Lambert W Function And It's Application
Вставка
- Опубліковано 10 лис 2023
- The Lambert W function otherwise known as the the "Product Log" is one of the unknown function/law to many students/learners that can be used to solve tough exponential math problems.
In this video, I will show you how to apply the Lambert W function to solving rhetorical exponential challenges in a step-by-step method guide leaving no stone unturned.
Here, I will also guide you on the use of some of the laws of indices and logarithms in addition to the Lambert W function to fast trick the solution to this math problem.
These and other tricks, tips and secrets to solving tough algebraic questions from International Math Olympiad questions other exam bodies.
Help to share this nice video tutorial to all that it may benefit within and outside your domain using the video link below.
• Lambert W Function And...
Also, kindly drop the answer to the numerical value to this math problem when you use the Wolfram Alpha calculator in the comment section but if you do not have access to the Wolfram Alpha calculator simply comment ''help me with the numerical value'' in the comment section and it will be given to you very fast.
Give this video a thumbs up if it has actually benefited you in one way or the other.
Here are the links to other videos where I applied the Lambert W function in solving some algebra. Watch and comment.
1. • Solving Exponential Eq...
2. • How To Solve Exponenti...
3. • How To Solve x^2=-5Inx...
A million thanks to all of our subscribers and viewers all over the world for your consistent support.
We all at OnlinemathsTV love you dearly ❤️❤️💖💖💕💕💕💕😍😍
Thanks! Thanks!! Thanks!!!
Cheers!!!🎉🎉🙏🙏🙏
#lambert #maths #algebra
Great video ❤❤
Thank you very much! I wish I had a teacher like that
Nice video from you sir.
By the way love from India 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳
we can solve this equation like this also:
To use the Lambert W function to solve the equation 2x+x=5, we can first rewrite it as x=5−2x. Then, we can apply the logarithm function to both sides and use the property logb(ac)=clogb(a) to get:
log2(x)=log2(5−2x)
log2(x)=xlog2(2)−log2(2x)
log2(x)=x−xlog2(2)
Next, we can divide both sides by log2(2) and multiply both sides by −1 to get:
−log2(2)log2(x)=x−log2(2)
Now, we can use the change of base formula logb(x)=loga(b)loga(x) to rewrite the left-hand side as:
−log2(x)=xlog2(e)−log2(2)
Finally, we can use the definition of the Lambert W function to write the equation as:
−log2(x)e−log2(x)=(xlog2(e)−log2(2))exlog2(e)−log2(2)
Therefore, the solution is given by:
−log2(x)=W(−(xlog2(e)−log2(2))exlog2(e)−log2(2))
To find the numerical value of the solution, we can use a calculator or a software that supports the Lambert W function
thank you
But the result is absurd or meaningless. What's it's numerical Value???
thanks for this video
Very nice video.
The way you make the equation transformation is amazing. Thanks
I'm glad you like it and find this video tutorial explanation facilitating.
Thanks and much love from OnlinemathsTV 💖💖❤️❤️❤️💕
Very nice my friend Jakes! This is what I tried and solved in my recordes video too haha Keep up the good work my friend👍👍👍
Wow! Thanks so much prof for always supporting this channel sir.
We all here love you dearly Dr....💕💕💖💖💖❤️❤️
Thank you sir for your most excellent instruction.
I learned that if the Lambert W function contains the ln2, then it cannot be treated as a factor. God bless.
Thanks for supporting this channel my good friend.
From my research so far, it can be treated despite the presence of the In2 sir but I will still do a thorough and detail findings from what you just observed so far sir.
Thanks for watching and dropping this observation sir.
Much love from everyone @OnlinemathsTV to you sir ❤️❤️💖💖💕💕
Thanks for this video tutorial master JJ
Always welcome my good friend and thanks for watching this video.
Much love...💕💕💕
hello, many thanks for your nice videos - would you mind to write your nme on the board? i fail to hear exactly
~1.7156
Very correct sir.
You the best,
Respect sir....👍👍👍
Which calculator should I use to input the lambert W function?
Thanks for this video sir but could you give the numerical value to this final answer because I don't have access to the Wolfram alpha calculator.
Sure, approximately 1.72.
You can the soft copy of the Wolfram Alpha calculator on play store for android and apple store for your iPhone phones, hahahaha.
Thanks for watching sir/ma.
Ok it was very nice.
Many many thanks to you for appreciating this sir
Отлично показано, как выражение следует приводить к виду X*e^X, но непонятно, как вычисляется численное значение W-функции из конкретно заданного числового значения аргумента. Какова формула вычисления? Вот это бы мне понять.В Википедии имеются алгоритмы, но на каком основании таковые строятся? Есть аппроксамации функции на определенных интервалах Х, но какие истинные значения они аппроксимируют, откуда они берутся? Если можно, пожалуйста разъсните по возможности доступно!
Thanks @williamspostoronnim9845 for the comment and the question asked. The approximation here is to 4sf(significant figures).
How to calculate lembert w function
Is there no easier method than this ?
You should not have an apostrophe in "it's" in the title. That's a contraction for "it is", not the possessive form for "it." That would be just "its", without an apostrophe.
🤓
You are too serious here.
What's it's NUMERICAL VALUE???
The numerical value is;
1.71562….
You can use Wolfram Alpha to calculate the lambert w function.
❤😂❤😂❤
There is terrible reverberation, echo, distorted sound or something like these in the film
Pls kindly check your system of the phone are using to watch this video because I can barely observe all you just listed on this video.
Thanks sir.
I learned from another video, in which gentleman derived a general formula for your problem 2^X + x =5 . It goes like this a^X + X -B is = to B - W(a^b lna)/lna
Don't teach me grammar and put forth the numerical value.