Nice, but...when we are talking about solving equation, it is necessary to find all of them or indicate that it's required find some of x. From graphs of functions easy to see that the equation has two solutions. One is : 1.15

@Iam-ix4he

6 ай бұрын

Ok RS Grewal

@seba8115

6 ай бұрын

I correct myself: the grade is nineth so the ecuation has 9 solutions)

@bikox4352

6 ай бұрын

Both functions have only one intersection point in the first quadrant. The solution shown is therefore completely correct.

@debikk4204

6 ай бұрын

​@@bikox4352They have two intersection points. One being in x=27, and second in x≈1.151, which was mentioned

@yuborthedominator687

6 ай бұрын

@@bikox4352yea, in the first quadrant maybe, what about the other ones?

Nice trick, but there's a flaw in the end when you deduce from x^(1/x) = 27^(1/27) that x=27. There's no warranty that it's the only solution, you'd have to prove that the function f(x) = x^(1/x) is injective for that. And it's not the case. It's the same mistake as when you have x² = 2², and you deduce from it that x = 2. You missed that x = -2 was also a solution.

@Iam-ix4he

6 ай бұрын

Ok RD Sharma

@DaBruhMe

6 ай бұрын

Interesting

@bikox4352

6 ай бұрын

@Altair705 Both functions have only one intersection point in the first quadrant. The solution shown is therefore completely correct.

@Altair705

6 ай бұрын

​@@bikox4352 Sorry but no, as stated by others there are 2 solutions: x=27, and x≈1.150825. If you just drew the functions you probably missed the intersection point at x=27 because 3^27 is a huge value. Even if there was a single solution what I said still applies: that method allows to find a solution but you have to justify that there's no other one.

@angelabdul3308

5 ай бұрын

Yeap, if you replace x for 27 on the third step you can tell there's something wrong already​@@Altair705

The two real solutions using the Lambert W function are x = 27 and x = 1.150825 (6 decimal places). There are of course and infinite number of complex solutions as well.

@tobiasketteringham7201

6 ай бұрын

You don’t use a real solution to the 6th decimal unless you are going to find all infinite possibilities

@davidbrisbane7206

6 ай бұрын

@@tobiasketteringham7201 Yes you do. You are obviously not an engineer!

@tobiasketteringham7201

6 ай бұрын

@@davidbrisbane7206 and you are OBVIOUSLY not good at english, but regardless I would love to see how you came up with this basic question without a calculator and calculated it to the 6th decimal? As stated this is a basic university level question.

@davidbrisbane7206

6 ай бұрын

@@tobiasketteringham7201 There's a series you can use to compute the Lambert W function branches W(0) and W(-1). By the way, I'd be interested to see how you calculate e^x without a calculator 🤣😂🤣😂.

@Foxxey

6 ай бұрын

​@@davidbrisbane7206 You, pen and paper are turing complete. Everything your computer calculates you can too. e^x can be calculated using a series as well.

The funny thing is the note at the bottom saying you should know this … pride comes before the fall when you are dealing with math geeks and there can be multiple solutions.

What I'm not happy about is that your method of solving required that you knew 27 would be the solution in the first place.

@littlewizard9138

6 ай бұрын

That is exactly the reason I hate this method, in the end he basically guess the solution

@davemiller401

3 ай бұрын

@@littlewizard9138 Yes I thought the same thing. I just kept yelling at the silent screen saying WTF?!!!

@turbosinaboy

3 ай бұрын

The only thing I can deduce from this and another video is you need to get a/a ratio. How dou you get this a? In this case: 3^a = 9*a In the other video they have 2 instead of 3 and 32 instead of 9. So, they have 2^a = 32*a. Being a = 8, one solution is 256. Its important to note that 9 is power of 3, and 32 is power of 2.

It's easy to see x > 0. Now, notice the given condition is equivalent to x ln(3) = 9 ln(x) or ln(x) / x = ln(3) / 9 = ln(27) / 27. Because the first derivative of ln(x) / x is (1 - ln(x)) / x^2, we know ln(x) / x is increasing on (0, e) and decreasing on (e, infty). Thus, observing that x = 27 > e is a solution implies that there is one other solution in (0, e).

@eobardthawne6903

6 ай бұрын

Can you please explain how you deduced ln(3)/9 = ln(27)/27?

@MrKrabbs2009

6 ай бұрын

@@eobardthawne6903you multiply fraction by 3/3 so you get (3 ln3)/27, then you put 3 in ln -> ln(3^3)/27, 3 cubed is 27.

@eobardthawne6903

6 ай бұрын

@@MrKrabbs2009 now I get it. Thanks.

