To my dear recent patrons, This video was uploaded 3 months ago (as an unlisted video) and I just published it today. So if you don't see your names on the outro card, I am sorry about it. I will be sure to include your names in my new videos. Thank you, blackpenredpen

Plz someone gifts this man a big whiteboard XD

It’s not a black pen red pen video without the Lambert W function 😅

5:45 "Because i like u." Lmao, nice! 😂😂

The Lambert W(tf) function again...

"Let u, because I like you, I like you guys." Aww, we like you, too. I wasn't expecting such an endearing moment in a math video.

I think I may speak for everyone that regularly watches your videos: We like you too and your enthusiasm you bring for solving math problems

You really must continue with the math for fun series 😍

1:30 approximating an 'approximately equal to' sign.

Teacher: talking about Lambert W function Me, not paying attention: 2 + 2 = fish

Spoiler alert (lnx)^(lnx) = 2 lnx*ln(lnx) = ln2 e^(ln(lnx))*ln(lnx) = ln2 ln(lnx) = W(ln2) lnx = e^W(ln2) x = e^(e^W(ln2)) x+lnx = 2 ln(e^x)+lnx = 2 ln(xe^x) = 2 xe^x = e^2 x = W(e^2)

Since blackpenredpen didn't go through too much details into proving that u has to equal 1, I will do it for those reading this! Looking back at the original equation, there is no way u=ln(x) can be negative otherwise the left hand side would be positive and the right hand side would be negative. This justifies taking the log of both sides in the first step and taking ln(u). Now, once we reached u² - u = ln(u), we can move everything to the left hand side to get u² - u - ln(u) = 0. Let the left hand side represent a function: f(u) = u² - u - ln(u) We want to prove that the only root of this is u=1. If we can prove that u=1 is an absolute minimum then that will be a much stronger claim than what was wanted to know. That's what I'll prove. Proof by Second Derivative Test The first derivative of f is: f'(u) = 2u - 1 - 1/u f'(u) = (2u² - u - 1)/u f'(u) = (2u + 1)(u - 1)/u Since u>0, the ONLY critical point is at u=1. Now, looking at the second derivative: f(u)= 2 + 1/u² This is clearly 3 at u=1, which is greater than 0, so we've proven that u=1 is a minimum. Further more, since u=1 was the ONLY critical point according to the first derivative, then we've shown it's an ABSOLUTE minimum as well. Conclusion: Yes, u=1 is the only way we can get u² - u to equal ln(u), but it also follows from our proof that if u≠1, u² - u is greater than ln(u). Filling in the last steps of that part will be left as an exercise to either the reader or blackpenredpen's calculus students

#2 is my fav because I'm 58 and had never heard of W(fish) function until I learned it here!

Q1: e^(e^(W(ln2))) Q2: W(e^2)

To show that u²-u and ln(u) only intersects once, I think it's enough to know that u²-u is convex and ln(u) is concave. That, combined with showing that they are tangent at u=1, is enough to show that this tangent point is their only intersection.

Before even starting the video I knew he Will use Lambert W and I stopped the video😂😂

Thank you for all your videos that you do, they're always really helpful for my maths even though I'm from the UK :)

This video reminds me of how much I love maths. I wish I had done more of these when in school, practicing helps to cement fundamentals.

Let u because I love u was THE line of this video. Great job

that was awesome. i like how all three had quite different solutions.

Wow, your math problems and the way you solve it really gives me inspiration to learn more about this subject, Math, that i used to hate a lot.

These videos are really helpful 😊. I love logarithm now 😀.

Here is a very interesting video, thank you !!!

Really fun and excellent way of solving

q2 also : xlnx = 2 take e to both sides --> x^x = e^2 take super sqrt --> x = ssrt(e^2)

Him: Let u, because I like you guys, be = ln(x) Us: It’s only *Natural* he says that about us

Sigue con esos videos resolviendo ecuaciones cortas y extrañas 👍🏼👍🏼

If you know what W function is it, both of 5 questions take just a minute. Thanks teacher, I have learned W function from you 😊

Thank you very much for your videos

I love this channel.

Dear bprp, First let me tell you that what you are doing is amazing and an incredible inspiration for me and most certainly for many others as well :) Furthermore, I want to point out that for all real numbers r the equation x(x-1)=r*ln(x) has a unique solution at x=1 : 1(1-1)=r*ln(1) => 1*0=r*0 and this is true for all real numbers r. But it could be very difficult, if at least one of the coefficients of the polynomial on the left side is not equal to 1 (and obviously if those coefficients are not equal to each other) or the parabola is shifted horizontally. Greetings from Germany 😊

Hey blackpenredpen i really love your math for fun videos. Try this math for fun question : In rectangle ABCD the area is equal to the perimeter and to the diagonal . find AB and BC.

BPRP IS MY FAVOURITE

Love logarithms!

