I Solved a Very Irrational Equation

[@SyberMath]

Comparing e^pi and pi^e: kzbin.info/www/bejne/oKmwlKFtnsiqbcU

[@Grundini91]

Our Calc teacher in high school made us remember e out to 15 digits. The mnemonic he taught us was 2.7, Andrew Jackson, Andrew Jackson, Right Triangle. This worked because the first 15 digits of e are 2.718281828459045. Andrew Jackson was the President that won the US election in 1828, and 45-90-45 makes an isosceles right triangle.

[@SyberMath]

Mind-blowing!!! 😍💪👏

[@ElChocoLoco]

why?

[@falconsloth5967]

​@@ElChocoLocotests are stupid

[@HoSza1]

how much have you profited from it?

[@audience2]

Andrew Jackson served 2 terms and was the 7th president of the USA, which takes care of the first two digits.

[@UTAngoid]

I had a go before watching this video and started by taking logs straight away: (x + pi).ln(e) = (x + e).ln(pi) But ln(e) = 1. Therefore, and expanding the right hand side at the same time: x + pi = x.ln(pi) + e.ln(pi) Collecting the x terms onto the LHS and everything else onto the RHS gave me x - x.ln(pi) = e.ln(pi) - pi x(1 - ln(pi)) = e.ln(pi) - pi x = (e.ln(pi) - pi) / (1 - ln(pi)) This is the same as the solution in the video, but written slightly differently; I could multiply both top and bottom by -1 to re-arrange: x = (pi - e.ln(pi)) / (ln(pi) - 1)

[@andy_in_colorado7060]

My steps were almost exactly the same as yours.

[@SyberMath]

very good! 🤩🤩

[@SanePerson1]

Exactly, but it would have made the video six minutes shorter.

[@kicorse]

I did it using the change of base, which also gave complex solutions. The imaginary part is (2n pi)/(ln(pi) -1) where n is any integer. The real solution is the case with n = 0, and the real part is (pi - e ln(pi)) / (ln(pi) - 1) in all cases. It fits with intuition that x should be a small positive value because e^pi is only slightly greater than pi^e, so it would only need a small increase in both exponents for the larger base (pi) to compensate for this.

[@SyberMath]

Wow!

[@gkotsetube]

What change of base, exactly?

[@kicorse]

@@gkotsetube pi^(x+e) = e^(ln(pi)(x+e))

[@القناةالترفيهيةالتعليمية-ج1ذ]

Apply the Ln in the two sides of the equation and it will be solved by itself

[@遠傳五華]

And less calculation.

[@EugeneKogan-e8z]

I appreciated that you showed essential properties of logarithms in this example. Thanks!

[@SyberMath]

You're very welcome!

[@benyasir423]

Et si on prend le logarithme ( ln ) des deux membres de l'équation dés le début. Où je me suis trompé?. Merci

[@oenrn]

By the fundamental theorem of engineering: e = 3 = pi Therefore the answer is all x € C.

[@rithvikmuthyalapati9754]

🤣

[@SyberMath]

😁😍🤩😜

[@alextang4688]

Take natural log (ln) and separate x. 😉😉😉😉😉😉

[@SyberMath]

Good thinking! 😍😍

[@roykay4709]

I just took the ln at the second step. Same result. I was hoping for a while, that you ended up with something more elegant.

[@manuelgonzales2570]

Excellent. Thank you!

[@SyberMath]

You're very welcome!

[@jacksonstarky8288]

I was wondering about a base of pi for logarithms right before the video asked the question. I'm not sure what it would be useful for, but I'm a recreational mathematician who hasn't taken a formal mathematics course in three decades.

[@tiptop_1997]

Check our channel for simplified maths videos like this. youtube.com/@elijahmathsclass626?si=-Y-M8HbaaD09oq4E

[@claudelebourlegat]

simple ! use (ln) directly

[@abhaygvasista9433]

this solution is miraculously close to the value of i^i

[@snowfloofcathug]

There was a Pi memorisation competition at school and we had multiple from my class of 14 memorising a few hundred, I got to 1200 by the third year (only really memorising more during the month leading up to the day each year)

[@sophiophile]

Are there complex solutions as well?

[@SyberMath]

@@shmuelzehavi4940 check @kicorse's comment

[@SyberMath]

check @kicorse's comment

[@benYaakov]

At first sight , I thought it wouldn't have closed form. But simplification works out smooth.

[@arantheo8607]

k is real and positive (e/pi)^x has as its domain the set R and the set of positive number as the range As ( e/pi) < 1 , the graph of f(x) = (e/pi)^x - the base is smaller than 1- slopes down as it moves to the right, but it is always positive. As it moves to the left, the graph grows tall very quickly One-to-one and onto

[@Mehrdad_Basiry]

Beautiful question...❤❤❤.

[@SyberMath]

Thank you! 🤩🤩

[@mbmillermo]

These days, and for years earlier, "log" usually means ln, and we use log10 to get the base-10 log, but this isn't how Google does it, for some reason. I think "log" used to mean base-10 log because of how we used slide rules with base-10 logs all the time. Those days are long gone.

[@bsmith6276]

What I see is that log without a specified base meaning log_10 vs ln depends on what level of mathematics work you are at. Highschool and up to basic calculus usually uses the convention that log means log 10 and higher level college courses use log means ln. Information that Google scrapes is dominated by the Highschool/basic calculus crowd.

