A Nice Exponential Equation, x²=2ˣ

[@SyberMath]

Hello everyone! I forgot to mention one of the solutions. 😂😂😂 You can see it clearly when you graph x^2 and 2^x together. I restricted the domain and then totally forgot about the case x

[@tambuwalmathsclass]

👍👍

[@ahmadmazbouh]

I would like to see a video about it talking about Lambert function with some details 🙃

[@Qermaq]

@@ahmadmazbouh Black pen red pen does a lot of those BTW.

[@Qermaq]

Hey maybe do a follow-up video where we learn how to find the third solution, which as I do not get the Lambert W function yet I shall call -23/30 plus one smidge.

[@ahmadmazbouh]

@@Qermaq i prefer to see em here cuz bprp's accent is a bit hard for me as a non English-speaker, I'll check it out tho, thanks

[@moeberry8226]

To start ln(x^2) is NOT equal to 2ln(x) but rather 2ln(|x|) . Also by using the absolute value and applying the Lambert W function you will get all 3 solutions.

[@haroldgamarra7175]

there are no 3 solutions

[@moeberry8226]

@@haroldgamarra7175 yes there is graph them and you will see an intersection point at x= to a negative value. As well as x=2 and 4

[@akikawa2540]

@@moeberry8226 I think so too. You are correct. Graphics of 2 functions are crossed 3 times. Solutions of x are "2", "4" and the negative number x=-a in which (a^2)*(2^a)=1. I cannot solve the third value, but it must be between -1 and -1/2.

[@moeberry8226]

@@akikawa2540 it’s not hard actually to solve it if you know the Lambert W function. You can solve this analytically and get the exact value. First take natural logarithm on both sides and you obtain 2ln|x|=xln(2) Note: there is an absolute Value symbol within the natural lot because x can be negative so you will divide this into 2 cases. Case 1 x>0 and Case 2 x

[@akikawa2540]

@@moeberry8226 Thank you so much. I'm not aware of the Lambert W function. I'm much moved to hear the negative solution can be solved analytically. Very interesting. I'll study it.

[@JohnRandomness105]

I saw this before, I think on "Mind Your Decisions". x = 2 is the most obvious, and I recall missing x = 4. For positive x, x² increases faster from zero than 2^x increases from one, and they cross. But 2^x being exponential, it crosses x² again and shoots up beyond. These are x = 2 and x = 4. When x decreases from zero, x² goes up while 2^x continues down from one. Again, they have to cross, but only once. Finding the negative x might need a numerical method such as Newton's method.

[@SmallSpoonBrigade]

X=2 is the one I got immediately from a bit of algebra. Right before I clicked play, I realized that 4 would also work. Mostly because you can write it as 2^2 and get 2^4 like that.

[@ffggddss]

Don't know whether it's on MYD, but bprp (blackpenredpen) worked this problem, and got all 3 solutions. The negative one (obtainable by using the Lambert W function) happens to be very close to -23/30. Fred

[@mathlove7474]

hello😍😍, see my math videos please 💐

[@abhirupkundu2778]

everyone knows that, not a big deal@@SmallSpoonBrigade

[@vitaliypolyakov8805]

There is also a third solution to this equation. This is clear from the graphs y = x^2 and y = exp(x). There is an intersection point at x < 0, but this solution is not expressed in terms of logarithms, it is x ~ -0.767

[@almanduku9043]

you're right {{x==-((Rationalize(766.664696))*(10^-3))},{x==2},{x==4}}

[@royfeer8651]

SyberMath, your videos are pretty good and helpful. I've been watching your channel for about 3 months and am gradually becoming skilled and insightful when solving math problems. Three months ago, I couldn't solve a question much easier than this one. But now, I solve this problem within 5 minutes! I am so excited and will keep watching your videos.

