Solving Exponential Equation 5^2x=7x Using Lambert Wn Function | Lambert w function introduction
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- เผยแพร่เมื่อ 5 ต.ค. 2024
- Solving Exponential Equation 5^2x=7x Using Lambert Wn Function is a well detailed video clip which explains clearly the hidden secrets to solving challenging exponential equations.
Here, Jakes gives the functionality of the Lambert W function and how it is applied in solving tough exponential equations.
Watch from the beginning to the end without skipping any parts of this video for the full detail.
Also drop in the comment section bellow any question or contribution with regards this solution.
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so ln25 not ln27 at last divide.... you missed it
5^2x=7x
raise both sides to 1/x power
5^2=25=7x^1/x
/7 both sides
25/7=x^1/x
using a formula i made,
(if x^1/x=y, x=e^-W(-lny))
=>x=e^-W(-ln(25/7))
Many thanks for this question.
Thanks Jakes, that is an elegant way to solve the problem. However, the solution’s argument for the W Lambert function (-Ln (25)/7) falls outside the domain of such function [-1/e, ∞). The curve 5^2x - 7x does not intersect the X axis, the closest value (y=-0.485) occurs when X=0.241.
Many thanks for this nice observation @RafaelJDiaz-uy1cs.
From your comment and other's, I will always make use of the graphical method using Desmos in subsequent cases before making any video in order to always give the best to this wonderful comment.
Again thanks a bunch sir.
Maximum respect....🙏🙏🙏
1- how did it become 27 in the denominator?
2- what's the numerical value of x?
Sir give us all ln and ex important equations
Great teacher thank you sir
Ok sir. We will drop more challenges on that and thanks for the comment, it is very encouraging and we promise to serve you better in this platform as far as mathematics is concerned by His grace sir.
Much love ❤️❤️💖💖💕
Jest nasz wspaniały matematyk.
Dziękuję.
😍😍😍 Thanks a lot sir.
We are glad you are gaining some values from our video tutorials.
We will put more effort to serve you better sir.
You welcome to our channel.
Much love 💕💕💖💖
Thanks 🙏 so much for doing great job.
Sir, which app do you use in editing your thumbnail especially geometric figures?
👍Well done.
Thx just needed a clear example for lambertW
Ok sir, I will do that in another video tutorial
Excellent sir thanks
You are most welcome sir.
Thank you for watch, and encouraging us.
More love❤️❤️❤️
Just brillant! I wish i could have solved it.
Thanks a million @Robert Roth. I like your name sir🙋
Thanks for encouraging us with this comment. You will solve even tougher mathematical challenges than this one so long it is in your heart to know.
Thanks for watching and commenting sir.
Much love💖💖💕💕
-(ln 25)/7=-0.459839 < -1/e=-0.367879 so according to Lambert function definition ,there should be no real solution at all.
😊😊😊😊😊😊😊😊😊😊😊
I am pleased. Thank you
You are most welcome sir.
Answer should be ln25 in the denominator. In 27×××
Thank you sir for your detailed explanation 🙏
Is the function that we need: -LWn[(-ln25/7)/ln25] not ln27 at the denominator, right ?
= -LambertWn((-0.459839)/ln25)
= -LambertWn(-0.142857142)
if we give this in "Wolfram Alpha"
we get x= 0,169192603
if we put this into our equation, we get:
5²*⁰,¹⁶⁹¹⁹²⁶⁰³= 1,723935.....
which is not equal to 7*x= 7*0,169192603 = 1,18434....... ? 🤔
b) Why did you put the minus of (-ln25) before the LambertWn function, is this a rule, I mean, can't we divide [-ln25/7]/(-ln25) = 0,142857142, so
LambertWn(0.142857142), here we get x= 0.125951132
This is not equal with : -LambertWn(-0.142857142), as far as I can see........
You are correct @Birol, it is a typo on my side sir. The denominator is ln25 not ln27.
I promise to be more careful next time when am solving math sir.
Thanks for the keen observation....
Respect👍👍👍
The Lambert function is not defined at -ln(25)/7 = -0,4598394036 .
The domain is [-1/e,+infinity[.
Therefore no solution.
Greetings from Germany
@@michaelkoch6863 Hello Michael, thank you very much for your feedback. He did not give the exact solution, just wrote the Lambert Wn Function: x= LWn[(-ln25/7)/(-ln25)]
= LWn[(-0,4598/(-3,2188)]
= LWn(0,142857)
x= 0,125951 which does not give the result !
If we write: x= LWn[(-ln25/7)/(-ln25)] and put the minus of the denominator before the function, we get:
- LWn[(-ln25/7)/(ln25)]
-LWn[(-0,4598/(3,2188)]
- LWn (-0,142857) so with the minus sign in the bracket, would give:
x= 0,169192 which also does not fit the equation, my question was and is, where is the mistake here ?