[@StevenMRSenior]

Log has meant base 10 and Ln base e for over fifty years to me. All the maths videos I have seen on KZbin use this convention and they are not only to high school level. I didn’t do a degree in maths but did do some degree level maths in the engineering degree that I did.

[@mbmillermo]

A few facts about software: GNU awk, GNU Octave, MATLAB and GNU R all use log() for log in base e. Python does this too, but there is an option for a second argument which is the base. Thus, in Python, log(2) would be the natural log of 2, but log(2,10) would be the base-10 log of 2. Both GNU Octave, MATLAB and GNU R have log10() functions, but in awk you must use log()/log(10) instead.

[@zawatsky]

@@mbmillermolog не пишется без основания. Log₁₀x правильно пишется как lgx.

[@mbmillermo]

@@zawatsky lgx? I haven't seen it, but apparently the ISO standard is that we should use lb, ln and lg for bases 2, e and 10, respectively. Thanks!

[@barryzeeberg3672]

Much simpler: take ln() of both sides, then you have a simple algebraic equation that can be directly solved for x, giving the same solution that takes 8 minutes in the video.

[@giorgiocanal1659]

You can solve it in a simpler way by changing the base on the right side π=e^log(π). Then you take the exponents and solve algebraically for x.

[@SyberMath]

good thinking!

[@barberickarc3460]

If we ln both sides right away then isolate the x we get the answer pretty quickly. x = [ ln(pi) (e - pi)] / [1 - ln (pi)] Try plugging it back in to the original equation, everything cancels so beautifully, very satisfying!

[@SyberMath]

Nice! You rock 🤩🤩

[@DanielFSmith]

That was the thing that surprised me in this video! (The solution was pretty mundane, to my disappointment.)

[@berkeunal5773]

Where are you from? What is your main language?

[@SyberMath]

TR

[@berkeunal5773]

@@SyberMath Telaffuzdan anladım 🇹🇷

[@NirDagan]

I applied ln on both sides...

[@levskomorovsky1762]

(x + π)ln e = (x + e) ln π x - x ln π = e ln π - π x (1 - ln π) = e ln π - π x = (e ln π - π)/(1 - ln π)

[@mcwulf25]

Similar to my solution. Didn't bother with the k.

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@davidpatterson5426]

With a calculator I get e^pi = pi^e, so e^x = pi^x, so x =0.

[@davidpatterson5426]

Oops, my bad… they’re not equal… must have miskeyed something.

[@ryanrahuelvalentine2879]

My Hero.

[@SyberMath]

🤩

[@audience2]

x = log_e/pi ((pi^e)/(e^pi)) looks nicer

[@juergenilse3259]

e^(x+pi)=pi^(x+e) ln(e^(x+pi))=ln(pi^(x+e)) (x+pi)*ln(e)=(x+e)*ln(pi) x+pi=x*ln(pi)+e*ln(pi) x*(ln(pi)-1)=pi-e*ln(pi) x=(pi-e*ln(pi))/(ln(pi)-1) I tink i came to the solution (except for the factor -1 on enumerator and denominator, which does not change te value) with less steps, because i first do ln on both sides.

[@alterherrentspannt]

I like your math puzzles.

[@SyberMath]

Glad to hear that!

[@SergeySvotin]

Wolfram Alpha was more precise ^ ^

[@SyberMath]

of course 😁😁

[@SergeySvotin]

@@SyberMath you can just ln both sides right from the start, it would be more simple and you'd get the result in wolfram alpha state)

[@RbWadim]

Bruh. Why not just log the initial statement, and after x+pi = (x+e)*ln(pi) get an answer in 2 steps => x(1-ln(pi)) = e*ln(pi) - pi => answer?

[@shlomobachar4123]

Who invents this kind of crazy problems? 😅

[@SyberMath]

Crazy people like me 🤣

[@shlomobachar4123]

@@SyberMath And I asked since I was in school who sits and creates these unsolvable problems instead of going to the beach? Now I found…

[@shlomobachar4123]

@@SyberMath Now I will invent a question: Solve x^i=i^x ?

[@marklevin3236]

Fake ln of b oth sides and we have a simpler linear equation. Although there are logarithmic coefficients...

[@martialversaux5746]

And why is it surprising ?

[@dominiquebercot9539]

(Pi)^(x+e)= e^(x+e)lnpi A partir de là, ça va tout seul!!!!

[@michaelyap939]

The question of this video is OK, not the best and not the worst. But pretty obvious you intentionally make the solution longer than necessary! You don’t have to do that, because such action bring down quality of your channel.

[@wes9627]

(x + π) * ln(e) = (x + e) * ln(π) ln(e) = 1 x + π = (x + e) * ln(π) [1 - ln(π)]*x = e * ln(π) - π x = [e * ln(π) - π]/[1 - ln(π)]

[@giuseppetornabene989]

ci volevano solo 3 passaggi; devi studiare MOLTISSIMO ed evitare di fare queste figure

[@tamilselvanrascal5956]

🎉🎉🎉

[@barakathaider6333]

👍

[@ThanhNhan_GiaSu234]

Hi❤❤❤

[@marceliusmartirosianas6104]

{{{ po versijos--}}}]e^Pix= e^Pix}}}=\ epix-ePix\=e^Pix-1=epix/1=epix = wpix^2=2 tada epix=1 e=o; pi=1'; x=1 ;

[@giuseppemalaguti435]

x=(π-elnπ)/(lnπ-1)=0,20655...

[@남이명준-s1b]

How can be press that button?