[@taimao2]

Interestingly I'm getting the answer x=1/supersquareroot(2^(-1/2)). I first was using the Lambert W function and I was like hmm that looks like it can be written using super square root. Since, the identity for super square root is: supersquareroot (z)=e^W(ln(z))

[@saravanann8246]

In general, the interval{1,e} mapped by y^x=x^y to the interval{e, infinity}. It is a one to one, onto map. And between 0 and1 , (0,1)mapped to(0,1) the numbers with its self

[@sanjuiyengar3587]

to find negative solutions write x=-n where n is positive we can then write n^2=1/2^n this implies a range of n is 0=-1 here try 1/2 on the rhs this gives value of 0.707 first iterating this value eventually gives lhs=rhs at n=0.766 or x=-0.766

[@rcnayak_58]

We can find out x in a more simplified way. We can rewrite our problem x^2 = (2^(x/2))^2. So that x^2 - (2^(x/2))^2 = 0. This is similar as a^2 - b^2 = 0 which is (x - 2^(x/2))(x + 2^(x/2)) = 0 (a product of 2 terms). Let us solve the first term, i.e., (x - 2^(x/2)= 0. From this x = 2^ (x/2), raising power both side by 1/x, we have x^(1/x) = 2 ^ (1/2), this gives one value of x = 2 ... (i) Again 2 ^ (1/2) can be written as (√4) ^ (1/2) = ((4)^(1/2))^(1/2) = 4^(1/4) which gives the 2nd value of x = 4 .... (ii). Now the second term which is x + 2^(x/2)= 0. So that x = - 2^(x/2). Raising power both side by 1/x, we have x^(1/x) = - (2 ^ (1/2)), this gives no positive real value solution for x because while the right side is always - ve , x^(1/x) is positive for any values of x > 0. Solving this using Lambart W function (productlog function) we get another value of x = - 0. 7667 ... (iii). Therefore, it has three values of x, i.e., 2, 4 and -0.7667 (approx.)

[@AinulMustafaHD]

nice

[@bitterlemonboy]

nerd

[@deltalima6703]

Thx, I found 2 and -0.76 by inspection and was looking forward to see his plan. It was a disapointment to see him miss it.

[@dmtri1974]

Nice, but still you have to solve rigorously x^(1/x) = 2 ^ (1/2), which is the same as the video shows....

[@alaincharoy953]

A good approximation of the third value is -0.766666

[@Anonim29122]

Broo, just draw y=2^x and y=x^2 and you’ll see there are 3 solutions!!!

[@astronull8589]

omg ty so much, I not thinking on that

[@dwm1943]

That was really good. I saw x = 2 immediately, but never even considered x = 4, which is 'obvious' when you're shown it. However, I plotted (mentally) y = x^2 and y = 2^x, and it was clear that there was another intersection somewhere in the negative x values and y < 1. So I think I will claim a draw in this match, 2-2.

[@BartBuzz]

There is a 3rd solution. Graphing y = x^2 - 2^x shows three x values where y = 0. Namely, x = 2, 4, and -0.767. x = -0.767 satifies x^2 = 2^x

[@KoffiBossoMath]

Hello

[@David_Sablatura]

(ln(abs(x))/x)=((ln2)/2) and you get all three solutions. From this you can conclude that the third solution must be between -1 and 0.

[@thomasselerland4354]

If you plot the graph of x^2 and 2^x you will see there are 3 intersections 2,4 and -0.766

[@sanesanyo]

Actually it's not so obvious to see this. Only if you use some program to plot these graphs. It's far easier to see this if you plot ln|x|/x

[@johnsimmons6624]

For x>4, 2^x > x^2. For 2

[@almanduku9043]

{{x==-((Rationalize(766.664696))*(10^-3))},{x==2},{x==4}}

[@DetCoAnimeFan]

2 and 4 are clear, but the third solution is where we miss out.

[@rafadarkside9029]

There is THIRD solution for x in (-1;0) Because there is no assamption about x, it should be written at 1:00 : 2*ln|x| =x*ln2 - there must be used an absolute value sign. And then there are 2 cases to consider (x0)

[@kinshuksinghania4289]

actually there are 3 solutions to x² = 2^x . You have already shown x=2 and x=4, however there is also one negative solution. If we consider functions y=x² and y=2^x, their graphs intersect for exactly one negative value of x in addition to the above stated two values. y=x² is a strictly decreasing function for negative values of x and gives 0 at x=0 and blows up to positive infinity as x approaches negative infinity whereas y=2^x is a strictly increasing function and gives 1 at x=0 and approaches 0 as x approaches negative infinity.