Q1: Ln(x)^ln(x)=2 Taking the natural log on both sides: Ln(ln(x)^ln(x))=ln(2) which simplifies to ln(x)*ln(ln(x))=ln(2) Let u=ln(x) u*ln(u)=ln(2) You can write u as e^ln(u) so, ln(u)*e^ln(u)=ln(2) Taking the Lambert W function on both sides: W(ln(u)*e^ln(u))=W(ln(2)) ln(u)=W(ln(2)) u=e^W(ln(2)) Since we defined u to be ln(x): ln(x)=e^W(ln(2)) x=e^(e^W(ln(2))) Q2: x+ln(x)=2 You can write x as e^ln(x) so, e^ln(x)+ln(x)=2 e^ln(x)=2-ln(x) Multiply by e^-ln(x): 1=(2-ln(x))*e^-ln(x) Multiply by e^2 so we can use W Lambert function: e^2=(2-ln(x))*e^(2-ln(x)) Taking W Lambert function on both sides: W(e^2)=W((2-ln(x))*e^(2-ln(x))) W(e^2)=2-ln(x) ln(x)=2-W(e^2) x=e^(2-W(e^2)) This was pretty hard to type so please let me know if I have made any mistakes.

Wow this lecture reminds me of my calculus class when i was in high school XD

I solved the first now you try question! First time i was able to solve an equation from your videos

Could you teach some tricks for optimization and real problems with them in the current life? Kisses from Spain and thanks for the video, as always you're excellent ❤️🤙

Pre-watch: (1) The first one appears to be the only one of the three that's soluble in "standard" functions. x^(ln x) = 2 - - - take ln() of both sides ... (ln x)² = ln 2 ln x = √(ln 2) x = e^√(ln 2) = 2.299184767... Curiously, if this value were to satisfy eq. 2, it would automatically also satisfy eq. 3 (just equate the LHS's of eqs. 1 & 2, and that is eq. 3). Alas, it does not. My guess is that eqs. 2 & 3 can be solved using the Lambert W function. (2) x ln x = 2; let y = ln x; x = eʸ then yeʸ = 2 - - - take W() of both sides ... y = W(2) x = e^(W(2)) (3) x^(ln x) = x ln x This one doesn't seem as cooperative ... let's see how you do it. Post-watch: Very nice! I missed the 2nd solution on eq. 1; it works, too. And for eq. 3, your graphical solution seems to be the only way to solve it. Graphical solutions usually don't cut it, but your argument (along with concavity) can make it rigorous. There can be no other (real) solutions. Fred

You sir are a legend

I am so greatful to you in this trouble so me time in sri Lanka!

When sacas la gráfica jeje, la vieja confiable

Best math on yt

Cool pockemon ball microphone and also some nice explanations!

For 2), I had xlog(x)=2 log(x)=2/x x=e^(2/x) 1=1/x * e^(2/x) 2=2/x * e^(2/x) W(2)=2/x x=2/W(2) And according to WolframAlpha this is the exact same value as e^W(2)?

3:25 i knew just looking at the thumbnail that good old W(x) was going to come up. If it was up to this guy, the Lambert W function would be built into every scientific calculator.

Super! Bravo!

Great ! some interesting staff: x intercepts as u^2-u=0 and u=0 and u= 1 then only ln_eX= 1 and then E=X.

Thanks

0:16 reminded me of WII (mariokart) and those joyful moments

Q2: turn it side into exponents of e, and we get e^(x+lnx)=e^2 This becomes x*e^x=2, and then x=W(2) Q1: Let a=lnx, we have a^a=2 .Take the ln of both sides and we have alna=ln2 -->e^lna *lna=ln2, take lambert W of both sides and get lna=W(ln2), so a=e^W(ln2), replace a with lnx and have lnx=e^W(ln2), so x=e^(e^W(ln2)) lna=W(log2)-->lna*e^lna=lna*a=ln2

3:33 why did this give me a better idea of what the lambert w function does than anything else

@blackpenredpen I had a math problem that I would love for you to do. I have had people tell me "it's impossible" and it "can't be simplified" but I know there must be a way. x^(x+1)=ln 3

The first one I got by taking log base x on both sides to get ln x= logX(2). Use change of base Ln(x)=ln(2)/ln(x) Ln^2(x)=ln(2) You get the idea 3rd one I just guessed e.

x^ln(x) = xln(x) e^[x^ln(x)] = e^[xln(x)] = x^x On the right, we have some variable raised to the same variable. On the left, we have something raised to something else. This means, by comparison, that the something and something else are equal. So, e = x^ln(x) 1 = ln(x)^2 +/- 1 = ln(x) e^(+/- 1) = x Only positive 1 works, so x = e

These are my answers: Q1: (ln(x))^ln(x)=2 We know the solution to 🤔^🤔=2 is 🤔=e^W(ln(2)) So if 🤔=ln(x), then ln(x)= e^W(ln(2)), so x= e^(e^W(ln(2))) Q2: x+ln(x)=2 ln(e^x)+ln(x)=2 ln(xe^x)=2 x=W(e^2)

"I feel the pressure" 😂😂😂

For the second question, can't you just solve it like this: ln(xlnx) = ln(2) ln(x) + ln(lnx) = ln(2) Let ln(x) = u; u = ln(2) - ln(u) u = 0.693 - ln(u) Finally, solve out for the common points of two sides of the equation (like in part 3). Great Channel btw.

