutungwong315]

In the Riemann Paradox and Sphere Geometry System Incorporated π = 2 So 3^π - π^3 = 3^2 - 2 ^3 = 9 - 8 = 1 3^π is Larger than π^3

[@jimv9210]

Maybe someone has already made this point, but isn't it sufficient to observe that the left side must be greater since a log varies more slowly than its argument? π/3 vs ln(π)/ln(3) It may not be quite as explicit as Math Window's demonstration, but if you're aware of that fact (and maybe need to come up with a fast answer) I think it's all you need to know.

[@michelebaffo5741]

not quite, try 2^3 and 3^2

[@jimv9210]

Yes, I see (another "intuitively obvious fact" down the ol' toilet).

[@davidmilhouscarter8198]

I haven’t watched the video yet. My guess is 3 raised to pi is larger because the exponent is larger.

[@quigonkenny]

Well, both numbers are greater than e, so it should be the one with the greater exponent, 3^π.

[@smithfrederick2]

cube root of both sides and dividing by 3 gives 3^0.1415.... compared to pi/3 3^0.1415 approx: but greater than 3x0.38=1.16 and pi/3 is approx but def less than 1.05, thus 3^pi > pi^3

[@Anonymous-zp4hb]

Might be overkill, but I just threw calculus at this problem immediately... f(x) = 3^x / x^3 f'(x) = (1/x^6)( (ln3)(x^3)(3^x) - (3^x)(3x^2) ) What are the roots of f'(x) ? x clearly can't be cannot be zero, so... (ln3)(x^3)(3^x) - (3^x)(3x^2) = 0 (3^x)( (ln3)x^3 - 3x^2 ) = 0 (ln3)x^3 - 3x^2 = 0 (x^2)( (ln3)x - 3 ) = 0 (ln3)x - 3 = 0 x = 3/ln3 Since 3^4 > 4^3 ( 81 > 64 ) And 3^3 = 3^3 ( 27 = 27 ) And 3/ln3 < 3 We can conclude that for all x in (3, +infinity) 3^x > x^3 Therefore, 3^pi > pi^3

[@davidseed2939]

Take powers of 1/(3π) 3^(1/3) ~ π^(1/π) if cam be shown that the function x^(1/x) has a maximum at x= e and is monotonic decreasing away from ths maximum. π >3 thus 3^(1/3) >π^(1/π)

[@bkkboy-cm3eb]

3^x > 3x (x>1) ∴3^(π/3) > 3(π/3) = π ∴3^π > π³

[@mzallocc]

3^π vs π³, take the cube root of both, 3^(π/3) vs π Since π > 3, π/3 > 1 and 3^(π/3) > π, cube each side, 3^π > π³

[@bkkboy-cm3eb]

@@mzallocc why 3^(π/3)>π ?

[@mzallocc]

@@bkkboy-cm3eb ok here we go: 3^π vs π³, take the cube root of both, 3^(π/3) vs π = 3(π/3) let x = π/3 3^x vs 3x. for x > 1, 3^x > 3x, and since π > 3, 3^(π/3) > 3(π/3) = π ie 3^(π/3) > π, cube both sides 3^π > π^3 Cedit to you, I just took your solution and did it reverse.

[@SSJBartSimp]

I just did the range of 3^3 to 3^4 is greater than 3^3 to 4^3, and pi is between 3 and 4 so 3^pi is greater than pi^3.

[@АртемФедоров-ю7б]

Substitute pi with 4 (why 4 and not 2? Because it should be greater than 3) and do the math

[@IITIAN_MOTIVATED]

Why cant we use log to simply solve it

[@nigelrafferty2545]

(3 + 1)^3 = 64 and 3^(3+1) = 81. Now let the 1 be reduced by small amounts to zero - would the inequality ever change sign? So 3^pi is greater. Why make it so complicated? And you don't have to remember any tricks!

[@divadbyzero2793]

Fun fact: If both a and b are greater than or equal to e, and a>b, then b^a>a^b always.

[@michallesz2]

3^pi = cln( pi*ln(3))=31,544... pi^3 = cln( 3*ln(pi))=31,006...

[@juanmacedo8816]

Solved this by closing a limit

[@hydraim9833]

Excellent explanation really. Instantly subscribed, thank you for this video

[@ParalyticAngel]

Before I watch. Off course pi other 3 is larger. Cause pi is larger than 3 and exponential growth is always winning, except the base is = 1.

[@auni4078]

I think taking log with base π would make this a pie

[@gregotterstein6773]

Not sure why you had to bring calculus into this. It can be explained algebraicly

[@Mmmyyyzzz]

3^4 means 81 and 4^3 means 64 so bigger exp wins

[@dustyoldduster6407]

I just used my scientific calculator. Lot less work.

[@fariesz6786]

i mixed up NAZ and ZAN again and got confused q3q

[@kevindegryse9750]

Great. I would have appreciate to have the justification of why the greater than symbol does not flip at any point (common mistake with inequalities) (multiplication by positive numbers, ln function continuously increasing). But really great overall.

[@keithdow8327]

Thanks!

[@mathwindow]

Thanks for your support!

