how is i^x=2 possible?

[@blackpenredpen]

Learn more complex numbers from Brilliant: 👉 brilliant.org/blackpenredpen/ (20% off with this link!)

[@ayoubbenchetioui6481]

pls Can 2^x=0 in the field of complex num ?

[@icecoffee3485]

@@ayoubbenchetioui6481 x is negative infinity

[@ChavoMysterio]

Please solve this equation: (-2)^x=2 Thank you

[@Mnemonic-X]

Where can I apply this useless knowledge?

[@govcorpwatch]

We need more τ. e^-(τ/4 + nτ).

[@debtanaysarkar9744]

I really love it when you say, " I don't like to be on the bottom, I like to be on the top"🤗🤗🤗🤗

[@b_atanassov]

🤨

[@thexavier666]

A very normal, totally not suspicious, comment

[@stefanalecu9532]

🤨📸

[@noreoli]

@@thexavier666 yea absolutely, no complications there right? 🤨

[@debtanaysarkar9744]

@@noreoli true, no complications 👌👌👌

[@AyberkDurgut]

Complex world is crazy.

[@zeroo8756]

I wil be find new world Name is fantastic number

[@Begubut2]

By saying crazy, you are underestimating the craziness of complex numbers.

[@mehmetalivat]

Gerçekten bu inanılmaz. Kuantum mekaniğinde de çok önemlidir.🙂

[@tusharjawane9056]

I am not even good in real numbers

[@stefanalecu9532]

Because the world is complex by itself

[@enderw88]

"Please don't say 90 degrees, we are all adults here"

[@vivianriver6450]

This one is fun. I actually found your e^(e^x) = 1 video first. I took a different approach to solving this. I tried computing the log-base i of 2, which using log quotient rule, is equivalent to ln(2) / ln(i), the bottom part of which I was able to solve by getting the angle (or family of angles) that has a cosine of 0 and a sine of 1, which is pi/2. Thanks!

[@blackpenredpen]

Awesome, thanks!!

[@redroach401]

​@@blackpenredpenhello

[@simonwillover4175]

Of course! In fact, a^b can always = c, for all Complex numbers {a,b,c} where {a,b,c} not = 0 or 1.

[@DroughtBee]

I’m scared to look up that proof

[@maximilianarold]

@@DroughtBee The proof is left as an exercise for the reader

[@TheEGod.]

@@maximilianarold I think there isn't really a proof. I think it's just an assumption. Like 5^3 will always be equal to a value. Lets call this value c. so 5^3 = c. As we know, c = 125, so 5^3 is 125. We knew what a and b were, but we didn't know what c was, but there was an answer. So what if we know the value of a and c, but not b. so i^b = 2. Last time we didn't know a varible, there was an answer, so there must be an answer again. That is my assumption on how he got it, but he might actually have some reason behind it...

[@19divide53]

Let a^b=exp(b*ln(a)), since exp: C -> C* is surjective, there is a complex number z such that exp(z)=c.

[@parikshitkulkarni3551]

I have a truly marvelous demonstration for this proposition which this comment section is too narrow to contain.

[@Saytome165]

"Complex number is a pathway to many abilities some consider to be unnatural." - Chancellor Palpatine said during his complex analysis lecture

[@srengp3805]

That was the exact line i thought of when first learning about complex numbers

[@terryjwood]

He's not wrong.

[@chrisjuravich3398]

Very nice explanation about why the +4n part is needed to complete the answer. i raised to any multiple of 4 will always result in 1.

[@beatrix4519]

this kind of math is so interesting to me I never took precalc or a calculus class just college algebra we only got a slight introduction to imaginary numbers so all of this baffles me glad I don't need calculus for my degree 😅

[@tomasgalambos3115]

I have been watching your videos for about 2 years now, now that Im in UNI, and im learning more and more about math, I can follow the videos much better, and I just love to see the progress, and the interesting things you show here on youtube

[@proximitygaming8253]

**For those who want the tl;dr explanation:** i^x = 2, so x = log base i of 2 = ln(2)/ln(i) by base-change. This is just ln(2)/(pi/2 i) = ln(4)/(pi * i).

[@mathcat4]

Hey, nice video, I've got a fun challenge for you: Determine all positive integer pairs (p, q) for that p^q + q^p is prime. Not what you usually do, but it has an interesting solution.

