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

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

Plz someone gifts this man a big whiteboard XD

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

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

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

You really must continue with the math for fun series 😍

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)

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

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

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.

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.

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

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

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

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

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

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

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.

Really fun and excellent way of solving

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

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.

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

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

Thank you very much for your videos

I love this channel.

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)

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

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 😊

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

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.

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.

BPRP IS MY FAVOURITE

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

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

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

You sir are a legend

Love logarithms!

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 ❤️🤙

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 😊

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

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

When sacas la gráfica jeje, la vieja confiable

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)?

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

Super! Bravo!

Thanks

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.

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

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

@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

Best math on yt

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

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

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

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.

for 2, you can directly apply W(xln(x))=ln(x)

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

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

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.

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)

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

When and where have you learned those things?

That was very fun

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)).

3 is my favourite

Q1 sol equals e^e^w(ln2)

Interesting video.

"I feel the pressure" 😂😂😂

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

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

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).

alternate to the first one: x ^ lnx = e ^ ln2 lnx / ln2 = log_x(e) lnx / ln2 = lne / lnx ln^2(x) = ln2 x = e ^ sqrt(ln2)

Do a QnA!!!

Is there a method to purely algebraically solve 3.?

3:33 Yeah, I got my fish back!

Can you find the integral int((1+x^2)^(-1/3))dx

good video 👍

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

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

In 2nd question, wouldn't taking Ln on both sides be more easier?

Reniel escopete R 11-humms B 2nd pandemic

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

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.

Ans that I found are a)e^e^w(ln2) b)1.552

Cool 😎

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

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

Q1 was easy: You can also use a substitution. Before that you take the ln on both sides and get u*ln(u)=ln(2). Just change that to ln(u)*e^(ln(2))=ln(2). Finally, you use W(x) and do the resubstitution and you get x=e^e^((W(ln(2)))) Q2: do e on both sides: e^(x+ln(x))=e^2. Then use properties of exponents and pull them apart. e^(ln(x)) simpplifies to x and you can use W(x) again, so x*e^x=e^2 x=W(e^2)

when did this channel become blackpenredpenbluepen?

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

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."

The good old Lambert-W-function.

For (2), why do I get a wrong value if I do lnx=2/x | d/dx; 1/x=-2/x^2; x=-2 ?

more lines/definitions for ex. 2