@Aleblanco1987

4 ай бұрын

My first thought was using logarithms

@yokkie2846

2 ай бұрын

​@@Aleblanco1987 Same! That seemed like the only way that made sense to me before watching the video. After watching the video though, I'm not disappointed with their method, but not satisfied either if you know what I mean, hehhe 😅 It was a nice & short way of solving it, but not a complete one

just cube each side and you get 3^(3x) = x^27. Then, (3^3)^x = x^27. Finally, 27^x =x^27 and then it's really easy to see x=27 is a solution. However, 27 isn't the only solution as another commenter already pointed out.

@foldingpapers5574

6 ай бұрын

yeah, my initial thought was "there's gotta be an easier method with cubing", thanks for writing it out!

@AstronxD

4 ай бұрын

Agreed. Solving a few steps and doing what could've been done on the first step... This looks more like a hit and trial method rather than solving.

@jshrekmaster

4 ай бұрын

You lost me at x^27=27^x. Lets say 27 is 3 and x is 2. You're saying 2^3=3^2. Therefore 8=9. And how do you already know from that that x=27? Is it because 9 and 3 are the only non-hidden values and through pattern recognition x is obviously 27? This math yt community is insane man

@foldingpapers5574

4 ай бұрын

@@jshrekmaster no, obviously not - the statement is that through these transformations you would reach the statement x^27 = 27^x, which could only possibly hold true for one value of x, namely, 27. How you extrapolated what you extrapolated is genuinely beyond me

@instinx9154

4 ай бұрын

@@jshrekmaster x is an arbitrary variable that can be replaced with any number. 27 is a number that can't be replaced without any algebraic justification. So, you can only replace x and not 27. when you set some random number to the power of 27 is equal to 27 to some random number, just looking at it 27 pops out as an easy solution since 27^27 visually is the same as 27^27. The reason why I cubed each side is because 9 and 3 are coincidentally factors of 27, and that 3^3 is 27, so by cubing each side, you transform a 3^x to a 27^x and an x^9 into an x^27 through properties of exponents. There is another solution using the Lambert W function in addition to infinitely many solutions using imaginary numbers, but those solutions are well beyond the scope of this video.

Using Lambert W function , I got the following solution: x=exp^(-W(-Ln[3]/9))=> x=1.15082 which is not the only possible solution. But, it does satisfy the equation.

You are missing solutions. If you try x=1 and x=2 you can see that there is another solution between these values of x.

doing this before the video x^-9 * 3^x = 1 x^-9 * e^ln 3^x = 1 x^-9 * e^x ln 3 = 1 (x^-9)^-1/9 * (e^x ln 3)^-1/9 = 1^-1/9 x e^-(x ln 3/9) = 1 | you might have noticed i only used one of the roots. The W function actually handles the rest of these roots. im using the principal for now. -(x ln 3/9) e^-(x ln 3/9) = -ln 3/9 Lambert W Function: W(-(x ln 3/9) e^-(x ln 3/9)) = W(-(ln 3)/9) -(x ln 3/9) = W(-(ln 3)/9) | W(xe^x) = x x = (-9W(-(ln 3)/9))/ln 3

@zyklos229

6 ай бұрын

The Lambert function is funny. Afaik it's just the numerical calculated inverse of the definition. So no closed form and actually cheating by claiming, the result was not estimated numerically 🤔

@zihaoooi787

6 ай бұрын

@@zyklos229 some people actually consider it a closed form because of its two representations of the principal: W(x) = Σ [from n=1 to ∞] [x^n * (((-n)^n-1)/n!)] and W(z) = z/2π * (integral [-π to π] [((1 - v cot v)^2 + v^2)/z + (v csc v) e^-v cot v] dv) but whether this is closed is debatable sooo yeah

@RunstarHomer

5 ай бұрын

@@zyklos229 The W function has nothing to do with numerical approximation. It is defined as the inverse of f(x)=x*e^x. Many functions are defined in this manner, as the inverse of another function. Logarithms, radicals, and inverse trig functions are examples.

@1:32 you already know the answer that's why you multiplied & divide by 3.

@hillarycastaneda

4 ай бұрын

That made me mad

@tristan583

20 күн бұрын

Exactly

正确的解题思路和方法是对等号两边同时取Log, 将复杂的指数形式变成对数形式，经运算后即可得到解。

I am starting to feel that random things appeared just to prove LHS = RHS

Great approach….👍 Thank you very much much Sir

Damn thats really interesting, i'd be stuck with "x/log x = 9/ log 3" 😅but yeah as others said, you kinda assumed it's got only one solution.

La fonction w de Lambert nous permet d'obtenir les deux solutions dont x=27

You have a logical fallacy when you say “you should know this”. Most of the regular folks clicking on this video don’t know ‘this’. 😊

Most questions usually accept the multiplication of numbers that are in the question as the answer. In this case 3*9, there can be many answers but an easy one

Why are you able to multiply both the radicant & index by 3? Just to get the equation to work?

