How to solve this unbelievable exponential equation

[@familyguytimes3664]

this guy makes me love math even more each time I watch him!

[@therealelectromagneticwave]

I wholeheartedly agree. I feel that my actions or the things that i really want to do during this summer get repeatable and I don't think that it's going to be beneficial anymore. In particular, i decided to challenge and refine myself by learning math from my past school year so that i would be more efficient at solving problems whether it is Algebra, Geometry or Trigonometry. Thank you Sir

[@hmwh4t]

@@bprpmathbasics hi I’m stuck on this question , I want to find derivative of y=(3xy)^4 , anybody please help 🙏🏿

[@shen144]

@@hmwh4t Solve for y first.

[@freemathacademy6632]

kzbin.info/www/bejne/ipqbeGqJiKufgKc

[@hmwh4t]

@@shen144 I am being asked to use chain rule

[@supramayro434]

at the end of the video he went from 😁 to 😒

[@enhace15anos.83]

🤣

[@the_magisterate]

when i first saw this question and tried to solve it, I divided by e^2x on both sides. Interestingly, after rearranging the variable terms, we can get the equation for sinh. idk if it's cheating but I just took the arcsinh(1/2) to get the same answer. I think it's pretty cool that seemingly unrelated mathematical concepts like hyperbolic trig and golden ratio can unexpectedly unify

[@WaluigiisthekingASmith]

Reminds me of how sin(54) =phi/2

[@pedrosso0]

​@@WaluigiisthekingASmith sin(54 deg) = phi/2

[@xGuardianAngelx]

@@pedrosso0 yes that is indeed what they wrote..

[@mokouf3]

So nice!

[@marcushendriksen8415]

Let e^x = y. Then the equation becomes y + y^2 = y^3, which is the same as y^3 - y^2 - y = 0. Factor out a y: y(y^2 - y - 1) = 0. Therefore y = 0 or y^2 - y - 1 = 0. The second equation leads to y being (1+sqrt5)/2 or (1-sqrt5)/2. Since e^x = y, y cannot be 0 or negative (since the real logarithm function is undefined for those), so x must be ln(1+sqrt5) - ln2

[@marcushendriksen8415]

Whoops! I should've said that non-positive values aren't in the range of the real exponential, not that it isn't defined for such values (because it's defined everywhere!)

[@dougiemckee4658]

thats what i did and got the same answer

[@januszjasniowski8890]

Dziękuję za wykład zw. z liczbą e. Wspaniały matematyk i wykładowca. Jeszcze raz serdecznie dziękuję. Oglądam pana wykłady z matematyki od kilku lat. Mam bardzo duży zbiór zadań z matematyki średniej i wyższej oraz z olimpiad na joutube.

[@jessetrevena4338]

Great video! Very satisfying solution.

[@radagast7242]

he is an expert at switching markers

[@andrewmccauley6262]

x^n = x^(n-1) + x^(n-2) is a fundamental property of golden ratio and related to fibonacci and lucas series, which I love. So I was on to this one pretty quickly, but still cool to see.

[@ScorpioHR]

I just wanted to ask "what's with PokéBall" on the newest video and now I see it's missing. So I have a following question: What happened to the PokéBall?

[@deez8600]

Your videos never disappoint

[@Walid_Kasab]

I just love this channel

[@sebicastian4573]

You are awesome, i thought i will resolve by my self, but now i realise that much terms in math that i need to know for get your level in math. Great video and im sorry if my english is not good for write :)

[@jedreksupski3272]

Where can i buy this t-shirt?

[@slothguy_]

that marker switch tho

[@activatewindows7415]

Beautiful solution.

[@maths3239]

The complex solutions are: ln((-1+√5)/2)+i*(2kπ+π) and ln((1+√5)/2)+2ikπ where k∈Z. Thanks to KasabianFan44 for the addition!

[@KasabianFan44]

There are more complex solutions than this! x = ln((1+√5)/2)+2ikπ (for any k∈Z) is also a valid set of solutions - these are the complex solutions to e^x = φ. In particular, when k=0, this corresponds to the real solution in the video.

[@maths3239]

@@KasabianFan44 You're right. I watched the WolframAlpha's solution and there are really 2 type of complex solutions. Anyways I don't know how the second solution comes out, so if you would be so kind as to describe it?

