Exponential Equations With Powers of X

[@TheOrganicChemistryTutor]

Exponents - Free Formula Sheet: bit.ly/4ehTaPN Final Exams and Video Playlists: www.video-tutor.net/ Next Video: kzbin.info/www/bejne/ml6oZpyplq53iqM

[@rickyjames97]

I think that this way to solve is for who already know the solution. You should decompose 729 before and then you play with x^x^3, comparing both. If you do like the video it's like throwing randomly hoping that it works

[@sabalsneh9315]

Moral of the story:- don't always take log . Wait and use other alternatives too .😁😁

[@user-ud6xw3js3r]

How could you've used log here ??

[@georgesadler7830]

MR. Organic Chemistry Tutor, thank you for the video.

[@josephmuriuki5147]

Your explanation is very clear,thank you very much.I was stuck with a problem of this nature thanks J.G.😅

[@georget8008]

We could transform the problem in the form u^u=729, and apply the Lambert W function to solve it. Couldn't we?

[@akinrotimidaniel2325]

Yes

[@bonbonenuranium5034]

I hope your comment will be seen

[@mikkel6183]

That would be the general way to solve this for all values instead of just 729, yes

[@joshuaquezada9363]

Srry for my ignorance, I just watched some videos about that Lambert function and how to solve exponential equations and I was wondering, is there a way to calculate de exact answer after applying it without having to use a calculator?

[@georget8008]

@@joshuaquezada9363 you can ask the same question for the ln(x) or the e^x functions. The answer is :no

[@Goku_is_my_idol]

So i tried to solve this ques from thumbnail Now right side is a power of 3 Take x= 3^(k/3) We get 3^(k*3^(k-1))= 729= 3^6 ==> k*3^(k-1)= 6 ==> 3^(k-2)= 2/k Now we shall compare their graphs to see how many solutions are possible Since in the negative k --- 3^(k-2) lies in the second quadrant and 2/k lies in the third quadrant they never intersect One of them is increasing and the other one is decreasing. So if they intersect that particular point will be the only point of intersection Its obvious that k=2 satisfies x= 3^(2/3)

[@TranNam-db6ov]

Totally interpretable

[@Goku_is_my_idol]

@@TranNam-db6ov thnk u

[@marcellom2292]

This case in simple because have 729, but how do you do if you have another number for example 5?

[@mikkel6183]

You can use the LambertW function for this

[@wisdomemeka6271]

@@mikkel6183 please how

[@rcnayak_58]

This is an alternative solution to the problem, maybe bit easier. The order of evaluation of any exponential number such as 2^3^2 is 2^(3^2) so that the answer is 2^9 which is 512. And not 64 as some calculate. Now let us proceed. Our question is x^x^3 = 729. Let x^3 = k. So that x = (k)^(1/3). Our left hand side term now can be expressed as x^k which is {(k)^(1/3))^k}. This equals to k^(k/3) which can be expressed as (k^k)^(1/3) which is equal to 729 (9^3). So that k^k becomes ((9^3)^3) = 9^9. Therefore k^k = 9^9 i.e, k = 9. In other words, x^3 = 9 which makes x = cube root of 9.

[@deutschnacharabischemgeschmack]

Find it more understandable. Thanks.

[@aritraj10b28]

Huh! I expected some integer and this turned out to be an irrational number. Moral: Never be the die hard fan of Pythagoras!

[@epicroyalegamer3272]

Abhi Beta Class 10 Mai Sirf Pythagoras Sab Kuch Lagta Hai, Jee Dene Ao Bete

[@Andrew-dh1ws]

Khan academy should hire you, no joke

[@Andrew-dh1ws]

DarConall khan academy could pay him better perhaps

[@桃胡-q9n]

From my own experience, this dude helped me a lot more in my academics than Khan Academy itself...

[@akashjadhav4829]

He should hire khan academy

[@haha-or9sz]

Bu iyiydi. Yeni bir bakış açısı kazandım. Teşekkürler

[@eliaselieza7991]

Nimeelewa sana ,let apply your logic concepts,

[@petrit8520]

Hello. Can you check again a to the power of two parentheses power of three?!

[@BinaHejazi]

actually (x^x^3)=729 can be written as (x^2)^3=9^3. taking cube root of both sides we have: x^2=9 and hence x=3 and thats the answer!! isnt that right?

[@axelnils]

No, x^x is absolutely not the same as x^2

[@jim2376]

If a^a = b^b then the bases are equal. So cube root of 9.

[@pratigyaberi2000]

From the equation we get x^x = 9 . As the answer say x= 9^1/3 which is equal to 2.08 And 2.08^2.08 = 4.57( APPROX ) not equal to 9

[@deutschnacharabischemgeschmack]

You probably cubed both sides bcs you know the cube of 9, but using another exponent like 4 is logical. What if the number after the equal sign is not familiar to us?