[@SyberMath]

That's right!

[@TheMrGoncharov]

That's right, the equation equalises if you put x=-0.766665

[@omsingh7683]

Hey man how y=x2 is stricly decreasing for negative value of x

[@sunacchi3640]

@@omsingh7683 x² is +infinity when x approaches negative infinity and it's 0 when x is equal to 0

[@melstone8590]

ok so, knowing that X = 2, 4 and .766665. Where's the real life application to this knowledge? Serious question

[@ilyashick3178]

Note 2*x in not equal 0 or x is (-~,0)&(0,+~).Domain can be Positive or Negative. Solution: takes log on base x in place; Step1: 2*log x=log 2+log x. Step2: 2*log x-lox=log x ,Step 3: log x= log2 means x=2. Verification 2^2=2*2 Means x-s equal 2 and in this case domain is positive only.

[@rossetto23]

I got the third negative solution by expanding the exponential by Taylor around -1/2. My approximation was -0.761.

[@gregwochlik9233]

Disagree. There are three solutions. Graph the exponential and quadratic and they intersect twice when x > 0. There is also an intersection between -1 < x < 0. I graphed it in excel using the "scatter plot" with two functions.

[@tom-kz9pb]

Saw the "2" solution at a glance, but there is a big difference between "seeing" a solution and actually solving for it, which is the real point, and also sometimes giving a more complete answer (like also "4").

[@Guderian0617]

Just by using common sense, 2*x tends towards 0 when x is negative, and x^2 tends towards infinity when x is negative. Since x^2 dips below 2^x at x=0, there must be another crossing point at a negative x value, of which I cannot be bothered to work out

[@justrandomthings8158]

There is

[@justinmplayz8809]

You right but I don't think that's "common sense" anymore

[@justrandomthings8158]

@daniel halachev what’s your point

[@levskomorovsky1762]

The function y = 2^x - x^2 is negative for negative values of x up to the value x = -1, and positive for the value x = 0. Hence there is another value x between x = -1 and x = 0, where y = 0

[@lucasvillani5484]

I was just about to study increasing and decreasing functions using derivatives, and this video is definitely a better lesson than whatever I will get lmao.

[@sanjuiyengar3587]

To find cap on integer solutions By inspection and symmetry 2 is a solution Now x is even integer write x=2n This means 4n^2=2^2n=4^n Once again n has to be even Put n=2k 16k^2=16^k This is true for k=1 and so n=2 and x=4

[@seroujghazarian6343]

2 and 4 aren't the only solutions, as a matter of fact, there are infinitely many complex solutions, of the form -2productlog(n,±ln(2)/2)/ln(2)

[@focus7477]

No there are precisely 3 solution. I have checked that on graph.

[@seroujghazarian6343]

@@focus7477 3 real solutions. Infinitely many complex solutions.

[@focus7477]

@@seroujghazarian6343 Oh. Right

[@TheTinkywinky3]

@@seroujghazarian6343 for example?

[@kryukovaleksey3940]

The equation has 3 solutions. One can plot graphs of functions 2^x and x^2 and the lines have 3 intersection points. And then we can use numerical method, for example.

[@WahranRai]

If you plot / draw the curve f(x) = x^2 - 2^x, you will see that the curve intersects the x-axis at 3 points : x1= -0.766664695962123, x2 = 2; x3= 4 Many online tools for plotting : Maxima, Geogebra...

[@SyberMath]

Read the pinned comment

[@WahranRai]

@@SyberMath I wrote my comment without reading your pinned comment. You are talking about graph of x^2 and 2^x : it is not clear, it is separatly, the sum or the difference ? I am talking about classical solution how to solve f(x) = g(x) which equivalent to solve f(x)-g(x) = 0 then find intersection with x axis of the function f(x) = x^2 - 2^x (one graph)

[@ThePiMan0903]

Nice video SyberMath!