[@zoomlogo]

oh hello there lol

[@Farid_Bang_Official_Channel]

Nice question. xD. Since you could say that p is equal to a number 2ⁿ-2 and q is 1 therefore p^q+q^p is 2ⁿ-1. Since i know that it isnt determined whether there is an infinite amount of mersenne or not, answering this question, would be either quite impossible or just impossible. A couple mersenne examples would be (2,1): 3 (4,1) : 5 (6,1) : 7 .... You could generally say, that there are infinitely many tupels where p is a prime -1 and q is 1 so that (p -1)^1 + 1^(p-1) = p So my answer would be that there are infinitely many tuples, as many as there are Prime numbers...

[@Farid_Bang_Official_Channel]

(2,3)...

[@damyankorena]

wlog assume p=2 and eval mod3 Very easy ngl

[@mathcat4]

@@zoomlogo lmao hi

[@SidneiMV]

The "secret" is always the same: put everything in base *e* (Euler's number)

[@Player_is_I]

This is my humble request to whomsoever is reading, please consider my problem::: By Euler's identity => e^iπ + 1 = 0 => e^iπ = -1 Square both the sides => (e^iπ)² = (-1)² => e^2iπ = 1 take natural log of both the sides => ln(e^2iπ) = ln(1) => 2iπ = 0 Please explain😢😢😢 By the way I have a couple more such demonstrations that kinda contradicts the identity which I am unable to recall rn, although I also have worked with this a lot, it feels not a peaceful identity at somepoints, But I do remember that the problem in the eq comes right after squaring both the sides. I am not sure with all this as I did this a long time ago but whatever I wrote is what I remember rn, sorry if I wasted any time, please consider atleast replying 🙏🙏🙏

[@ianzhou3998]

Here is what you wrote: Square both the sides => (e^iπ)² = (-1)² => e^2iπ = 1 ^^This is not true. For complex inputs, a^b^c is not always equal to a^(bc). The rule of complex analysis dealing with log and exponent branches says so. Or else, one can prove anything we want. Here's one of my personal favorites with this fallacy: Suppose we have a real number a. Then, a = e^(ln a) by definition. It follows that, a = e^(1 * ln a) = e^[(2iπ/2iπ)*(ln a)] = e^[(2iπ)*(ln a/(2iπ))]. Applying our fallacy, we see that the above expression equals [e^(2iπ)]^[ln a/(2iπ)]. But e^(2iπ) = 1 by Euler's identity. Thus, we get: a = 1^[ln(a)/(2iπ)]. But 1^x = 1, (unless you use the same fallacy!) so all a = 1. Thus, all real numbers are 1 (obviously not true). BPRP actually did a video on this a while back, seeing if 1^x = 2 is possible. Long story short, that equation had no solutions. But, there was a solution to the equation 1 = 2^(1/x). Raising both sides to the power of (1/x) caused some issues with domain and range restrictions, so the solutions obtained were technically "extraneous."

[@Player_is_I]

@@ianzhou3998 Thank you very much for correcting me or rather teaching me the actual reason. I just took a look at some articles, I didn't noticed the actual law which was a^b^c = a^(b×c) for these elements should belong to the Real world. Well I was on the right path to find my mistake on my other such demonstrations that had the same mistake, when I said "the problem in the eq comes right after squaring both the sides". Half knowledge is more dangerous than no Knowledge Anyways thank you for explaining that much and telling about the video of 1^x=2 and sorry If I were to be silly. 😊

[@rogierownage]

@@ianzhou3998 This is not quite right according to Microsoft Copilot. What goes wrong is that 1 actually has infinite imaginary representations. Following Euler's formula you get 1 = e^(n2iπ). When n=0, this leads to the real representation e^0 which is ofcourse 1. But if you substitute this for 1 in the earlier equation, you get: ln(e^(2iπ)) = ln(e^(n2iπ)) 2iπ = n2iπ Divide by 2iπ on both sides: 1 = n Which makes sense because n is any integer.

[@Fillex5000]

I've seen the explanation below and it's not quite right. Power to multiplication rule (a^b)^c = a^(b*c) is always true and the problem here is dependent on the branch of the ln(x). See, ln(x) has multiple branches we can choose from. Similar to how a square root of 4 can theoretically be 2 or -2, we just assume the positive branch when using the symbol √. Many functions in the complex plane have this property. When doing ln(x) in the complex numbers it's best to first express the x in its polar form re^[i(θ+2πn)] and then do the logarighm. Then you see that (e.g. in an equation) there can be multiple solutions from different branches depending on the n you choose (keep in mind that n has to be an integer).