[@AkashKumar-cz9gm]

x^(4*x^2) = 1/sqrt(2) How do you solve this equation?

[@tombufford8659]

Looking at this I would try 81 x 9= 810-81 = 729. 81 = 9x9 = 3 ^ 3 and then 3 ^ 3 ^ 3 = 9 ^ 3 = 729. However this would be (x^3)^3.

[@TheLordrain12]

can we find all x^x^n = k equalities with only using algebra logarithm and stuff? (of course not with this technique)

[@sarbaripanja6979]

We can, we have to use the Lambert W function....you can check blackpenredpen for this function

[@TheLordrain12]

@@sarbaripanja6979 I'm going to check that, thank you.

[@ananeakwasi6]

Great work

[@kamranshadkhast5035]

(x^x)^3 is different with x^(x^3) which one do you choose to go ahead and why?? when it is not specified by prantesis.

[@DavidCorneth]

The latter. x^x^3 is defined to be x^(x^3). The former would be a but dull as it's just x^(3*x).

[@Nerdwithoutglasses]

I thought he would use Lambert W function. Anyways, this is nice

[@oenrn]

Scroll down to read 100 comments by people who don't know how to use exponents.

[@mikejackson19828]

🤣 So true!

[@attila3028]

4:09 intuitive but needs a proof

[@Modemon57]

Yeah.. I'm stuck here..

[@masterdegree5513]

We can use log

[@tomaskovarik1]

much easier to use the Lambert's W function

[@TheFlax33]

good lesson tks

[@TrueBagPipeRock]

thanks

[@syedmdabid7191]

Valde facilis, x=cube root of 9=9^1/3 responsi

[@buildandmakewithnila]

On step no. 3, how did you bring it out, am confused

[@evacameron8670]

amazing

[@Артьомдругартем]

x=9^(1/3)

[@saktishvikmend]

hi.. i need your help pls. 3x + 3^(x+1) + 3^(x-1) = 13. Whats x?

[@Shreyas_Jaiswal]

Is the first term 3^x?

[@DavidCorneth]

Maybe use inspection. trying x = 1 gives 3*1 + 3^(1+1) + 3^(1-1) = 3+9+1 = 13. So x = 1 is a solution. As each of the terms of the LHS is strictly increasing x = 1 is the only solution. (yep I came late to the party)

[@桐谷星爆]

OK, How can you know the third power will help you? The step 1 is really hard to think.

[@theamazingone5217]

It's so you can substitute x^3 and simplify it. It may look hard to apply at first, but after you do some exercises, it comes to mind very easily. In this exercise he could've substituted x^3 with y (or whatever letter you usually like to use) and would've got a much easier to deal with expression. Since y^y is more common to deal with.

[@桐谷星爆]

@@theamazingone5217 Thanks, I should do more exercises to improve myself.

[@satyendratripathi4684]

Can anyone explain or give mathematical proof that when (x³)^x³ = 9^9 Then, x³ = 9

[@abdellahaitouahmane1593]

not corect. the answer not clear

[@Shreyas_Jaiswal]

You can write this way by comparing both sides.

[@mikkel6183]

Rewrite (x³)^x³ = 9^9 as u^u = 9^9 where u = x³ Take the natural log on both sides: ln(u^u) = ln(9^9) => u * ln(u) = ln(9^9) Rewrite u as e^(ln(u)): e^(ln(u)) * ln(u) = ln(9^9) Take the LambertW function on both sides W(e^(ln(u)) * ln(u) = W(ln(9^9)) => ln(u) = W(ln(9^9)) Raise both sides to e's power: e^(ln(u)) = e^(W(ln(9^9))) => u = e^(W(ln(9^9))) => u = 9 Substitute u with x³: x³ = 9

[@Kelvin-yw3dp]

@@mikkel6183 omg. It makes me more confuse while seeing your explanation. why need to take the natural log and what is LambertW function?

[@sirisaksirisak6981]

3 how to 9,8,1,3 so reverse power x=3xxxxx=9.9.9=729

[@tahsintahsinuzzaman781]

Nice one, keep it up

[@MRATtasa]

Things can go complicated further when your r.h.s number can’t be expressed in term of (symmetric) exponents.

[@nataliexelena6256]

What app/website/thing do you use to write with?

[@dhanvin4444]

Same question

[@bidarvandi]

any country education math has a webside and all things about math is there i am sure you find it .i am not from us so if you are find it is usa eduaction math system.

[@bakangsbakho8852]

Wait, so no Lambert W function ???

[@abdellahaitouahmane1593]

the last step is not cleari can resolve it by using Log or Ln

[@Modemon57]

(x^3)^(x^3) = 9^9 => x^3 = 9, it sounds not so good.. Maybe a little proof?