I did 3 like that : X^lnx = xlnx e^ln^2(x) = (e^lnx)*lnx Flipping the e to the powers to the other sides we get : e^-lnx = (lnx)e^-ln^2(x) Multiply both sides by -lnx we get : -lnx(e^-lnx)=-ln^2(x)*(e^-ln^2(x)) And now, all that is left is to take the lambert w fucntion on both sides to get : -lnx = -ln^2(x) a simple equation with the solution x=e. Also gives x=1 but we check to see it dosent work for 1.

This is cool

mans gotta make a video on the lambert W function, like an explanation

That was very fun

I feel so happy that I solved the second equation at the end🔥🔥 i basically did: x+lnx=2 ln(e^x)+lnx=2 sum of logs = log of the product =ln(xe^x)=2 xe^x=e^2 so the answer is the Lambert W Function of e^2

3:33 Yeah, I got my fish back!

9:08 Q1 let lnx = u u^u = 2 u*lnu = ln2 u = W(ln2) lnx = W(ln2) x = e^W(ln2) Q2 x + lnx = 2 e^(x+lnx) = e^2 e^lnx * e^x = e^2 x*e^x = e^2 x = W(e^2)

Interesting question, is there the chance to solve with Lambert W function the equiation x^x = 2x to find second root?

Please make a full course on logarithms if you already made then please give me it's link

Interesting video.

Q1 the sol: exp(exp(W(ln(2)))) Q2 the sol: exp(2-W(exp(2)))

heres what I got for 9:09 questions 1) x = e^[w(2)] 2) x = w(e^2)

good video 👍

Do a QnA!!!

3 is my favourite

Cool 😎

I thought it was interesting that you used substitution on the last one I had used it on the other ones too when doing it in my head while in bed, I didn't get the last one done in my head though sadly

How does one solve u^2-u=ln(u) without graphs?

The good old Lambert-W-function.

Only got the third one right before watching, as a 14 year old. Btw I love your videos, keep it up!

1st HW Question's Answer x = e^W(2)

In the second one you can still take log on both sides and solve the quadratic

I did the first one in my head before watching and i found x = √(2e) as a good approximation (~2.33). As some teacher said someday: "I dont know, but you used the wrong formula and got the correct answer."

600 subscribers, I think it's time you get a full size white board.

good !

5:43 I'm definitely not using that on someone ;)

I got the first and second but failed with the third :-( Thank you !

Q1 sol equals e^e^w(ln2)

Q1:x=e^e^w(ln2)

9:12 just a bye? no "thats it"? :'( Anyway im happy to see the fish back 🐠

An easier way to solve the 1st log problem would be to rewrite each side so they have the same base so then we would rewrite the x in x^ln(x) as e^ln(x), so then we would have e^(ln(x))^2, then we would rewrite the 2 as e^ln(2), thus we have e^(ln(x)^2=e^ln(2), we can see they have the same base so then we have the equation (ln(x))^2=ln(2), we then take the square root (positive and negative) on both sides to get ln(x)=sqrt(ln(2)), then we exponentiate it to find that the solutions are x=e^sqrt(ln(2)), and x=e^-sqrt(ln(2)).

Cool vid but never in my life have I heard of the W function.

Please make a video about Numerical Methods sir, topic under Roots of polynomial: Mullers method. Thanks in advance sir!

INCREDIBLE!!!!!!! Edit: adding my name at the last is impossible and you know why.😁

Can u do a video on how to draw the gamma fun.

Reniel escopete R 11-humms B 2nd pandemic

Here's how I did the u^2 - u = ln(u) step without the graph: -Take e to the power of both sides. e^(u^2 - u) = u -Differentiate both sides to get e^(u^2 - u) * (2u - 1) = 1 -Divide both sides by (2u - 1). -Now you have the identity e^(u^2 - u)= u = 1/(2u - 1) -Using the quadratic formula you can get: u = 1 or -1/2 -Since ln(u) is not defined for negative numbers we discard -1/2 and say that u = 1.

this is 2 cool sol.: BTW can u solve (*) y' =-ay⋅(1-y)^2, i.e., WTF y(x) s.t solves (*) for a>0? I Think LambertW function is involved. maybe W(0,x) or W(-1,x).

Another good one: x^ln(y)=y^ln(x). Looks really complicated, until you realize...

The solution to the 3rd equation, X^(lnX)=XlnX, is an interesting one.

This is the fourth time you uploaded a previously shot video.but how did doctor peyam find your lost video

btw for the 3rd problem: using properties of logarithms (ln(X))^2 - ln(X) = (ln (X/X))^2= ln(1)^2=0 so you have 0=ln(ln(X)) 1=ln(X) and finally X=e