[@narfharder]

5:09 "Check this out" watch the word "note" at 144p, trippy.

[@gietie1694]

you could also take the ilog of 2 and rewrite it as ln(2) / ln(i) = ln(2) / 0.5ln(-1) = 2ln(2) / pi i if im not mistaken

[@bijipeter1471]

Thank you, sir

[@narayanchauhan7541]

Just asking how did you become so good in maths? I saw your previous videos for doubts and you make questions easy.

[@1tubax]

Practice solving problems, read college textbooks, participate in competitions, watch and learn proofs. Over a few years you'll rack up so much intuition and knowledge if you practice right snd consistently.

[@JB-ym4up]

I took log base i on both sides and got x=logi(2)

[@tanvec]

Please don't say 90 degrees, as we are all adults now...I think I laughed a little more than I should have lol

[@blackpenredpen]

😆

[@stephaneclerc667]

I just discovered your channel and it IS GREAT! Your enthusiasm is amazing, I had a math texher like that 25 years ago, you really remind him. (his name was Gustave😂) I'm 38, love maths and I stopped at this level (high school math option in my country) But because of life and the obstacles on the way, I never was able to pursue in polytechnic. But I always kept a close link with mathematics and especially analysis. I litteraly do integrals during my free time, it's so beautiful.. My favorite is to trick arrogant people in the STEM fields with a "simple ∫ 1/(x^4 + 1) dx Most people fall in the trap. Anyway, I love your content and I'm gonna be watching a lot of it to stay sharp! Thank you

[@blackpenredpen]

Thank you very much!!

[@DPME820]

What trap?

[@stephaneclerc667]

@@DPME820 people think it's an Ln

[@vowing]

4:17 me too dawg glad we got one thing in common 💯

[@math4547]

Very nice video wow I'm a really huge math fan and keep it up !

[@blackpenredpen]

Thank you!

[@donaldmcronald2331]

I'd add a simplification. You can pull the 2 inside the ln as an exponent, so 2*ln(2) = ln(2^2) = ln(4). It makes the result a little prettier :D

[@govcorpwatch]

Make π τ again.

[@abhirupkundu2778]

and using ln(a^b)= blna, u can take the minus sign inside to make it ln(4^-1)= ln(1/4), to make it look even prettier

[@MikeyBarca02]

I'm so prone to clickbait thumbnails when it comes to maths, usually they don't work on me but your thumbnails always get me😂

[@eitancahlon]

please upload more, I really enjoy your videos

[@jaypeebeats141]

4:17 "i dont like to be on the bottom, i like to be on the top" xddd

[@Tetraverse]

does he know

[@Qbe_Root]

As x increases, i^x rotates around the origin in the complex plane. So you just have to make x increase fast enough, and the centrifugal force will stretch the unit circle out to the point that i^x = 2

[@gheffz]

Great video... much appreciated. Your info shared and your style... and your nice manner.

[@AbdulBasitWani.]

You are the best ❤

[@بشاررمضان-ع2ت]

You are great teacher

[@hayn10]

4:17 🤨 4:18

[@jesusnoagervasini8207]

The answer is obviously log(i)2

[@awoomywang]

HI STEVE, CAN I GET A PROMO CODE FOR YOUR LAMBERT W FUNCTION PURPLE TSHIRT BECAUSE I WANT TO ORDER 100 OF IT AND THE SHIPPING PRICE IS 800 DOLLARS

[@seanwang6716]

Yes so true

[@seanwang2635]

i agree, the lambert w function shirt looks so good haha

[@spoojy7881]

bprp merch ftw

[@pooface432]

makes you feel smart

[@blackpenredpen]

Could you send me an email [e-mail] and let me know why you are ordering 100 t-shirts? I will see what I can do for you.

[@dm319-j5y]

I just checked this on a 42 year old HP-15c, and I'm super impressed it got -2.