[@SyberMath]

Glad you think so!

[@michaelsadovsky936]

Wrong. You've missed one more root determined by the constraint x^2 = \frac{1}{2^x}. x here is negative, obviously. However, it is irrational and I am too lazy to calculate it. Nonetheless, this equation has THREE roots. You've missed the third one when changed the original equation for the analysis of the function \frac{\ln x}{x}; that latter is defined for positive x's only. So, you've lost the negative root. To see that just plot two graphs: y_1 = x^2 and y_2 = 2^x, and you see all three intersections of the curves.

[@SyberMath]

That's right

[@vcv6560]

I had this one shared with me 40 yrs ago by my college era GF, I approached the 3rd root by successive approximation using a programmable calculator (TI-57) after bounding the solution by graphing it, seeing somewhere between -1 and 0 being the answer. The irrational -0.766664696 rings in my head even today. While we didn't last I did later get a graduate degree in CS. Cheers

[@georiashang1120]

Well just get the 2 and 4 by guess,also knowing that there is a answer when x

[@tranqulity39]

By taking natural logarithm, you confined the positive only solution. Actually, the full solution is x=-2W(-0.5*ln2)/ln2, where W(.) is the Lambert W function.

[@XJWill1]

Actually, that only gives x = 2 and x = 4. It is missing the third real solution. You need to add x = - W(y) / y , where y = ln(2) / 2 if you want to include the negative real value for x. So the complete set of real solutions would be x = - W(-y) / y or x = - W(y) / y , where y = ln(2) / 2 Note that the first one gives two values for x since W(y) is multi-valued for y < 0

[@tranqulity39]

@@XJWill1 W0(.) and W-1(.) represent different branches.

[@XJWill1]

@@tranqulity39 I am aware. It does not change the fact that your comment is incorrect and missing the third real solution.

[@KhairulamriBahari]

2² = 2² Therefore, x=2 4² = 2⁴ Therefore, x=4 We are getting two values of x, 2 and 4...

[@TheMrGoncharov]

The equation has 3rd solution, which is x=-0.766665, I've found this solution by building the graph before watching the video, it was interesting how 3rd solution could be found, but video doesn't answer

[@lumri2002]

For practical solution, a) Why not just solve by letting x=2 as first possible value which will render the main equation of the problem to result 4=4? b) And then why not solve further by letting x=4 as other possible value which will now render the main equation of the problem to result 16=16? Using trial numbers in this problem are easier and faster to arrive at the correct answers.

[@Catilu]

By applying ln you are only considering the positive solutions, so obviously what you get is 2 and 4. But this way you're missing possible negative solutions, and there is one, which you can see by graphing x² and 2^x together

[@iemtruongvan7316]

Consider the function g(x)=2^x-x^2, this function is continuous on R and has g(-1)=-½; g(0)=1 so g(x) =0 has at least 1 solution in the range (-1;0). when x>0; Taking the neper logarithm on both sides, we have xln2=2lnx (lnx)/x=(ln2)/2 Consider f(x)= (lnx)/x-(ln2)/2. with x>0 f'(x)=(1-lnx)/x^2. f'(x)=0x=e. Setting up a table of variations of f(x), we can deduce that f(x)=0 has at most 2 solutions. also with x=2; x=4 are satisfied. so x=2; x=4 is the solution of 2^x=x^2.

[@peterpan408]

Sometimes it helps to draw a graph before going down a mathematical rabbit hole. Successive approximation also eliminates any need for calculus.

[@kharnakcrux2650]

for situations like these... use GrafEQ i've used it for years... it graphs difficult relations. without those limitations. second... get VERY familiar with the Lambert W function. it bridges complex domains with algebraic. you're involving different kinds of symmetry groups, you need more than the usual fare.

[@carloshuertas4734]

Another great explanation, SyberMath! I figured the numerical values of x in 5 seconds. To me, figuring the x values out in this math problem was easy. Thanks a lot!