You may use the Wolfram Alpha Website, to check it.
@@Birol731 Hmm and my answer is in my message. Nice day....
ohk wanted to comment on the ln27 but I guess my colleague already did, thank you
Bravo!
2^x=x+2 solve for x
Excellent explanation
Thanks for watching and commenting sir.
Great solution
Thanks a million sir.
We are glad you gained some values from our videos sir.
X = A/ln25. Not ln 27.
Yes, mind-drift(typo)
What's the numerical value of wn(2) ????.........
well, as it is a complex number I thought I could see that complex numerical result I don't obtain by algebraic means...
must be division by ln 25 in the final answer
It suppose to be In25 but you wrote In27 or am the one lost here?
Pls lecture my ignorance.
Hi the problem I just gave you the i is not a complex number it just stands for interest. As in compounded interest.
🤣🤣😂😂😍😍
Funniest and most wonderful comment ever.
Your comment is a great medicine to me because it made me laugh uncontrollably this moment sir.
You are more than a comedian and thanks for dropping this unique comment boss.
Loving you like mad!!!💕💕😍😍😍💖💖❤️❤️
typo on your QED ln25 and not ln27
If we plot on graph, we could see 5^2X{(0,5),(1,25)} will not touch 7x {(0,0),(1,7)}, so I don't think there is a real X to be the correct answer
Yes, if you input the final answer into your Wolfram Alpha calculator what will come out are imaginary roots.
What calculator did he recommend to evaluate for x?
Wolfram Alpha calculator sir.
Please solve this one
8^x +9^x=10^x
So sorry for the late reply, I just saw your question/request now. I will make a video on this soonest sir.
Thanks.
El ln 27 no va, es ln 25. Y la solución es solo para Reales
Sorry for that error, the denominator is In25 and not In27 pls.
It is a typo sir....I will be more careful next time sir.
Regards....
Can you show me how to solve this problem? p times (1+.01r/365)^365 times [(1+.01r/365)^(365t)-1]/[(1+.01r/365)^(365)-1] -p times t = i solve for t THANKS
I can't get the actual form of your question sir.
Kindly drop it on my whatsap handle via +2348037118469
Thanks
5² ^x = 7x
Log 5² ^x = log 7x
2X * log 5 = log 7x
2X * log 10/2 = log 7x
2X *( log 10 - log 2) = log 7x
2X *(1-0,3) = log 7x
2X * (0,7) = log 7x
X* 1,4 = log 7x
Thanks for the work through/procedure sir. It is a nice approach sir.
So, what is the value of x sir or how do we solve for x from here sir?
@@onlineMathsTV olá! Professor! Eu sou do Brasil. Realmente, eu tentei resolver usando logaritmo, mas não soube continuar a resolução, por isso parei neste ponto. Obrigada pelo retorno.
O máximo que conseguiria continuar seria:
X* 1,4 = log 7x
X * 1,4 = log 7 + log x
X = (log 7) /1,4 + log x /1,4
X = 0,845/1,4 + log x /1,4
X = 0,6 + log x / 1,4
X - log x /1,4 = 0,6
Log 10^x - log x = 0,6
Log 10^x/x = 0,6
10^6/10 = (10^x)/x
10³/⁵ = (10^x) /x
?????
Professor! Eu não sei continuar.... o Sr. poderia ajudar-me?
Hi, from where you get Ln 27
Mind-drift from ln 25 to ln 27. Just a typo. You are sharp.
Mind-shift sir. Very sorry about that, I promise to be more careful next time sir.
X=ln×*(0)
Could you please check your solution again. How did (ln 25)/7 become (ln 25/7)?
Typo pls
❤
Thanks sir.
Over 25 not 27
Yes boss, it is a typo......thanks for the observation and I promise to be more careful in subsequent time sir.
Thanks a million sir.
you sure ln(-ln25/7) exist?
nope
Anyone with some basic knowledge of maths can do like what you have done, by just copying the nonsensical notion of lambda function to arrive at such an undefinable or unverifiable solution
The most basic correct way of solving such highschool probs always includes checking whether the found solution is correct or not, by just substituting the found solution in the original equation. But you have never ever done so, and just jumped to a baseless conclusion . that it's the solution
All solutions which contain any variable, without a specific value are invalid
нужно провести исследование функции
Ok sir
Niêu là tham gia thì cho chính xác là một con a b miền Nam
Make me to understand your point for clarity sake sir.
Chữ xấu ,giải khó hiểu
So sorry about that madam. Am actually working on my accent so that people like you and others who may find it difficult to comprehend will get exactly what am passing out.
Thanks for the observation.
Much love💕💕💖💖
Hahaha.. amazing solution. NdasQ musmet 😂🤣
Thanks a million mistress
25 not 27
Yes, typo. So sorry about that, kindly pardon my manners pls.