Solving 3 Weird Logarithm Equations
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- čas přidán 19. 06. 2024
- 0:00 Hello, try the problems first
0:25 1. solve x^ln(x)=2
2:30 2. solve x*ln(x)=2
4:20 3. solve x^ln(x)=x*ln(x)
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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
Please make a video on integration xsquare-1/lnx dx from zero to one
The answer was ln3
My professor did it by taking a function integration x power b-1/lnx from 0 to 1 and then differentiated with respect to b and then at last taking b=2 then answer came ln3 is there any other approach
hey, what is the lambert W function?
it feels like mathmaticians just define functions that solve certain equations and say thats it.
like lets say I want to solve x^x =2 for x, I will define G(x^x) = x for any x taking the G function on both sides you get x = G(2) wow I solved for x genius wow.
I hope you get the point, we can't just define a function as the inverse of another function to help us and say that this solve our equation, letting the computer do the rest of the work, this is stupid.
@@oximas bprp defined what the function does, but its strict formula would seem unnecessary as its so complex. Here it is if you want to try solving with it by hand :) images.app.goo.gl/q6txbn3FwC7X5Mrk8
@@md.faisalahamed5202 well all of these is the game of orientable and non orientable surface
@@oximas functions are defined not just when the need for them arises but they become all too common in mathematics like equations for conics etc.
Plz someone gifts this man a big whiteboard XD
*cries in 3 wooden planks*
Brpr is capable of writing the whole solution on a single sticky notes and you rae saying small whiteboard is not enough??
@@tanmayshukla5330 *bprp
@@plislegalineu3005 What does PL stand for that is legal in the EU?
@@carultch 🇵🇱
It’s not a black pen red pen video without the Lambert W function 😅
Name a more iconic duo
And at least one green or blue pen.
Fred
What does the W mean in the LambertW function?
@@carultch
Wambert
5:45 "Because i like u." Lmao, nice! 😂😂
"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.
Unexpected Wholesome Moment
Ow, this was like breaking 4th wall.
The Lambert W(tf) function again...
*Cries in elementary functions*
You really must continue with the math for fun series 😍
What kind of series? P-series?
Ok... bad joke xd
@@juancruzftulis3249 amen bro
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
Thanks
Nice algebraic proof!
The left hand side CAN be negative for abs(u)
@@seroujghazarian6343 Ah sorry, my mistake! Thanks for catching it! :)
Hmm what about solutions on the complex plane?
1:30 approximating an 'approximately equal to' sign.
#2 is my fav because I'm 58 and had never heard of W(fish) function until I learned it here!
I am 59, and I never heard about it either. And I am a mathematician-computer scientist. Never too old to learn. Cool function!
This is simply a name for the inverse of x*exp(x), which can be proven to be a bijection, that's why W exists.
Basically this is just naming some inverse function. The more traditional method would consist in proving that the function we are trying to invert is bijective and thus that the solution is unique.
@@skycocaster xexp(x) is not always a bijection though, at 0>= x >= -1/e the inverse function forms a second branch
@@dexter2392 Yes, but we have ln(x) in the equation, thus restricting the considered domain to R+*, on which the function is bijective :)
If x*exp(x) was not a bijection on some considered domain, W wouldn't be a function.
Teacher: talking about Lambert W function
Me, not paying attention: 2 + 2 = fish
Have you been watching Flippin' Physics? They call alpha the fishy thing.
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)
Awesome
@@aizek0827 :))
Q2 - I got the same anawer but I did it in another way:
x + lnx = 2
e^(x + lnx) = e^2
e^x * e^lnx = e^2
e^x * x = e^2
x * e^x = e^2
W(x * e^x) = W(e^2)
x = W(e^2)
@@damianbla4469 That's how I did it, too. Both methods are great though.
Ahah spoiled!
For the 2nd equation, you could have started with the fact that exponential transforms sums into products. Btw, same answer.
x+lnx=2
e^(x+lnx)=e^2
e^x.e^lnx=e^2
xe^x=e^2
x=W(e^2)
Same as @damian bla
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.
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.
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.
Thank you for some actual proof
Q1: e^(e^(W(ln2)))
Q2: W(e^2)
That’s what I got too
yup
After applying the W function add 2 and exponentiate both sides so answer will be (e^(2-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 :)
Before even starting the video
I knew he Will use Lambert W
and I stopped the video😂😂
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
Here is a very interesting video, thank you !!!
These videos are really helpful 😊. I love logarithm now 😀.
Really fun and excellent way of solving
Thank you very much for your videos
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 👍🏼👍🏼
I love this channel.
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.
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 ❤️🤙
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 😊
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 😊
You sir are a legend
I solved the first now you try question!
First time i was able to solve an equation from your videos
Love logarithms!
Wow this lecture reminds me of my calculus class when i was in high school XD
BPRP IS MY FAVOURITE
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.
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.
very cool.....your answer is 100% true
@@mahmoudalbahar1641 Thanks!
I did Q2 like this:
x+ln(x)=2
e^(x+ln(x))=e^2 Notice: a^(b+c)=(a^b)*(a^c)
(e^x)*(e^ln(x))=e^2 Notice: e^ln(x)=x --> the exponential and logarithm cancel each other
x*e^x=e^2
x=W(e^2)
so x is approximately: 1.55714559899761...
plugging this in for x in x+ln(x) yields 2
@@imirostas4920 very smart!
On closer inspection of my final answer (e^(2-W(e^2))), it is indeed equal to yours (W(e^2)) by the properties of the Lambert W function. Following is a proof of this for those who are curious:
Rewrite e^(2-W(e^2)) as (e^2)/(e^W(e^2))
Remember the definition of the LambertW function as the inverse of xe^x, so W(x)e^W(x)=x which can be rewritten as: e^W(x)=x/W(x).