[@Giannhs_Kwnstantellos]

x= log i (2), taylor series, easy

[@aguyontheinternet8436]

Before watching video i^x = 2 x=log_{i}(2) x=ln(2)/ln(i) ln(i)=iπ(1+2n)/2 for any integer n x=(2ln(2))/(iπ(1+2n)) x=(-2 i ln(2) )/( π (1 + 2n) ) focusing on the principle value, we have -2i ln(2) / π Edit: shoot I didn't notice the extra answers with raising i to the fourth power

[@beginneratstuff]

I did it this way too before watching the vid lol I also didn't notice the extra answers, also I think you made a slight mistake, ln(i) = iπ(1 + 4n)/2 after you combine the π/2 with the 2πn into a single fraction and factor out the π. It didn't affect the principal answer tho

[@JonnyMath]

Your videos are amazing, thanks professor!!! 🤗🤩🥳

[@st3althyone]

Yes, it can. Take the natural log of both sides to get the exponent out, then divide by the ln i. X=ln 2 / ln i.

[@jessejordache1869]

The term in the denominator is sort of kicking the problem back at you. 'e^x = i' is not at all clear to me.

[@hoteny]

0:57 why do you guys hate 90 degrees and it always has to be pi over whatever?

[@stratonikisporcia8630]

Degrees are arbitrary, radians give the arc length

[@puremage0]

​@@stratonikisporcia8630every measuring unit like metres, kilogram, seconds, ampere, volts are all made up for convinience.

[@rokaq5163]

Radians are much more convenient mathematically, albeit less intuitive than degrees. If you go for function over form, radians are the way to go. And math is basically all function and zero form, so there's that.

[@gswcooper7162]

I would love to see you go further than complex numbers and solve an equation with quaternions instead - maybe something like x^x = 2?

[@Cubowave]

(4×ln2)/2πi is also a solution

[@Ноунеймбезгалочки-м7ч]

i^x=(-1^0.5)^x=-1^x/2, and you can only get -1 & 1 out of powers of -1, should be impossible if I got it right

[@mm0691]

Is that mean we would need to have a integer m & n to show every situation of the solution, where the n comes from the angle (2 pi n) and the m comes from i^4 ? ( i^(4m) ) That is A LOT of solutions

[@روائعالقرآن-ز1ن]

Can you do a video on how to change the pens in your hand? Thanks for your wonderful videos

[@_adityaacharya_8550]

we can use ln(z)= ln(|z|) + i(2npi+theta) too

[@Red-Brick-Dream]

"We are adults now, so say 'pi over 2.'" Thank you for this. From the bottom of my tired heart.

[@fizixx]

i^i. . . . .my favorite Very kewl video....love the info

[@hk4587]

Please make a video on how to solve any kind of ∑ problem... I need to learn..

[@zetadoop8910]

its a pleasure watching you. thanks

[@mathmachine4266]

ln(2)/ln(i) = ln(2)/(πi/2) = -2ln(2)i/π i^(-2ln(2)i/π)=2 Although, actually, the numerator can be ANY natural logarithm of 2 (ln(2)+2πNi)/ln(i), where N is any integer So i^(4N-2ln(2)i/π) = 2 That should hopefully make sense, since i^4=1, whatever power you raise i to, you could also add or subtract any multiple of 4. x = 4N-2ln(2)i/π, where N is any integer

[@its_eoraptor99]

Bro you look so much better without a beard, no kidding

[@gmjackson1456]

Great job!

[@wolfelkan8183]

Request: is there a complex number x such that 2^x = x?

[@gamerpedia1535]

2^x = x e^(x ln 2) = x x e^(-x ln 2) = 1 -x ln 2 = W(-ln 2) x = W(-ln 2) / -ln 2 Now there's a couple cool things of note here. Any number such that n = W(-ln x) / -ln x can be represented as an infinite power tower. I'll post the proof in a separate reply underneath this one. The other cool thing is that for 2^x = x, you can do 2^(2^x) = x or 2^2^x and following the chain, you're solving for the infinite power tower of twos

[@gamerpedia1535]

n^x = x xln(n) = ln(x) ln(n) = ln(x)e^(-ln(x)) W(-ln(n)) = -ln(x) e^(-W(-ln(n))) = x Identities of the W Lambert function tells us now that x = W(-ln(n))/-ln(n) For n^x = x So that's pretty cool, it's a shortcut to solve any convergent power towers.

[@coolcapybara111]

Broo this is insaneee 😵

[@michaelbaum6796]

Great, that is fascinating 👍

[@digitalfroot]

this was such a fun video lol i love how happy you get

[@SAGEmania-q8s]

lovely. Thank you. I will visit here whenever I got freetime like now.