[@santer70]

Why didn’t you take a calculator for 5€…?

[@grqb_tg2478]

@@santer70 ... Solo los brutos dependen al 100% de una calculadora .

[@johnhallett5454]

@@santer70 exactly what I did... and mine was only £1 in a Charity Shop...😂

[@santer70]

@ johnhallett5454 👍

[@ParadoxumNoobTDS]

@@santer70 well how do you think your calculator calculates math for you? You think it's magic? NO, PEOPLE HAVE TO USE THIS MATH TO DERIVE A FORMULA FOR CALCULATORS TO CALCULATE THESE EXPRESSIONS BRO

[@rolandpokorny2813]

3^3,14 is greater

[@noni345]

The circle is larger

[@soolsoolman]

let pi=4 , game set.

[@Leonhard-Euler]

pi^3 = 3^3*(1 + (pi-3)/3)^3 = 3^3 * (1 + (pi-3)/3)^(3/(pi - 3) * (pi - 3)/3 * 3) = 3^3 * ((1 + (pi - 3)/3)^(3/(pi - 3))^(pi - 3) < 3^3 * e^(pi - 3) < 3^3 * 3^(pi - 3) = 3^pi

[@denzilgounden4044]

Let pi =2 then 3^pi more

[@mathwindow]

@@denzilgounden4044 fun method!

[@RefurbIshment-z7l]

How about just taking a guess 50/50

[@SFefy]

Sorry but ln(π^3) is not equal of 3×ln(π). ln is the natural logarithm so the base number is e. π based logarithm is logπ(x)where π is the base number of the logarithm function.

[@bruhifysbackup]

What are you on? This is a property of logarithms.

[@yurenchu]

For any real number a>0 and any real number b, a^b = (e^ln(a))^b = e^(ln(a) * b) = e^(b*ln(a)) and hence ln(a^b) = b*ln(a) This also holds true when a = π and b = 3 , and therefore ln(π^3) = 3*ln(π) There is no mention or implication of a "π based logarithm" in the video.

[@ParadoxumNoobTDS]

@SFefy @yurenchu Here's a complete proof using fundamentals for the expression: ln(π^3) = 3 × ln(π) We will define the natural logarithm of π^3 or ln(π^3) as such: I. For any logarithmic function, we define as: f(b, x, y): b^x = y OR: log b (y) = x we generally read this expression as: "the logarithm base b of y is equal to x" so the natural logarithm is unique, the base is by default Euler's number e ~ 2.718 now we define a function for a natural logarithm by plugging in base b = e: f(e, x, y): e^x = y OR: log e (y) = x OR: ln(y) = x so if you plug in y = π^3 f(e, x, π^3): e^x = π^3 OR: ln(π^3) = x (1) Again, ln(π^3) = x is exactly the same as: log e (π^3) = x which reads as either: "The natural logarithm of π^3 is equal to x" OR, "The logarithm base e ~ 2.718 of π^3 is equal to x" II. We should know: the square root of a number *n* can be expressed as: √n OR ²√n OR n^(1/2) similarly, the cube root of n is: ³√n OR n^(1/3) III. Now, if we cube root both sides of the expression e^x = π^3 It will become: (e^x)^(1/3) = (π^3)^(1/3) we know that (a^b)^c = a^b^c = a^(b×c), therefore: e^(x × 1/3) = π^(3 × 1/3) e^(x/3) = π So now if we plug y = π into f(e, x, y): f(e, x, π): e^(x/3) = π OR: ln(π) = x/3 (2) Notice that: ln(π^3) = x (1) ln(π) = x/3 (2) we can rewrite expression (2) as: 3 × ln(π) = x (3) Combining (1) and (3), hence: ln(π^3) = 3 × ln(π)

[@kw4093-v3p]

I guessed and got the right answer

[@romank.6813]

Compare these two: sqrt(6)^pi and pi^(sqrt(6)).

[@-wx-78-]

Коварство запредельное: √6 и π лежат по разные стороны от e. Хотя можно и усугубить: √7

[@romank.6813]

@@-wx-78- Неа, лучше (е-1/е)^π и π^(е-1/е)

[@-wx-78-]

@@romank.6813 Месье знает толк в извращениях. 😉 P.S. В который раз вижу под русским текстом “Translate to Russian”, сподобился нажать - и вместо объяснения про константу Эйлера получил замену «Неа» на какой-то “her”.

[@bas1c941]

Решил по теореме "Степень пизже основания"

[@yurenchu]

Let f(x) = x^x . Then f'(x) = (x^x)*[1 + ln(x)] , and f'(x) = 0 ==> x = 1/e . From there, we can show that f(x) is continuously decreasing on the interval (0, 1/e) . Since 1/π and 1/3 are on that interval and since 1/π < 1/3 (because 3 < π), it means that f(1/π) > f(1/3) . ==> (1/π)^(1/π) > (1/3)^(1/3) ... take the reciprocal of both sides, which flips the "greater-than" sign (since 1/x is a continuous and monotonously _decreasing_ function for positive real values of x) ... (π)^(1/π) < (3)^(1/3) ... raise both sides to the power of 3π ; since x^(3π) is a continuous and monotonously _increasing_ function for positive real values of x, the "less-than" sign is unaffected ... (π)^(3π/π) < (3)^(3π/3) π^3 < 3^π

[@angusmackaskill3035]

3 to the power of pi

[@AnanthNat]

In general, smaller number raised to bigger power is greater than the other way around. 2^8 > 8^2 I hope based on this logic, we can conclude that 3^π > π^3

[@zihaoooi787]

this is only by heuristic, though. 3^2 > 2^3

[@robertveith6383]

Your post fails.