[@KasabianFan44]

@@maths3239 When you say “the second solution” I assume you mean the ones you mentioned in your original comment? (If you meant the other one then let me know and I’ll explain that one too.) Your set of solutions actually comes from the other half of the quadratic formula: i.e. e^x = (1-√5)/2. BPRP ignored this solution in the video because it has no real solutions for x; however, complex solutions do exist! To find them, we first need to remember two very important facts: e^(πi) = -1 and e^(2πi) = 1. From here, you can keep multiplying these together however many times you like, to see that e^(nπi) = -1 for any odd n - or, to write it another way, e^(-(2k+1)πi) = -1 for any k∈Z. (The negative in front of 2k+1 is not necessary, it will just make it easier to simplify later.) Now, like I said earlier, e^x = (1-√5)/2 has no real solution because (1-√5)/2 is negative… BUT what we can do is multiply both sides by -1: e^x • -1 = (√5 - 1)/2 Now the RHS is positive so we can take logs (later). But first let’s replace that -1 with e^(-(2k+1)πi) from earlier: e^x • e^(-(2k+1)πi)= (√5 - 1)/2 This simplifies to: e^(x - (2k+1)πi) = (√5 - 1)/2 And NOW we take logs of both sides: x - (2k+1)πi = ln((√5 - 1)/2) Finally: x = ln((√5 - 1)/2) + (2k+1)πi exactly as required. Note that this solution is never real for any integer k.

[@maths3239]

@@KasabianFan44 Thanks for the detailed answer, I also solved this in another way. By "the second solution", I meant the one I didn't write down (I couldn't figure it out). Thank you in advance for your answer!

[@ciberiada01]

Wow! An elegant problem and solution! I prefer avoiding the fraction and write it: ln(1+√5) - ln(2) As √5 ≈ 2.24 and √5 + 1 ≈ 3.24, so ln(1+√5) > 1 and ln(2) < 1 let's say ln(1+√5) = 1 + δ and ln(2) = 1 - ε, where δ and ε are < 0.5 So 1 + δ - 1 + ε = δ + ε < 1

[@phamnguyenductin]

Well, the solution is golden.

[@Dodegaw]

solved the task without starting video, just watching preview

[@Sg190th]

2:46 "we're not going to... yet" haha

[@EngNALrashed]

Love ❤️ it, thank you.

[@amankumarsharma8989]

Amazing solving exponential eq through Quadratic eq properties. Maths is so beautiful 🥰

[@sureanddraw]

Great logic

[@hartmutlorentzen9659]

He speaks like my teachers

[@user-wu8yq1rb9t]

Thank you dear Teacher ❤️

[@dinuwarabinudithdesilva5464]

That is the fabulous element, the gold(I mean the golden ratio)

[@slytherinbrian]

I just saw a video asking for x when 16^x + 20^x = 25^x, and the solution is almost exactly the same, just with an extra factor of ln(5/4) thrown in.

[@superioroakmc942]

My brain is confused.

[@josefstr]

Incredible

[@undsamuel]

I have a question (out of real solutions). If we replace e^x = u, then we would have u+u^2 = u^3, does that mean that the first equation has 3 solutions? 1 real solution and 2 complex solutions? Because when I try to solve all the 3 solutions by factoring I get an impossible solution: e^x = 0, which is not possible. So does that mean that, even if you have an equation of grade 3, maybe you don't get 3 solutions even in complex numbers?

[@zohar99100]

No need for this. Just solve. And what will be, will be... (Like the old song...)

[@mohammedislamnaddari1364]

I don't think it's a polynomial . In polynomials of degree 3 we always get 3 solutions ( at least one of em is real ) . Substituting u=e^x doesn't change the nature of the equation ( being exponential ) because you are still going to solve for "x" and not "u"

[@undsamuel]

@@mohammedislamnaddari1364 But that's not always true. If I had the equation (x^6)/(x^3) + x^2 + x = 0, then you also have 2 solutions and not 3. So I thought that maybe you don't always get 3 solutions on a 3 grade equation.

[@joeyhardin5903]

If you write it that way, there would technically be 3 roots but one of them is a repeated root

[@undsamuel]

@@joeyhardin5903 Which one is repeated? x^3+x^2+x=0 is x(x^2+x+1) = 0, where one "solution is x=0" which is not possible due to the original equation. The other 2 solutions come from "x^2+x+1=0", both complex solutions. So, I don't see any solution repeated.