In our case:
e^/e^W(e^2) = e^/(e^2/W(e^2)) in which the term e^2 cancels out and so does the double fraction so what you’re left with is just W(e^2).
Edit: upon rereading this, i realized there was a much easier way to do this and I actually feel kind of dumb for not seeing this earlier:
Remember the definition of the LambertW function as the inverse of xe^x, so W(x)e^W(x)=x
Multiply by e^-W(x) to get xe^-W(x)=W(x)
In our case:
e^(2-W(e^2))=e^2*e^-W(e^2) which precisely fits our above proven property and thus equals W(e^2).
I am so greatful to you in this trouble so me time in sri Lanka!
Super! Bravo!
0:16 reminded me of WII (mariokart) and those joyful moments
Cool pockemon ball microphone and also some nice explanations!
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
Thanks for the answers, Fred! Hope you have been well : )
@@blackpenredpen Thanks! I can't complain. Except about the weather and the 'lock-down,' of course.
And thanks for continuing to do these, and I hope you (and your piano-playing gf) are well, too! (You seem to be...)
Fred
@@ffggddss Thanks, we are : )
When sacas la gráfica jeje, la vieja confiable
Him: Let u, because I like you guys, be = ln(x)
Us: It’s only *Natural* he says that about us
US United States? XD
3:33 why did this give me a better idea of what the lambert w function does than anything else
oh wait you did a whole video on this lemme check it
Thanks
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.
@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
I had got two answers, can you tell correct answer for this question please
1st answer-0 and;
2nd answer-3.
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)?
Unsurprising, since that would mean 2/W(2) = e^W(2)
Then 2 = W(2)e^W(2), which fits the definition of the W fn, since you can then take W of both sides and get W(2) = W( W(2)e^W(2) ) which = W(2)
@@jakemoll very true!
Yeah it's a property of Lambert W
let y×e^y=x
y=W(x)
Hence
W(x)×e^W(x)=x
Guys, Why did you try so hard?what you do is actually half a line.
x=2/ln x →x=2/ln (e^W(2))=2/W(2)
Done!
@@jakemoll simpler than you actually do..
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
This is cool
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
You have wrong solution for Q2. You have e(x+lnx)=e^2 -> xe^x=e^2 , so if you take W-function from both sides you will have x=W(e^2).
I got the same answer with logarithms: x+lnx=lne^x+lnx=lne^2 ->ln(x*e^x)=lne^2 -> x*e^x=e^2 -> x=W(e^2) = 1,557 .
Now 1,557+ ln1,557 = 2
@@d4slaimless yes looks like I messed up. I put x+lnx in the exponent on the left side but they should be multiplied not added
Best math on yt
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)
Please make a full course on logarithms if you already made then please give me it's link
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.
That was very fun
Interesting question, is there the chance to solve with Lambert W function the equiation x^x = 2x to find second root?
mans gotta make a video on the lambert W function, like an explanation
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
Please make a video about Numerical Methods sir, topic under Roots of polynomial: Mullers method. Thanks in advance sir!
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.
Same
Interesting video.
Cool 😎
"I feel the pressure" 😂😂😂
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
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)).
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).
Can u do a video on how to draw the gamma fun.
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."
3 is my favourite
In the second one you can still take log on both sides and solve the quadratic
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.
How would you ever find the common point? It's not nice like 1 its a irrational number.
good video 👍
Do a QnA!!!
3:33 Yeah, I got my fish back!
The good old Lambert-W-function.
good !
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)
This is the fourth time you uploaded a previously shot video.but how did doctor peyam find your lost video
When and where have you learned those things?
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)
In 2nd question, wouldn't taking Ln on both sides be more easier?
Is there a method to purely algebraically solve 3.?
On problem 2 you said that for calculating lambert w function you need wolfram alpha or mathematica. But thats not true. You need just a calculator and use Newton-Raphson or Hailey iteration formula.
Q1 sol equals e^e^w(ln2)
heres what I got for 9:09 questions
1) x = e^[w(2)]
2) x = w(e^2)
I think that first one is wrong
@@joshmcdouglas1720 Yep, i guess it has to be e^(e^(W(ln2)))
@@Veefencer yep!
Veefencer that's what I got, too
600 subscribers, I think it's time you get a full size white board.
INCREDIBLE!!!!!!!
Edit: adding my name at the last is impossible and you know why.😁
How does one solve u^2-u=ln(u) without graphs?
@@stapler942 e^(u^2 -u) =/= u for all cases so you cant generalise it
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.
Nice.
I am sure you can't use differentiation this way. For example if you have equation x^2=x the obvious answer is 1. But if you differentiate both sides you will have 2x=1 and it gives you x=1/2.
Derivatives for functions are not necessary equal at the same point where functions equal. They are equal where tangents are the same.
Ans that I found are a)e^e^w(ln2) b)1.552
Reniel escopete R
11-humms B
2nd pandemic
Can you find the integral int((1+x^2)^(-1/3))dx
(1-x^2)y'=1-xy-3x^2+2x^4 would you solve this differential equation i think it's non exact
Q1 the sol: exp(exp(W(ln(2))))
Q2 the sol: exp(2-W(exp(2)))
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)
You've wrong answer at Q1 bcs you assumed that u was equal to W(ln2) when it was equal to ln(W(ln2))
I see
your explanation is garbage of course, but I see the error
W is for u*e^u form, not for u*lnu, so I need extra exponentiation
u*lnu = ln2
let lnu = w
e^w*w = ln2
w = W(ln2)
lnu = W(ln2)
u = e^W(ln2)
lnx = e^W(ln2)
x = e^e^W(ln2)
@@NoNameAtAll2 Thanks for calling my explanation garbage, very heartwarming.