My suggestion when trying the entrance exam is the following: 3^x = x | * -ln(3)*e^(-ln(3)x) -ln(3) = (-ln(3)x) e^(-ln(3)x) | productlog W(-ln(3)) = -ln(3)x | /-ln(3) -W(-ln(3))/ln(3) = x
@@moebadderman227 That's the operation used for the next equivalent transformation . Here it is a multiplication with the factor (-ln(3) 3^-x), to be able to apply the product log for the following equivalent tranformation.
This question has a much shorter answer by using calculus You can prove there is no number that evaluates 3^x = x by computing the limit on infinity, and since infinity isn't a real number (arguably not even a number , just an expression) and by proving that it is the only solution, then you can derive it in just below 5 minutes (the actual math,proving that 3•3...•3 diverges is just stupid math that needs to be done)
@@FanisBartzis Don't get me wrong, I am good at math, and it energizes me. But physicists increasingly have problems with "mathy math" after about sophomore year in college. When I see 3^x = x, I can see it has no real answer. So I write (e^(ln 3))^x = x. Then let u = e^(ln 3) and I get u^x = x. I will write x = a + b i and go from there. But this youtube teacher is known for writing many steps rather than getting down to (what I would call) business. To each his or her own.
@@pidigi ok, formally when we divide by X we should split the problem and solve it separately for the X=0 case and the X!=0 case, but the former is trivial by inspection
@@Apaximatic_Play Я хотел сначала написать по-англ, что у уравнения нет вещественных корней. Но посмотрел комменты, :- неск. уже это отметили. Пока просматривал, наткнулся на ваш вопрос. Решил ответить. Успехов вам! PS Только, правильно говорить не 'возвести в логарифм', а 'взять логарифм от обеих частей'
К чему такие заставки? Стала модно в заставки внедрять фото известных гениев, ученых, артистов, видимо для привлечения подписчиков? Ну-ну. И это "мода" во многих сайтах - эпидемия. А задачи хорошие.
You showed that there is no real solution, which also is obvious by plotting. But stating that we are dealing with a complex root doesn’t provide a ‘checkable’ solution. What is the solution, or can it be that there isn’t a solution, neither real nor complex? By the way, the Lambert W function is a nice.
@thunderpokemon2456 Means no real solution, if we graphically search y=x^2+1 for an intersection of y=0, it appears no solution, but technically x=-i, i would both have y=0. This means that just because y=x^2+1 and y=0 don't appear to intersect, 0=x^2+1 can't be automatically assumed for no solution. Through this, using proof by counterexample, if two equations have no real intersection, they may have a complex intersection.
At a glance, your 'l's kinda looks like 'e's....jus saying. My Math Prof would've flunked me for not writing a little bit clearer owing to the nature of the answer!
@@coldair9632 thanks, that’s what I got, but that is a complex solution. Wolfram Alpha shoes one real solution. I think taking the ln of both sides, evaluating for +/- absolute value, will provide real solution.
3^x = x e^(Ln(3^x)) = e^x e^(x*Ln3) = e^x Ln 3 = ln 3 + 2*pi*k*i, k in Z e^x * e^(ln 3) = e^x, because e^(2*pi*k*i) = 1, for any k in Z So e^ln3 = 1 3 = 1 easy
Where's the 'lie'? The title is a question, not a statement, so isn't a true/false statement. If the question does not come from a Yale admission test, then there's a problem.
a real point is still that at 7'25 the teacher says And writes down that W(-ln 3) does - not - exist , staying visibly on screen until the end . so who would like to make the question more complex .
Oh well,User.user.user....!! I follow the mathematics but the pronunciation is an Irritation for people with musical ears.And I do not see why this program cannot find people who pronounce language properly.It would enhance the quality. Have you got it ?
Er... 3^x > x for negative x, and 3^x > e^x > x for positive x. So: no solutions.
Plot twist: you only learn about the Lambert W function after entering Harvard!
Or you watch videos from an oriental gentleman.
@@choiyatlam2552 i love math
Plot a few points of the curves y= 3^x and y=x, and you'll see that there are no real number solutions. The curves don't intersect anywhere.
Not that it would help you in the Harvard entrance exam, but the complex answer has real value = 0.2297501 and imaginary value = -1.2664477.
My suggestion when trying the entrance exam is the following:
3^x = x | * -ln(3)*e^(-ln(3)x)
-ln(3) = (-ln(3)x) e^(-ln(3)x) | productlog
W(-ln(3)) = -ln(3)x | /-ln(3)
-W(-ln(3))/ln(3) = x
I got it like this
# " -ln(3)*e^(-ln(3)x)"
wat
@@moebadderman227 That's the operation used for the next equivalent transformation .
Here it is a multiplication with the factor (-ln(3) 3^-x), to be able to apply the product log for the following equivalent tranformation.
This question has a much shorter answer by using calculus
You can prove there is no number that evaluates
3^x = x
by computing the limit on infinity, and since infinity isn't a real number (arguably not even a number , just an expression) and by proving that it is the only solution, then you can derive it in just below 5 minutes (the actual math,proving that 3•3...•3 diverges is just stupid math that needs to be done)
as a PhD physicist, I agree it is that simple and I never saw anything like it. I suppose if x is imaginary there is some solution?