[@帝释天-z1w]

讲的真好 虽然不懂说了什么 但看一遍就明白了

[@mmoncure11]

Very pretty problem and solve

[@holyshit922]

ln(lim(F_{n+1}/F_{n},n=infinity)) where F_{n} is Fibonacci sequence

[@MathPhysicsFunwithGus]

Another beautiful video! I often used this method (u=e^(ax)) during my undergraduate physics courses!

[@rageprod]

If we take it to the complex game (which we should, because it's fun), we have: e^x = φ, therefore x = lnφ (as shown) but also, e^x = (1-sqrt5)/2 = - 1/φ (demonstration left as exercise :P), therefore x = ln(- 1/φ) = ln(-1) + ln(1/φ) = ln(-1) - lnφ (by the property of logarithms of exponents) but what the heck is ln(-1)? recall e^iπ(2k+1) = -1, for k in Z; we take the natural logarithm of both sides: ln(-1) = iπ(2k+1) therefore: x = iπ(2k+1) - lnφ, for k in Z So we have one real solution and infinitely many complex solutions, because of the complex logarithm's periodic property! Very nice. Edit: sign mistake, oops As corrected by Michael Vogel, e^iπ2k = 1, so ln1 = iπ2k for k in Z. lnφ = ln1 + lnφ, so iπ2k + lnφ for k in Z are also complex solutions!

[@Mmmm1ch43l]

actually, you forgot that in the complex world, not only ln(-1) but also ln(1) can take infinitely many values, so you also need to incorporate these (and also you made a sign mistake in your fourth line) so the full solution would be: x = iπ(2k) + lnφ or x = iπ(2k+1) - lnφ, for k in Z

[@rageprod]

@@Mmmm1ch43l Oh yeah, you're right. Thanks for the correction!

[@mattikemppinen6750]

haha nice coincidence, was just twiddling around with another problem with phi a moment earlier, came to check youtube and saw the thumbnail for this and solved the equation and almost couldn't believe it :D (although now, in retrospect, looking at this one you almost feel that the solution has got to have something to do with phi. like you said in the end "golden exponential equation" =) )

[@quevineuxcrougniard2985]

Ce chinois est tout a fait compréhensible grâce à son tableau en dépit de son baragouin imbittable ! Alors je m'abonne et je le remercie.

[@Faxbable]

"Ce Chinois"... 🙄🤦‍♂

[@LearnWithFardin]

x=sinh-¹(1/2) My approach: e^(x) + e^(2x) =e^(3x) 1+e^(x)=e^(2x) (e^x)[e^x -1]=1 e^(x) - e^-(x) =1 2sinh(x)=1 Therefore, x=sinh-¹(1/2) [I know ,this is an algebra channel but anyway]

[@ccbgaming6994]

What’s the connection between e and sin here

[@robertveith6383]

You are needing grouping symbols for certain exponents: e ^x + e^(2x) = e^(3x), etc.

[@LearnWithFardin]

@@robertveith6383 Thanks yah! Re-Edited!

[@ccbgaming6994]

@@andrewf46b763b Oh, I just completely missed the h notation, thx

[@marcushendriksen8415]

Very clever

[@ahmedelmarrikh7491]

Vous êtes excellent mon professeur,tu m'as bien convaincu, merci

[@Mohammed-oh8rs]

The best youtuber in lofe

[@underpipse9846]

I'll come back when this gets over 1 million views

[@kashoot4782]

x= lnphi, that’s beautiful

[@pneujai]

expected it to be quadratic but didnt expect golden ratio :o

[@ZANGETSU.4]

U can use "ln" though to both sides from the equation

[@NotBroihon]

Elaborate.

[@marcushendriksen8415]

Yeah and then you end up with the ln of a sum on the LHS and just 3x on the right. Hardly the simplest way to proceed

[@Robert_H.]

log(1 + e^x) = log(e^(2x))^= 2x log(1 + e^x) = Log(2) + x/2 + x^2/8 + ... x^2/8 - 3x/2 + Log(2) = 0 x = 6 - 2 sqrt(9 - log(4)) = 0.48141118... Comparision to exact solution: x = log(phi) = 0.4812118... So in case you haven't seen the substitution u = e^x, you do come up with a very accurate solution using a Taylor approximation.