[@anestismoutafidis529]

In case of the derivation of y= i^x to y´=x*i if x=2: i^2(Y)= 2(y´)

[@Cosmomaths]

Respect, i wish i’ll have you as math teacher when i’ll go to college 😂

[@jaii5955]

Our genius is back we amazing questions 😀

[@1224chrisng]

my man's hoarding whiteboard markers like they're Hagaromo chalk

[@blackpenredpen]

😂

[@sardine_man]

I got the same answer as wolfram alpha by simply thinking that x=a+bi and then from there rewrite i^x as e^(pi/2*i(a+bi)) and then distribute and match the e^(pi/2*a*i) part to match the angle 0+2pi*n and then I matched the e^(-pi/2*b)=2 so in that way you match 2 written in the form r*e^xi, 2=2*e^(2pi*n)

[@PennyLapin]

Is there a use to solving this for the more general form of e^i(pi/2+2npi) and show that you can produce an x that satisfies all infinitely many integer values of n? I looked at that expression and rewrote the theta value as (pi+4npi)/2 to make it more comfortable, then I set (e^i(pi+4npi)/2)^x = 2 and followed the same process to isolate the x from the equation. (e^i(pi+4npi)/2)^x = 2, n ∈ Z x * i(pi+4npi)/2 = ln(2) x = 2ln(2)/i(pi+4npi) x = -2iln(2)/(pi+4npi) Someone could ask "but what if we don't start with e^i(pi/2)? What if we start with e^i(5pi/2), or e^i(13pi/2)?" I asked that while watching, but I realized that other polar values of i can still be raised to a power that makes i^x = 2. Edit: Thanks to another discussion I thought about possible problems here and the reason why the video's focus on the principal value was there, based on where ln(x) isn't well-defined. If anyone has input on where this does and doesn't work, that would be appreciated!

[@neoxus30]

If you wanna make it work for non-positive complex numbers, just change the 4npi part. The solution of i^x = -2 is 2n - 2iln(2)/π

[@bettyswunghole3310]

I shudder to think of the complexity of any maths problem that would require the use of a green pen in addition to the red, black and blue!😄

[@armanavagyan1876]

Stunning proof👍👍👍

[@lawrencelawsen6824]

I love this channel

[@spoon_s3]

i^x = 2 x = ln(2)/ln(i) x = ln(2)/(i*(pi/2 + 2n*pi)) x = -iln(2)/(pi/2 + 2n*pi) Where n is all integers (arguably) more simple solution

[@carlopaternoster5878]

I tought the same, but it seems this is not the same as what is in the video. I do not know how to pass form one to another, they should be the same

[@ClukrFG]

BPRP: x=4n-(2iln2)/π My mindset: x=log_i(2)

[@randomboi550]

this is the first time I attempted the question on the thumbnail and actually got the answer before watching the video. I was excited at first but now I realise that this may be the beginning of me not needing to watch his videos to get my mind blown, which is pretty sad. I loved the mystery and the elegance of the solution, it would only blow my mind if I couldn't figure it out myself or needed a lot of hints. But now that I can do this myself, the magic of math is gone. Now I'm just sad.

[@kaisson8]

I didn't get why we have to write i^4n ? thank you !

[@scottleung9587]

I got the principal value for x just fine, but for the general solution I somehow ended up with 4n in the denominator.

[@darrenxavierjohan861]

Same :(

[@ezio99ez]

Can you have exact result for x, in x^i = i^x ?

[@japanpanda2179]

Yes. x is either i or -i. Sorry if this is disappointing.

[@Yaash-ph1pi]

Can u do a video where to use blackpen and redpen

[@jmlfa]

The rule (a^b)^c = a^bc … IS NOT VALID FOR IMAGINARY NUMBERS. If it were, then it is easy to show that -1 = 1: Simply replace 1 by e^(2pi*i) in the right part of the equation 1 = sqrt(1).

[@donovanknutson5128]

-iLn(2)/pi

[@Harrykesh630]

Professor please make a video on tricks used to solve limits

[@MC_Transport]

Can you do 100 related rates please

[@xinpingdonohoe3978]

Consider a≠0≠b as complex numbers. Then a^x=b can be solved. Such is the power of the complex plane. And then, if one of a or b is equal to 0, the other must be as well in order to be solved.

[@simonwillover4175]

Or a = 0 and b = 1, then x = 0, since 0^0 is (typically) defined as being 1.

[@xinpingdonohoe3978]

@@simonwillover4175 yes, if that computational convention is followed, then that is the exception.

[@souvikroy3584]

I have known a lot about complex from your video ✨

[@multilingualprogrammer3154]

@Blackpenredpen , look up this book called " (Almost) impossible integrals, sums, and series" and do a video on it.