[@sergiokorochinsky49]

What about the solution x=-0.766... Did you also get that solution just by looking at the equation? :-)

[@carloshuertas4734]

I got 2 and 4.

[@SyberMath]

Great job!

[@michaelsadovsky936]

@@carloshuertas4734 However, it is the partial solution. Not the general one. So, you've sold in 5 secs another problem: solve in natural numbers the equation XXX

[@tg12367]

ln(x^2)=2ln(|x|) (not equal to 2ln(x)) . So, there are 2 cases : case 1 x>0 and case 2 x0 X=2 and X=4 1 negative solution when x

[@eduardvalentin830]

who else remembers bprp video about this equation?:)

[@hungminh0212]

Me

[@angiulillo1968]

Me.

[@YURI-u3u]

Amazing 🤩

[@Mathskylive]

Một phương trình thú vị, giải bởi nhẩm nghiệm đạo hàm và bảng biến thiên. Cảm ơn.

[@MrMousley]

X^2 = 2^X X = 2 was the most obvious answer before I even clicked on X^2 = 4 and 2^X = 4 and then quickly realised that 4 works as well ! X^4 = 16 and 2^X = 16 and all of this without having to 'work out' anything

[@alexestefan7521]

When taking the ln, you restricted the domain to x>0. There is another negative solution you did not find

[@luaiderar6600]

your pfp though...

[@chandan.m84]

X^2=2^× Log on both sides 2×logx=x×log2 Log2=0.301 2logx=x×0.301 Logx=x×0.1505 Differtiate on both sides X=1/0.1505 X=6.6622251832

[@sergiokorochinsky49]

10:12 ...nope, this equation has 3 solutions.

[@SyberMath]

Yes

[@BacMN]

THANKUUIII

[@dariosilva85]

There is also a negative solution ~ -0.76666

[@Perririri]

…és meg a negyedik komplex szám!

[@lyrimetacurl0]

Apparently there are loads of complex ones (infinity of them probably) so let's stop with the 3 real ones.

[@panos1435]

Hey syber! If x

[@KoffiBossoMath]

Great video and well explained. Thanks for sharing

[@SyberMath]

Glad it was helpful!

[@henrysiegertsz8204]

Brilliant explanation!

[@SyberMath]

Thank you! 🥰

[@田舎の爺さん]

別解　(1) x = 2^pとおくと、与式は、(2^p)^2 = ( 2)^(2)^p. (2). 指数の関係から、2p = 2^p これを解くと、p=1 p＝2 をえる。(3) p＝1のとき、x=2^1=2. P=2のとき、x=2^2=4が解になります。

[@jorgepinonesjauch8023]

La ecuación tiene 3 soluciones dos son triviales x=2 y x=4, la otra es una solución que puede obtenerse por la función de L'ambert y da un resultado cercano a -0,76666.... con otros decimales más

[@iemtruongvan7316]

Phương trình còn có 1 nghiệm nữa nằm trong khoảng (-1 ; 0) nếu dùng hàm số liên tục f(x)= x^2 -2^x . Ngoài ra dùng hàm thì chứng minh được khi x>0 thì phương trình f(x)=0 có nhiều nhất 2 nghiệm bằng cách tính đạo hàm và dùng bảng biến thiên của hàm số f(x)=2lnx-xln2 (do x>0) nên lấy lôgarit neper cả 2 vế). Khi đó chỉ ra 2 nghiệm là: x=2; x=4.

[@Abuushaymaatz]

What if you could find the horizontal and vertical asymptotes and use them to sketch the graph?, Since the function is not continues as x tends to zero there's no Y-intercept, but the x-intercept is zero ie the graph will test to pass through the origin at (0,0), then to know if it's increasing or decreasing we will simply use iteration method to plot those points starting from x=1,2,3,4,4,5... And so on I really see also this can be used to show the nature of this graph, for someone who is not familiar with calculus. Generally you did well, congrats

[@azmabatoolasif67]

Loved it thoroughly. Thank u

[@cyruslavian7098]

Excellent,-I-learned-a-lot.Specially-you-are-mixing-it-with-calculus.Question:Re-Y=X^X,it-is-1,at-both-X=0,and-X=1,the-minimum-Y-at-(1/e),(thanks-to-Desmos-Scientific-Calculator),why?What-is-the-name-of-this-curve?,and-what-is-the-value-of-Y-when-X-is-Negative?can-you-talk-about-that?