[@yoavcohen120]

Got it right before watching the vid

[@p12psicop]

When I was in college I majored in albejra.

[@tolvajtamas8567]

This is very awesome, tho Ive never heard of the golden ratio. 😅

[@yashdaga2566]

Graph could also have given a good idea about the answer(you couldnt obv get the exact ans but u could verify it using drawing graphs of e^2x and e^x+1

[@PragmaticAntithesis]

Here's my solution. I'll see if I'm correct after watching the video. Let y=e^x Let g be the golden ratio (1.618...) y+y^2=y^3 This has three solutions: y=0, y=g, y=-1/g However, because y is an exponential, y>0. Thus, y=g. Solving for x: e^x=g, so x=ln(g), which is about 1/2.

[@user-orvhnxr]

Ваші відео мені подобається. Особливо там де Ви вирішуєте комплексні рівняння.

[@dracokinerek]

what I did: e^x+e^2x=e^3x | take ln ln(e^x+e^2x)=3x | use ln rule: ln(a+b)=ln(a)ln(b) ln(e^x)ln(e^2x)=3x | simplify x2x=3x | simplify 2x^2=3x | solve x=0, but e^0=1 x = -3/2

[@sivasubramanian2056]

What does that symbol mean which is on your t shirt.

[@johanndohmann1281]

sooo cooool!

[@niceguy999918]

I hate math. This guy is brilliant. "It's not so bad" it's impossible!!! How do you take nothing and create something? How you get a square too of 5? Where it come from? I just added all the top numbers and got e with the little number thing over the top of the e. I got 5 on top of e.

[@HimadriSaha99]

Woah with this example I just realized that golden ratio (phi) equals e^(arcsin(0.5))

[@fxexile]

wow this guy looks like black pen red pen guy

[@JonathanMandrake]

-infinity would also be a solution if allowed

[@muhammadhussainsarhandi9928]

Sir at 0:45, you said e to the x can never be equal to 0, but what if x approaches negative infinity? i.e. e to the negative infinity, which equals 1 over e to the positive infinity, which is equal to 1 over infinity, and finally which is equal to zero, Respected Sir, please correct me if I am wrong,

[@KW-wr3xv]

He's saying that as a basis so you can solve this equation. Otherwise, you would be dividing by 0 which would mean that the equation is undefined and has no solution.

[@JordHaj]

We say that the limit of it equals zero, which means e^-x goes to zero and(in this particular case) it's never exactly zero There are some cases where the expression has limit zero yet still can equal zero at some values(eg lim [x->oo] sin(x)*e^x = 0 , but sin(x)*e^x equals zero at x=pi*n), but due to the fact that e^x is never ever exactly zero(yes even complex x) we can divide by it and not get extraneous or lose existing sol's Also, if we let x->-oo, the LHS will be much larger that the RHS(lim [x->-oo] (e^x+e^2x)/e^3x = lim[x->-oo] (e^-2x+e^-x) which goes to +oo, which means that e^x+e^2x is bigger than e^3x for very big negative values of x(it's true for all x < ln(phi), in fact), so there cannot be solutions to the equation

[@oenrn]

"Approaches" is not the same as being "equal to". As x approaches - infinity, e^x approaches zero. But since x can never actually BE - infinity, then e^x can never BE zero.

[@Faxbable]

Actually, if infinites _would be_ acceptable solutions (however they cannot be acceptable as one has to solve this equation over real numbers, which doesn't contain infinities), then infinities would be solutions. A better way to show that for -infty is not dividing initial equation by exp(x) but put variable change X := exp(x). As a result, we get X + X^2 = X^3, that may rewrite X^3 - X^2 - X = 0, or X(X^2 - X -1)=0, which has the three solutions : 0, phi and (1-sqrt(5))/2. So that X=exp(x)=0 implies x=-infty. But there is not "algebraic method" to find out x=+infty, which suggests that it is very questionable to accept infinities as possible solutions. (About X = (1-sqrt(5))/2 that is a negative number, this is less questionable to find out complex solutions but quite difficult as complex function EXP is not injective: there are actually an infinity of complex solutions.)

[@crabbydood933]

Just graph both sides and get the rest intersection.