[@lucidx9443]

Nice. Can you please also cover possible applications? I find it easier to remember that way!

[@esajpsasipes2822]

One application is to exercise your math skills...

[@lucidx9443]

@@esajpsasipes2822 That's given right he's an effective communicator. I'm just asking for actual applications.

[@esajpsasipes2822]

@@lucidx9443 I was more wanting to say that there might not be any (not sure), but even if there aren't any direct applications, it at least serves as an exercise (or entertainment i guess).

[@ryanjackson0x]

More generally, quantum mechanics uses this type of math

[@prg_prashant]

Sir, what's about taking log both sides...

[@jimnewton4534]

It is not clear to me that (x^y)^i = x^(iy). Clearly such is true if you look at an integer exponent, but it is NOT TRUE for general exponents. For a counter example consider f(t)=e^(i π t), which is clearly not identical to 1. However consider for x>0, f(x) = e^(i π 2t/2) = (e^(iπ))^2^(t/2) = ((-1)^2)^(t/2) = 1^(-t/2) = 1. This shows that sometimes it is false that x^(yz) = (x^y)^z.

[@19divide53]

It should be exp(z)^w=exp((z+2kπi)w), but general exponents a^z is defined by a^z=exp(z*ln(a)). In the video bprp is considering only the principal branch so it simplifies to exp(zw)=exp(z)^w.

[@ガアラ-h3h]

My take i^p = 2 => ln 2 = p*ln i => = ln 2 *2 = p * ln -1 => ln2 * 2 = p * ipi => p = ln2 *2/ipi which can be written as -ln 2 * 2i/pi what a beautiful result

[@dataweaver]

i^x=2 is the same as e^(½πix)=e^(ln2). So x=ln2⁄(½πi), or −i⋅ln4∕π

[@AC-zv3fx]

I got everything but why on 5:54 we don't multiply everything on the right side by 4 instead😅

[@Ramp4ge28]

Complex numbers are like cheating, you can have everything with they

[@kiza_l1247]

can u do vid solve equation erf(x) = 2 find x?

[@olahalyn4139]

I wrote the answer as ln((2)^(2)(-i/pi)). I wonder if the whether general answer of i^x = a, would always be ln((a)^(a)(-i/pi)). It looks quite nice as well.

[@MyNameIssaSimon]

Try an induction proof

[@olahalyn4139]

Tried it lol. Doesn't work. I did find after starting again that the actual general form is x = ln(a^(-2i/pi)) which I could prove by induction. Thanks for the hint. @Simon N I don't get to do maths much these days as I have left sixth form and uni course doesn't have any advanced maths in it really alway fun when bprp uploads.

[@person1082]

just take log_i of both sides

[@amrz1990]

Why wolfpharm alfa give the answer with log not ln ?

[@VSP4591]

Splendid.

[@varun3282]

yep solved x=(2/i*2n+1*pie)ln2 n belongs to Integers.

[@MathsKaDar]

Is there is any solution for √x+1=0 I wanna know how u gonna solve it

[@stratonikisporcia8630]

√x + 1 = 0 √x = -1 By definition, i² = -1, hence: x = i or x = -i

[@Andy-ju8bb]

Is there a particular reason for writing 2ln2 in the solution rather than ln4? I assume since that's the answer on WA, there must be a reason.

[@Engy_Wuck]

it's the standard form, because it's "more beautiful" to have the lowest possible number in the ln. Otherwise you could also write "ln 4^i"

[@Andy-ju8bb]

@@Engy_Wuck Thank you.

[@General12th]

Hi Dr. Pen!

[@raghavdhyani5739]

brother i took both side to the power i then replaced i^i by e^-pi/2 then after something i got x = i^-4lni/pi pls explain how much wrong i am

[@Toxic__rl]

Hey, so i had an exam today, and 'this woman o_o' gave us a quiz online, everyone could get different thing to solve... I got to do an integral of x^100 * sinx dx... I've looked at it and was like "wtf...". Is there any 'clever' way to solve this or my only choice was doing integrating by parts / D I method (not all the way ofc o_o... I've made 5-6 multiplications and then + ... + C xD)

[@Toxic__rl]

@blackpenredpen HELP (if it's even possible) XD

[@ANTI_UTTP_FOR_REAL]

At school Teacher: Whats your favorite number? A random kid: 3 Another kid: 7 This guy: *i*