[@bitterlemonboy]

Is your space key broken?

[@extralives8693]

I hope I am just a kid but since it is said x^2=2^x then rule a^m=a^n Therefore m=n because if both the base are same (or said to be) then x=2

[@dushyanthabandarapalipana5492]

Thanks!

[@SyberMath]

You bet!

[@ernanifantinatti7043]

Pretty cool!

[@jwang734]

y = x^2, and y = 2^x has 2 intersections. (2, 2) and (4, 16) are the 2 intersection points.

[@iwd1705]

@1:40: How is it "more" obvious" at this point than at the start? It's just not. x = 2 is the most obvious solution from the start. And there must be a third solution for some negative x-value (whatever size it has got), because 2^0 = 1 but 0² = 0. 2^x will continuously decrease with growing negative values of x. x² will start to continuously increase with growing negative values of x. So they WILL have to meet somewhere between x = 0 and x = -1. ( (-1)² = 1 and 2^(-1) = 0,5; so the exponential is below the quadratic again and will remain there forever going to the left).

[@nasrullahhusnan2289]

Hallo SyberMath, something is missef when you take the logarithm of x²=2^x. Both sides are greater or equal to 0 for any real x. But x²=(-x)². Taking the logarithm gives log(-x)²=2log|x|. Simplifying yields 2log|x|=xlog2 For x>0 --> 2logx=xlog2 For x 2log(-x)=xlog2 Using W Lambert's function we will get 3 solutions: x={2, 4, -.766665}

[@paulbytheriver4976]

thanks, that was fun 😊

[@SyberMath]

Glad you enjoyed it!

[@YTSparty]

Just plug/chug get 2/4 fast. Notice that after X=4, 2^x blows past x^2. The irony is you notice how log works. You realize X*X = 2*2*2*....2. So X must be a multiple of 2. X=2 is kind of obvious, 2*2 = 2*2. X=4 or 4*4 = 2*2*2*2 = 2*2*2*2. So X must be equal to the number of 2's in X*X or twice the numbers of 2s in X. So X = 2*LOG2(X). Using numbers with multiples of 2, 2,4,8,16,32, etc. Gets 2 and 4. And you notice again, over 4, X increases faster than LOG2(X).

[@kriskris4534]

There is a third solution approximately -0.7667, or exactly -(2*lambertw(0, log(2)/2)/log2 Where lambertw is the Lambert W function.

[@tonyscott1658]

Oh please: just plug x=2 into x^2=2^x and you get 2^2=2^2=4 and so by inspection x=2 satisfies the equation. Since x^2 = ...m you can be sure there might be an additional root.

[@suk-younsuh7322]

How about graphing y=ln(x) and y=x. It is easy to see that they are interesting only two points w/o resorting differentiation.

[@fernandoangulo1960]

Fantastic, i used Geogebra to make graphical solution too, Thanks for sharing & teaching 👍🏻

[@SyberMath]

Glad it was helpful!

[@raatkedhaibaje4765]

x=4 (hit and trial method) A beautiful and easy method btw

[@raghugottapu_79]

x=4 is the perfect and only solution

[@guidogiuliani8096]

What program are you using?

[@georgelaing2578]

There is another solution between -1 and 0. If you graph each expression, you can see the intersection.

[@Alex-qk5ei]

if you enter on wolframalpha (-lambertW(ln(2)/2)*2/ln(2)) you'll find the negative solution

[@markwalter1050]

Use any base you want -- as long as it's a constant: 2*log_b(x) = x*log_b(2) [2*log_b(x)]/2 = [x*log_b(2)]/2 log_b(x) = [x*log_b(2)]/2 log_b(x) = x*[log_b(2)/2] log_b(x)/x = (x*[log_b(2)/2])/x log_b(x)/x = log_b(2)/2 => x = 2 log_b(x)/x = 2*log_b(2)/2*2 log_b(x)/x = log_b(2^2)/4 log_b(x)/x = log_b(4)/4 => x = 4

[@seroujghazarian6343]

ln is preferable to apply lambert W

[@omarchiheb8641]

Please, can you tell me which app you are working with?