[@alfahentriza5571]

In your example the coefficient exponential (e) is a natural number , What if coefficient in exponential (e) is a less than 1 and greater than 0 ?, for example e^0.2X + e^0.8X = 1. Or what if one of the coefficient is the negatif, for example e^(-0.2)X + e^0.1X = 1 How to solve for X in that situation .. Thanks..

[@robertveith6383]

You wrote those wrong. Those exponents must be inside grouping symbols: e^(0.2X) + e^(0.8X) = 1, etc.

[@redtoxic8701]

Same method, just different substitution. In first equation you can note e^0.2x = t so it becomes t+t^3=1. In the second equation e^0.1x = t so 1/t^2 + t = 1 t^3-t^2+1=0

[@alfahentriza5571]

@@redtoxic8701 Ok .. thanks

[@somedieyoung879]

Why we can't just write x=0

[@SuperYoonHo]

awesome

[@D7mh76]

I just missed the factoring, i did write it as -s(s²︎-s), i forget to put -1 in “s” place. s=e^x

[@rosiekan6308]

by dividing e^x, would we lose a root?

[@kazuoshimahara182]

we would lose e^x=0, but that has no solution, so it’s not a problem

[@louhumphreys9363]

Let y = e^x Then eqn becomes y + y^2 = y^3 y cannot be 0 as ln(0) isn’t a real number Dividing by y and rearranging yields: y^2 - y -1 = 0 Then just solve and find corresponding x values

[@awakely45]

Very nice!

[@fe9666]

Oh goodness, I thought this was going to be a simple one but I underestimated this. Now I have to review this and review on the golden ration. Applied math loves natural logs and e, so thank you for posting this video to challenge mathmeticians.... well, rusty ones like me.

[@reinerwilhelms-tricarico344]

I don't get the argument that you will admit only the solution with the plus sign. If you assume x is complex, you can just do the grind, and get that one solution is x1 = ln((1+sqrt(5))/2), which is real (about 0.48121), and the other is complex, x2 = ln((1-sqrt(5))/2), which then turns out to be x2 = -x1 + i pi. which is equal to -0.48121+ i pi. But for that one needs to do a trick that is only possible with the golden ratio, namely: if m1 = (1+sqrt(5))/2 (golden ratio) and m2=(1-sqrt(5))/2, then it turns out that m2 = -1/m1 . In taking the logarithm, make use of log(-1) = i Pi, and one gets: x2 = log(m2) = log(-1/m1) = log(-1) - log(m1) = -log(m1) + i Pi

[@marcushendriksen8415]

The argument follows from the assumption that we are dealing with the reals, not the complex numbers. Unless otherwise stated, it's usually the default assumption

[@fsbioranucci1083]

👍👍👍

[@elremingtonrodo4678]

💥🇲🇽💥🇲🇽💥🇲🇽💥🇲🇽💥🇲🇽💥Exponenciales Vía Ecuaciones. Cuadraticas. Resolución,......

[@toml3267]

My first guess just by looking 5 seconds at the thumbnail was exp(phi) 🙃

[@Faxbable]

5 extra seconds would have been necessary 😄🤓

[@alejrandom6592]

Fun fact: ln(φ)=arcsinh(1/2)

[@seroujghazarian6343]

solution: ln(phi)

[@الربيع-ذ3د]

اول مره بحياتي احس ان الرياضيات سهله 😁 هههههه اعتقد بالمتوسطه ماخذينها ليش دروسهم اسهل؟!!

[@cosmicvoidtree]

Personally I’m kind of getting tired of the golden ratio. Don’t get me wrong, i think it’s a nice number but it’s used as part of a solution so much now that it’s no longer cool, but feels forced

[@marcushendriksen8415]

I kind of agree. I'd like a video exploring the other metallic ratios

[@GodbornNoven]

x= ln ((1+r5)/2)

[@kvmii7760]

At the end of the video "What do you think?" Going from 😁 to 🙁

[@CharanSaiAnnam]

3/2

[@Faxbable]

Qwerty

[@Walid_Kasab]

I would like to see just calculas channel

[@real_michael]

He has one already

[@oahuhawaii2141]

Yeah, I got e^x*((e^x)^2 - e^x - 1) = 0. e^x = 0 has no solution. (e^x)^2 - e^x - 1 = 0: e^x = (1 +- sqrt(5))/2, but keep only the positive solution. e^x = (sqrt(5)+1)/2 = Phi x = ln(Phi)

[@weonlygoupfromhere7369]

Got this in a min pog

[@aliali-i2z5q]

😁😁👋😁😁 thanks

[@Randomperson_100i]

Shouldn't you factor e^x instead of dividing it so you get e^x(1+e^x-e^2x)=0 and get the solution x=1?