@@stevenknudsen7902 OMGGG
@@FanisBartzis hey, I'm 59 years old, almost ready to settle into a beach chair and let others do the mathematical lifting!
@@stevenknudsen7902 Sir i wanna become a mathematician,do you know how motivating this moment feels like?
@@FanisBartzis Don't get me wrong, I am good at math, and it energizes me. But physicists increasingly have problems with "mathy math" after about sophomore year in college. When I see 3^x = x, I can see it has no real answer. So I write (e^(ln 3))^x = x. Then let u = e^(ln 3) and I get u^x = x. I will write x = a + b i and go from there. But this youtube teacher is known for writing many steps rather than getting down to (what I would call) business. To each his or her own.
When you divide by x you need to assume x!=0
Thanks
We know x!=0 because 3^0=1 not 0
@@mjj29 i’ts a formal fact: until you have solved the equation you don’t know the value of x so you nee a priori declare x!=0
@@pidigi ok, formally when we divide by X we should split the problem and solve it separately for the X=0 case and the X!=0 case, but the former is trivial by inspection
Are you using a factorial sign or something else?
-ln(3) lies outside the domain of the Lambert W function.
Replacing a functional expression with another functional expression. It's like a dog running after its tail.
but it encourages you to graph it and provides some insight. Otherwise one can just estimate.
=9 3^2 3^1 (x ➖ 3x+1)
As usual excellent explained... for me (73) sometimes a bit fast...
The domain of the Lambert W function is from -1/e to infinity.
"-ln3" is less than "-1/e". Is it correct to use it?
Yes.
Note that -log(3) < -log(e) = -1 < -1/e.
I don't know the details but Wolfram Alpha still gives an answer, the answer is just imaginary.
There are multiple branches of Lambert W function
x est complexe = 0.2298 + i * 1.266 ou 0.2298 - i * 1.226
il y a plusieurs solutions .
la fonction de Lambert ne sert a rien .
3^x = x
take log3 both sides
x = log3(x)
so.... 3^x = log3(x)
raise 3 to them
3^(3^x) = x
so, 3^(3^x) = 3^x!??
no real solutions imo....
All I heard is Lambo and VW function. Harvard here I come!
Is there anything Lambert W cannot do???
Freeing you from the shackles of having to say "natural log" twice for some reason.
It cannot make a sandwich, checkmate
head
@@sazob 💀
@@vieilatome2257 🛌
You keep saying "nature log nature log 3", when it's just "nature log 3".
In fact it is "natural log 3"'
@@marzipanhoplite17 Correct but I thought one complaint at a time would be challenging enough.
разве можно 2 части уравнения возвести в логарифм, может это неправильно. Если X = Y. Не значит что ln(X) = ln(Y) ?
Можно, если обе: X,Y>0.
И равенство сохранится, причём работает в обе стороны, тк логарифм и экспонента [ему обратная]) - функции взаимнооднозначные
@@Alexey_Pavlov ясно, прикольно, что кто то ответил, даже по русски)
@@Apaximatic_Play Я хотел сначала написать по-англ, что у уравнения нет вещественных корней.
Но посмотрел комменты, :- неск. уже это отметили.
Пока просматривал, наткнулся на ваш вопрос.
Решил ответить. Успехов вам!
PS Только, правильно говорить не 'возвести в логарифм', а 'взять логарифм от обеих частей'
К чему такие заставки? Стала модно в заставки внедрять фото известных гениев, ученых, артистов, видимо для привлечения подписчиков? Ну-ну. И это "мода" во многих сайтах - эпидемия. А задачи хорошие.
wouldn't newton's approximation method solve this is seconds?
No
you did a try .
?
no real solution here...
@@wolfberlin put i as the starting number
L0Lno
I know that you do not read our posts but why not set the question to have a real solution?
Example: (1/3)◌ͯ = x , solution: x = 0.548
a^x = x has no real solution for a>1
Wasn’t this in series 3 of young Sheldon?
They do not teach this in American high schools.
It's not the Harvard entrance exam that determines high intelligence, but exclusion from the football team. ;)
Thanks i'm the best mathematics grade7 in my school
I saw Jack Lambert in a bar one night
freaking awesome!!!
So what's the answer? I was hoping to see a number at the end.
How find the complex solution?
I'm pretty sure that Harvard University does not have an entrance exam. 🤣🤣😂
Don't do this way, it's unpolite, say "the entrance exam is a complex number".
@@marcoantoniofalquete557 😅😂👍 Yes, much better!
@@marcoantoniofalquete557 Damn, a W( way to put it )
@@ericmiller6056 Agreed
Please explain what you mean by a complex solution in this case.
It as "i" which is -1^1/2 which is not possible for existence in our dimension, any number in a+bi form is known as complex no.
@@thunderpokemon2456
i wasn’t in the solution.