Logarithm left the chat

Thanks for sharing!

Thanks for this. It makes me think.

Sorry, but you cannot just use as condition that base and exponents must be equal on both sides. It is not the only solution of the problem

@Iam-ix4he

6 ай бұрын

Ok Ramanujan

@seba8115

6 ай бұрын

The solution is logic

@bikox4352

6 ай бұрын

Both functions have only one intersection point in the first quadrant. The solution shown is therefore completely correct. The result of 3 power x is only positive for real numbers. The result of x power 9 is positive for positive real numbers and for negative real numbers is it negative and if x=0 is the result zero.

@King1Z7

6 ай бұрын

​@@bikox4352bro is really spaming the same wrong answer under every comment lmao

@seroujghazarian6343

6 ай бұрын

​@@bikox4352 x^(1/x) is not injective

Wow! Never seen this before!

Take x=3^t, now by comparing both sides you will get 3^t=9t, take t=3^y -eq(1) and now the equation becomes 3^t=3^(2+y), from which you get t=2+y -eq(2), now find intersection of eq1 and eq2 you will get that eq2 is a tangent of eq1 at y=1, then by putting y=1 you get t=3, which makes x=3^t=27 and this x satisfies our original equation, hence solved. 🎉

Let x = 3^t , so 3^t = 9t Let t = 3^s, so 3^s = 2 + s → s = 1 So t = 3¹ = 3, then x = 3³ = 27

@andrewhone3346

6 ай бұрын

Correct, and I think this method is better than the one in the video. However, you are implicitly assuming that x is a positive integer (which the video also assumes), but there is another real positive solution.

@kristir1262

5 ай бұрын

I like this method of solving best! Substitution all the way!!!😂

@Shirlippe

2 ай бұрын

This substitution is justified if you assume x is a positive integer. Just use unique prime factorization. From 3^t=9t you could have found 3^(t-2)=t, which has a single solution: t=1.

Cool video, please continue to make more

Could you tell me your pen's brand and also its model? Thanks

@rutamupadhye1828

3 ай бұрын

Pilot V7

I would avoid posting incorrect math. next time don't assume injectivity.

@King1Z7

6 ай бұрын

It's not incorrect math though... next time tey thinking before writing a comment....

@joshdeconcentrated2674

6 ай бұрын

there's an unmentioned root at around x=1.16. the injectivity assumption in the math doesn't find this root, and instead only finds a single answer@@King1Z7

@joshdeconcentrated2674

6 ай бұрын

@@King1Z7 from x^(1/x) = 27^(1/27) that x=27 you'd have to prove that x^(1/x) is injective, which it isn't.

Excelente. Incrível.

Nice trick with adding 3/3👍

You can use log to solve it quicker. Nice one. 😊

Replacing x with 3^n, I was able to reduce it to 9n=3^n, but then I was stuck (even though I could see that 3 was a solution).

Isn't there a more, how do I put it, layman solution. Since we are talking about exponentiation, it's evident that x is a power of 3. Just input values and reduce to check if it fits. Put x as 3, 9, 27 and so on. Try lower powers as well.

Much easier to use logic. If we stay w whole numbers, X has to be root 3 and larger than 9. First up is 27. Falls into place.

Wonderful......!🎉🎉

You need to use lambert W function for another x

Thank you...clever

Now I can sleep better knowing this is solved.

Fantastic

And he did it all in pen. Honor mode achieved.

We can take log on both sides and then multiply 3/3.

Interesante y buena la explicación, gracias por compartir. Mi hija y mi persona estamos muy agradecidos.

Thanks

Nice job

Is this really a trick to solve the problem or just something that's straightforward once you already know the solution? It's like showing someone how to solve to a maze as - "just turn here then here then here!"

@hillarycastaneda

4 ай бұрын

When he changed the 1 for 3/3 it was very clear he already knew the answer 🫠

@rutamupadhye1828

3 ай бұрын

practice automatically makes you know such techniques

Quite impressive

Very Nice

Formidable video, explicas super, apoyo mutuo like

We can take log3 that is log with base 3 both side and try to solve it, finaly we will get X/log3X = 9 and then finally we will get to know x=27

Lovely question

What pen are u using???

my try： make a^3ln3=9aln3 😊😊

Very nice

I have another easy method to slove it । 3^x= X^9 We can write in this form 3^1/9 = x^1/x After that we can follow the methods which mention in the video but we can save more step

Wow that’s awesome

The equation has two roots, not one. The other root is approximately 1.15 and can be expressed by means of the ""product log functtion".