[@ESOMNOFUONLINEMATH]

I am a Maths teacher. Make me your student, plz. Much respect. I teach Maths here too.

[@ahmidouaouladhadj7576]

جميل

[@michaelz6555]

Dividing by e^x results in an omitted degenerate solution of x=-infinity. In fact you should be able to plug that in to show trivially that 0+0=0. Substituting y=e^x and solving for y algebraically as a cubic equation will make that clear. Assuming, of course, you don’t divide everything by y 😆

[@professorpoke]

I know I am getting better when I saw the question and instantly knew that golden ratio is somehow involved.

[@schizoframia4874]

The golden ratio +1= ( golden ratio)^2 which seems related to this equation

[@gordonglenn2089]

I love this video! #GoldenRatioFan

[@pranavamali05]

Thnku

[@78anurag]

I simply took e^x as y and solved the cubic easily

[@isolatedpotato5757]

But why can't you remove out the e on this equation but you can do it with e^(2x)=e^(3x)

[@oenrn]

Because you're not "removing the e", you're applying ln to both sides. ln (e^2x) = 2x ln (e^3x) = 3x but ln (e^x + e^2x) is NOT = x + 2x

[@Physics011]

Can't we solve using natural log at first

[@NotBroihon]

No, the log of the left side is kinda useless. You simply it afterwards but the way shown in the video is easier.

[@雅音蒼]

It's just obvious to replace exp(x) with a variable t, and solve the quadratic equation of t (1 + t = t^2)

[@Skank_and_Gutterboy]

I would strongly caution against teaching this method. I solved this by substituting u=e^x, then solved the trinomial u^3-u^2-u=0. It has 3 distinct roots, u = (0, (1+sqrt(5))/2, (1-sqrt(5))/2), where you use back-substitution to get a solution in terms of x. Your method works here coincidentally because zero can be eliminated as a solution since ln(0) is undefined. You very quickly glossed over why your method would work for this solution. I don't think a student seeing this for the first time is going to catch why you can do it for this special case but not others. I think the teaching method here is going to inadvertently teach bad practice for the simple reason that, in the world of polynomials, x^3-x^2-x=0 and x^2-x-1=0 are NOT equivalent and a graph of the two functions readily shows this. A much better way to go is to factor out a u so that you have u(u^2-u-1)=0, then show why u=0 can be eliminated as a solution when using back-substitution to get an answer in terms of x (you probably want to show that the equation e^x=0 can be converted to ln(e^x)=x=ln(0), which is a no-go because ln(0) is undefined; just saying e^x cannot be zero is less straightforward, less likely to be remembered, and doesn't really hit the crux of the problem).

[@Mmmm1ch43l]

nah, you're wrong it's perfectly fine to divide the whole equation by a non-zero factor and he does in fact mention that e^x is always non-zero, so I see absolutely no problem with this approach ("Your method works here coincidentally because zero can be eliminated as a solution since ln(0) is undefined." lol ok dude, that's equivalent to what he said)

[@Skank_and_Gutterboy]

@@Mmmm1ch43l What I said is that this approach teaches bad habits to students. It's a shortcut, and a very poor one.

[@Mmmm1ch43l]

@@Skank_and_Gutterboy yeah I understand what you said, but I disagree it's a perfectly legitimate and sensible "shortcut" and is explained adequately for example: your point that "x^3-x^2-x=0 and x^2-x-1=0 are NOT equivalent" is extremely overblown sure, these polynomials *look* different, but they have the exact same solution set (apart from x=0 of course) so reducing the former problem to the latter is 100% fine

[@canaDavid1]

ln(phi)

[@bublik20]

Мощность )

[@davido3026]

e**×(1+e**x)=e**3× Extracting ROOT X, We have e**x=e**2-1 LN (e**x) = LN(e**2-1) X= LN (e**2 -1)

[@axildavei.esplana-11stemc74]

Who is legit? Blackpenredpen or just algebra?