You showed that there is no real solution, which also is obvious by plotting. But stating that we are dealing with a complex root doesn’t provide a ‘checkable’ solution. What is the solution, or can it be that there isn’t a solution, neither real nor complex? By the way, the Lambert W function is a nice.
No intersection means no soloution thats what is shown in graph
@thunderpokemon2456 Means no real solution, if we graphically search y=x^2+1 for an intersection of y=0, it appears no solution, but technically x=-i, i would both have y=0. This means that just because y=x^2+1 and y=0 don't appear to intersect, 0=x^2+1 can't be automatically assumed for no solution. Through this, using proof by counterexample, if two equations have no real intersection, they may have a complex intersection.
This is a Harvard Exam? It’s pretty easy.
Mr. Lambert enters the room.
no, he just left the room
@@SanderO-v9c You just the math room, d stick.
Can u pass Harvard??
Shows 3 to power of x = x
Isnt this highschool algebra?
Monomial, binomial, and polynomials
What logarithm of 3 equals the power you multiply to get that logarithmic answer
How many times do you multiply 3 by itself until the number equals your answer
Dumb down version
At a glance, your 'l's kinda looks like 'e's....jus saying.
My Math Prof would've flunked me for not writing a little bit clearer owing to the nature of the answer!
xln3=lnx
ln(x^x)ln3=lnx
ln(x^x)/lnx=1/ln3
x=1/ln3
where did i go wrong?
line 2 and line 4...
x = ln(e^x), not ...x^x
and ln(x^x)/ln(x) is not x...
@@wolfberlin oh yes mb, thank you
Can you please do e^x = x^2
I think it would be:
e^x = x^2
(e^x)/e^x = (x^2)/e^x
1 = e^-x * x^2
(1)^(1 * 1/2) = (e^-x * x^2)^(1/2)
1^(1/2) = e^(-x/2) * x
1 = xe^(-x/2)
1 * -1/2 = xe^(-x/2) *-1/2
-1/2 = -x/2 * e^(-x/2)
W(-1/2) = W(-x/2 * e^(-x/2))
--[Flip]--
W(-x/2 * e^(-x/2)) = W(-1/2)
-x/2 = W(-1/2)
-x/2 * (-2) = W(-1/2) * (-2)
x = -2W(-1/2)
Therefore x = -2W(-1/2)
@@coldair9632 thanks, that’s what I got, but that is a complex solution. Wolfram Alpha shoes one real solution. I think taking the ln of both sides, evaluating for +/- absolute value, will provide real solution.
Is this for bachelor's entrance? Is all the functions used here are taught in US highschool?
If you’re in the top math levels, yes.
I thought complex roots came in pairs.
Isn’t that with quadratics?
I believe it is true for roots of a polynomial with real coefficients but not for every equation.
3^x = x
e^(Ln(3^x)) = e^x
e^(x*Ln3) = e^x
Ln 3 = ln 3 + 2*pi*k*i, k in Z
e^x * e^(ln 3) = e^x, because e^(2*pi*k*i) = 1, for any k in Z
So e^ln3 = 1
3 = 1 easy
nobody can understand speech going as fast as light
It’s not that fast
does not excist? ...
прям раздражает как равно неправильно ставиться
For master degree?
Like my freshman calculus professor. Impossible to understand rapid heavily accented speech
No
Wow! I asked you not to lie a week ago and you are still lying in your title! Do you really think you have to go so low to get attention? BE TRUTHFUL!
Where's the 'lie'? The title is a question, not a statement, so isn't a true/false statement. If the question does not come from a Yale admission test, then there's a problem.
When you have no real solution, solving is a waste of time
HELL, no!!!!!!!!!!!! Imaginary numbers have their uses.
Only if you don't believe in complex numbers - where there might be a solution.
Its useful for converting units to other dimentional quantity
there are people who believe that complex numbers are not unreal .
a real point is still that at 7'25 the teacher says And writes down that
W(-ln 3) does - not - exist ,
staying visibly on screen until the end .
so who would like to make the question more complex .
Bad video!! You should explain Lambert W fuction. In particular, you are using W out of it's domain. What a shame.
soooooo slow
3×X=X
3×0=0
0=0?
Marollco luf barn there
Excist
W e get,e L ungdomar barn, e wrong ,luf barn oppositionen,
Thank you for watching! Have a great day!❤❤❤
STOP PRONOUNCING " natural " as " NAYtural "....!!!....😮...
@@irenehartlmayr8369 oh well, Irene, irene, irene...
Oh well,User.user.user....!! I follow the mathematics but the pronunciation is an Irritation for people with musical ears.And I do not see why this program cannot find people who pronounce language properly.It would enhance the quality. Have you got it ?
@@irenehartlmayr8369 yes, Irene, i‘ve got it. And i love math, you too right! My real „user“ name is Sander. Bye 😉
You can't just divide both sides by x .. that's not mathematically correct
You can by stating that x is not equal to 0.
You can by stating thar x is not equal to 0.
You can by stating thar x is not equal to 0.
You can by stating thar x is not equal to 0.
Of course you can 😂 I guess you didn't study mathematics beyond middle school... you shouldn't discuss a subject you blatantly ignore...