Lovely ❤

Nice Video Please continue uploading new videos

Can even do the same sum by logarithms right?

I took the log of both sides , base 3, that made it look better, got 27 in about 2 mins...

27, степень переносится, в знаменателе оказывается 27, вычисляется Х=27. 20 секунд на задачу )). Хм...1.15.. тоже решение.

I'ts the same to use one solution in equations of 2° degree when b^2-4ac#0

Elegant solution. I have an ugly solution with logarithms. Log_3 of both sides You get x = 9log_3(x) Call this * Let log_3x = y, therefore x=3^y. Call this ** From * you have x = 9y From ** plug in for x and get 3^y = 9y. Compare this to the original equation. Right side no longer has exponent, so it's a bit better. It's clear that y = 3 is a solution. Plug this into equation x=9y and you get x=27 Other solutions? Don't know.

@seroujghazarian6343

Ай бұрын

No, you need to use ln

Can we take log to solve easily if calculater is allowed.

Great

I laughing i don't know why, thanks for make my day better

Seeing the X’s being written like that physically pained me.

@unarmed_civilian

4 ай бұрын

"Physically"

@him050

4 ай бұрын

@@unarmed_civilian I felt it in the deepest depths of my inners.

@haarp9069

3 ай бұрын

Sounds like pedantism to me.

@him050

3 ай бұрын

@@haarp9069 not at all. It’s improper mathematical notation

@haarp9069

3 ай бұрын

@@him050 Definitely pedantism.

Beautiful

Just by looking i guessed 1.1

Just when I thought I escaped advanced math it shows up in my recommended. 😂

What song did you use for this video?

You scored 50 out of 100. This is why we learn logarithms & the natural logarithm (ln). Moreover, this is not at the level expected for an Olympiad. 😑 In Korea, power & exponential equations are covered in SOPHOMORE year of HIGH SCHOOL.

I learned this in elementary school and never ever needed it in real life.

Superb

I will look at mcq and substitute x with each of them to find answer

First off where do all of these other numbers come from? It’s like they came out of the blue out of thin air. Also, what would be the best place for me to learn math wether it be a website or a book or course etc.

すごいね🎉

You can rahter easily use logarithmic function

Pensar que dicho ejercicio es de nivel básico o intermedio aquí en Perú ☠️. Saludos desde Perú y excelente video.

World 🌎 International Math Olympiad 2019 - Algebra - Find f(x)?! Boost Your Confidence 😆 kzread.info/dash/bejne/f2yBr8hxoNCWdrA.html

Before I watched the video, I tried it. I took the 9th-root of each side and got 3^(x/9) = x and started plugging in at x= 0 and got up to 9, realized the pattern, so then I did x=18 and then x=27 and had the answer 3^(27/9) = 27 --> 3^3 = 27 --> 27=27. It was quicker, but maybe it's not as easily applicable to every type of problem like this? I dunno.

Myslím vhodné pro použití Lambert W funkce.

Jesus I felt the fear of maths again after twenty five years 😂

I guessed and got 27.

Nice

just multiply 3 and 9 , easy

Why didn't you simply take log on both sides ?

Can We can take log both side and solve ?

어차피 직관으로 풀거면 처음부터 3의 고작3승인 27정도는 생각할 수 있으니 27구할려고 방정식의 형태를 바꾼게 별도움은 안되는 듯..... 이건 풀이라고 할 수 없죠...

Definitely should have taken log of both sides first

The limit of x^(1/x) when x come to +infinity is ~1, so easily to notice that there are 2 answers😢, and when the student can find only one, the teacher will said that "you were trapped" 😂😂

*Wow* 🥰

There is another solution , i calulated it numerically, about x=1.14 can you show the calculation? Thank you

I was taught never to question when someone multiplies both sides with absolutely anything😂 and it is somehow acceptable

@danielderias4773

6 ай бұрын

By the looks of things everything he did here was correct except at the end when he assumed x was 27

I have been thinking about this for a long time, for me the second solution of approximately 1.15 makes perfect sense. But 27? I really can't understand how 3^27 could equal 27^9. Every time x gets bigger it 3^x and x^9 grow further apart. I've tried putting them in a graph, sadly the shear size of the numbers makes it hard to read.

@smalin

4 ай бұрын

Just consider expanding the exponents. 3^27 is 3 times itself 27 times. 27^9 is 3*3*3 times itself nine times. Either way, you have 27 threes in a row.

Amazing！！！！！！！！！！！！！！！！！！！！！！！！！！！

Easy way to solve can be Taking log both side with base 3

@seroujghazarian6343

Ай бұрын

ln*

Isn't this just "guess and check" with extra steps?

Take log both sides…why to do so much!