[@sbusisolinathi2494]

x^x^x=lne^2 can you guys help with this one above

[@joexuan6739]

From equation, it will be x^x=9. And then x is not equal to 9^(1/3).

[@mohmedzakymohamedzanatyzan4568]

243=

[@kerbicz]

Sorry buddy, but your solution is not complete. You found x=9^(1/3) as a solution, but you did not ask whether it is the only solution. Actually it is, but the question demands clarification. The key point is that the mapping t->t^t is strictly increasing in the domain t>1; please fill in the details yourself.

[@EmmanuelAmetotobariOlakada]

😅 never expected that

[@LAtomeAZZAZProDuNeu9-]

Derrières playlist : Exercices par paqués de dix en partent des unités avec un chiffres desNumérateurs dans les plus grand nombres de Chiffres Exponentiellement et Univers kzbin.info/aero/PLlne-T0J57yHCjAiv2QcN1INFVOYyhWed 3 Exercices par paqués de dix en partent des unités avec un chiffres numérateurs et leurs multipliés kzbin.info/www/bejne/hJeqnYevZamHbtU... 2 Exercices par paqués de dix en partent des unités avec un chiffres numérateurs et leurs multipliés kzbin.info/www/bejne/sHzFoZRnqpWne8U... Exercices par paqués de dix en partent des unités avec un chiffres numérateurs et leurs multipliés kzbin.info/www/bejne/gnrTaH97esaSh7s... Les ensembles des Numérateurs & leurs multipliés dans 9... kzbin.info/aero/PLlne-T0J57yGi8KDp5SixNePc1_RL6Lup Facebook ; Rabih AZZAZ facebook.com/NEU9f /45*

[@cescosulpesco]

Well, nice method, but if we accept more solution this equation can be pretty easy, because 729 = 27² = 3³ to the 2, if we're apply the commutative property of multiplication to the exponents, we'll have x to the x to the 3 = 3 to the 2 to the 3, finally we'll have 3⁶ = 729 that is verified, so other possibile solutions can be x1 = 3; x2 = 2

[@KrystalSquirrel]

How is it possible? Check your answers. 1^(1^3)= 1, not 729. Also 2^(2^3)=2^8=256, not 729

[@huyhuynh6771]

can someone tell me what grade level is this.

[@sanopann1896]

just grade 6

[@grupomarce2010]

I really don't know the usefulness of this video, because it was solved because you had the luck of x^3 be an integer with very particular properties. The problem is very biased. Better learn the Lambert W functions in order to solve it for any value and not just for 729. What is the point on solving it for 729, with a procedure that is useless for nearly any other value. Just google about Lambert W function and you will be able to solve it for the general case: x^(x^n)=k The result: x=exp(W(n ln k)/n). You can check that for n=3 anf k=729 gives you the same result.

[@original6hockey402]

OMG so what is the answer?

[@ananeakwasi6]

X^x^5=(100)^5 I need help

[@RedEyesBlackDragon0]

That was a very long path for a very short process.

[@ТАР-ю4ю]

It’s nice

[@merc340sr]

Ouch! Can we use the ln for this problem...

[@redeemerbadagbor1673]

Please can you help with this : x squared =16 raise the power x

[@samibaheru4029]

X is not the cube root of 9. X is 2.45...check your answer.

[@marcellocristiano7453]

7 2 9,,,, inverti la marcia.!

[@mintusaren895]

Desertion theory

[@MK-el1gn]

Bellísimo...

[@fabricecropsal7202]

How can you have x^3=9 when obviously x^x=9. Approximatively 2.08 for the first one and 2.45 for the second one that is the correct answer. Your demonstration is wrong. x^x^3^3 = x^9x when (x^3)^(x^3) = x^(3x^3)

[@tharuneinstiein7669]

My suggestion is don't write on your words over maths notation or assignment....

[@incognito731]

My guess before the video start: 3

[@معینکسرائی]

هر چند خیلی وقتها حرفم را جدی زده ام و با دوربین کار و ایل و تبارش شوخی ندارم

[@aashsyed1277]

What???????????

[@Solbashio]

I looked at the thumbnail like wut

[@xdstriker]

I thought x=3 💀💀💀

[@toanpham4110]

👍

[@felixdonamaria5779]

Otra pajeda

[@ZeroAndKeto]

2.0888888

[@junaidck2310]

3

[@patrast5315]

Looked interesting, but you speak so slowly that I got bored after live 40 secs

[@binamahadani3267]

X=3

[@kvcl123]

🥉🎖️🏅👍

[@tonyfong2330]

👍👍👍👍👍🙏🙏🙏🙏

[@ziadmosad6474]

Like

[@satyendratripathi4684]

Dis?

[@simonc9702]

👎

[@olatunjisodiq8504]

mathematically wrong

[@fvcostanzo]

Geez, could you do this any slower?? Ugh...