[@joeytian]

Or just use the calculator

[@jackcraftsolar]

3^pi > pi^3

[@Kriorem]

3 is less than pie...

[@jimf2525]

Advice: Don’t rattle off the chain rule. Show your work. I gave you a thumbs up expecting you’ll improve.

[@allasar]

Disagree. I would say most viewers of this channel know how to derive a fraction. Would be tiring to go over each step everytime, already knowing the result of the step. Basically, we skip steps all the time (for instance, we do not need the step by step derivation of lnx to get to 1/x). The question is: which steps do we skip? Personally I would not draw that line at the derivation of a fraction. Most of us have done it so many times, we do not need to see (f'g - fg')/g^2. Nor do we need (1/x × x - lnx × 1)/x^2 as a step, since we already calculated that in our heads before it is written down. There is no clear line, but writing out each step would get old really fast. We are solving an Olympiad question, not filling in a math exam.

[@jimf2525]

@@allasar We’ll agree to disagree, but not even Michael Penn skips that much math in 1 step. Though I admit he has, but in those cases his videos were quite long. Also, I like that ur willing to fight youtube’s algorithm on length. Content creators are making their videos for too long these days. They constantly stretch one or two minutes of information into 15. So, I probably shouldn’t have complained. 😀

[@valeyard00]

I'd rather she explained the chain rule than write out ln (a^b) = b ln(a). If you need to spell that out for someone then they aren't gonna understand chain rule off the top of their head either

[@caseywood9781]

Pi times three

[@joanignasivicente2012]

Great!❤

[@toshogme]

dude...lol too much. This is good practice for all you know....

[@ugurboyac1114]

BİR ÇUVAL İŞLEM YAPTIN!😀 ÜSSÜ BÜYÜK OLAN BÜYÜKTÜR. 😄😄😄😄😄

[@ikonikgamerz3853]

For positive bases & positive exponents, I use the rule of abba or a^b () b^a If a > b then a^b > b^a If a < b then a^b < b^a If a = b then a^b = b^a

[@octalbert7280]

This rule just doesn't work like if a = 10 and b = 2, a^b < b^a and not the opposite

[@SirRebrl]

@@octalbert7280 It didn't work for the problem in this very video, either, since 3 < pi and 3^pi > pi^3

[@prasoon7916]

This is pure trash

[@prasoon7916]

for positive numbers, to compare a^b and b^a, then if a>b>e, where e is the euler's constant (around 2.71), then a^b < b^a, and if e>b>a, then b^a > a^b, if a and b are on the other sides of e, then it can go both ways and we need to find something else

[@robertveith6383]

​@@prasoon7916-- Write a sentence.

[@elgoogssie3969]

The evidence is much simplier. What's greater? 3^2 or 2^3? No need to do much more calculations.

[@Chiavaccio]

👏👏👍

[@victorchoripapa2232]

Pi^3 is greater than 3^Pi

[@drbonesshow1]

This person keeps writing smaller and smaller. In the limit thereof this analysis equals zero. Poof!

[@marceliusmartirosianas6104]

3^Pi=Pi^3=12 ACADEMIC MARCELIUS Martirosianas 12 may AcademiC universita della Florida et Semi-protectet edit filter on 9 may 2o22 AcademiC Hcm Bon 11 augusre 2o2o Felds Medalist ACADEMIC Universita di Humboldey General Doctor Expert 2o17 -----2o23 17 moksliu atradejas. Nobel prize study in China -2o22.

[@justinlloyd3]

Pulls out calulator... 3 to the pi is larger. No need to watch further. But here is a comment.

[@juaneliasmillasvera]

Uff so tired of this type of videos... a^b > b^a if b>a. END.

[@Shyguy5104]

@@juaneliasmillasvera not quite right 2^3 < 3^2 and 3>2

[@juaneliasmillasvera]

@@Shyguy5104 Mmmm, by bad, so there is not a rule for every case of exponents changing variables? or there is a minimun diference between the two variables since what we can approximate the problem in a generallizate way? =)

[@joshualee1595]

@@juaneliasmillasvera well I’m not sure if there is complete generalization for whether a^b or b^a is greater, intuition should tell you that the larger the base number, the more exponent size dictates the overall value of the number. Think 2^3 and 3^2 versus 49^50 and 50^49. Intuition tells me that 49^50 is the larger of the two, but 3^2 is the larger of its pair. The question is where this relationship flips, I’m sure someone has already calculated that limit, and it could be a cool exercise to try out for yourself.

[@bbo1707]

@@joshualee1595 Actually the video already solves most of that by finding the maximum value at e. Assume ab^a, if both a and b are smaller than e, a^b