[@SyberMath]

Sure. Notability

[@miroslavstevic2036]

I remember showing this to my roommate. He was studying to become an architect. His solution was "True". I said how did you get that? He simply crossed x and exponent X with a pencil and said "it's true -> 2 equals 2". I immediatelly called him Gauss, the nick got traction and everybody in the campus started calling him Gauss. It is his nickname to this day.

[@SyberMath]

😄

[@mandaparajosue]

There's only one more (and negative) solution. But studying the function lnx/x is good to know better the problem, like to ensure that there is only one solution for e^x = e^x.

[@virabhadra2]

The equation e^x = e^x for sure has one solution, but it's the range: [-inf, +inf]

[@mandaparajosue]

@@virabhadra2 Well done! 👍

[@gantajayashankarjayashanka9660]

Thank you so much brother 💞

[@SyberMath]

You're welcome!!!

[@Kumurajiva]

i love your voice!

[@SyberMath]

Thank you! 💖

[@chudson5901]

Does the first step taking Ln eliminate the negative roots ? Are there any roots

[@DrLiangMath]

Very nice question and excellent explanation! 👍

[@SyberMath]

Glad you think so! 💖

[@Mathematics21st]

Wow great job syber

[@SyberMath]

Thank you!

[@JagadeshRaoThalur]

Can you also please tell me where do you find these kind of questions and how do you arrive at the answer before solving it?

[@AKBARCLASSES]

The great explanation. Love from 🇮🇳

[@SyberMath]

Thank you! 💕

[@AKBARCLASSES]

@@SyberMath I’ve also recently made a video of this problem but the audio is in Hindi. I’ve discussed in detail for the negative solution. Please visit AKBAR CLASSES KZbin Channel if possible.

[@unknownworld8238]

wasn't simple to: (I) 2*2*....*2*2=2^x (II) x*x=x^2 From (I) we see that 2 is the base From (II) we can see that 2 is the exponent. So it has to do with power of two and perfect square numbers. What are the powers of 2? 2^0=1 2^1=2 2^2=4 2^3=8 2^4= (2^2)^2 = 4^2 = 16 2^5=32 2^6=64 = (2^2)^3 2^7=128 2^8=256 = (2^2)^4 2^9=512 2^10=1024 = (2^2)^5 .... From here we can clearly see that we can write 2^4 as 4^2 therefore x=4

[@Sumitchaudhary5525]

By using hit and trial method x=4 i.e :- 4²=2⁴ 16=16 Again by ht x=2 2²=2²

[@carlbrenninkmeijer8925]

thanks!!

[@SyberMath]

You're welcome!

[@bobbybannerjee5156]

Excellent working out. May I ask what software (app) were you using to write with your electronic (stylus) pen?

[@SyberMath]

Sure. I make my thumbnails using googledocs and canva. I use an iPad, an Apple Pencil and the Notability App. I record the videos using screen recording.

[@HoppiHopp]

I’m totally lost. Why do you what to reach which solution?

[@patrick-8068]

x(p2)​=2p(x) log​(x)​ 2=log​ x 2log(x) 2= 2log x 4x(2)=4x 2x=x​ Answer Equ​ poly.

[@irinamladova]

1:00 2ln|x|=xln2 Abs! Good video)

[@marklevin3236]

For every x >1 except for e there exist a unique y not equal x such as x^y=y^x

[@jiezhou3296]

I am expecting the 3rd solution and you proved there only 2!

[@doubledee9675]

I know next to nothing about mathematics, but in under 10 seconds got x=2 and x=4 - did not think of negatives but they were just being invented when I went to school.

[@skxre9881]

i love those videos lol

[@georgesadler7830]

Thank you for the video.

[@SyberMath]

You're welcome