Entrance examination to Stanford University
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- Опубліковано 11 січ 2025
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@higher_mathematics
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1. Stanford does not have an entrance exam. 2. Stephen Hawking did not go to Stanford, nor did he teach there. 3. If you're going to use that much machinery for such an obvious question, at least prove that there are no other solutions.
He made the nearly exact same video with the same thumbnail with Harvard instead of Stanford.
Easily one of the worst math channels on UA-cam.
Clickbait should be a capital offense
How can a university not have an entrance exam?
Stanford and Harvard both have entrance exams which consist of one question as seen above.
@@haggismcbaggis9485 🤣😂
Sixteen lines of maths too many. It was blindingly obvious the answer was x=4 three seconds after looking at the question.
There's a reason why you are always supposed to show how you arrived at the answer...
@@fredpagniello3267
We could just notice that 2^x + x is an ever increasing function and clearly has no solution for x < 0.
Next notice that when x = 0, then 2^x + x = 1. This fact coupled with the fact that 2^x + x is an ever increasing function means that there is one and only one real x, such that 2^x + x = 20.
By observation, we see that 2⁴ + 4 = 20.
So x = 4 is the one and only real solution.
@@davidbrisbane7206Your explanation can be reduced to a very simple "abracadabra here's my answer." Yours works for simple problems with an obvious answer only. The purpose of working through a simple problem is to develop the acuity and facility to tackle those of greater difficulty. Essentially, you are building up skill from the inside out, not the reverse. The latter, by the way, is a standard for the approach to languages; the student learns the equivalent in their native tongue and therefore thinks in it and translates into the target language, exactly the opposite from how the human brain operates (that is, direct form function relationship, association of the basic structure that allows for acquiring more complex ones).
No credit without showing work
@@abefroman7393 How do you show the working if there is none to be done?
After the stage where you multiply by 2^20, split the right side into 2^4x2^16. This becomes 16x2^16.
Both sides are of the form Ax2^A=Bx2^B so A=B can be equated giving 20-X=16 at a much earlier stage and no Lambert W function required!
Friggin' brilliant! Thank you.
Hey what's the logic with which you can say that A = B from the equation you provided.
I think this very nice idea can be clarified additionally for example this way.
Consider equality Ax2^A=Bx2^B.
We have value B as positive (equals to 16). So if A is negative, then equality can't be acheived (Ax2xA is negative, Bx2^2B is positive). Suppose both A and B is positive. In this case if they are different the equality is also cannot be true because the function X*2^X is monotonic for positive X (you can take first derivative and check its sign).
wow
@DukeofEarl1961 sure, but that reduces the opportunity for adsense.
You can guess 4, and since 2^x + x is a strictly increasing function, it has at most one solution.
wgat about complex numbers?
You can guess because x happens to be a integer
If it were = 21 it wouldn't work
@@pierrekilgoretrout3143 2^x isn't well defined if x isn't real, it has infinite values (not every number, but infinite numbers)
Assumption: X must be a real number. 2^x+x=20 2^x=20-x analysis: 2^x is strictly increasing, 20-x is strictly decreasing. Conclusion: there can only be one solution. Now you can use newtons approximation methode or just start with small whole numbers as that is always a good start. one, nope, two, getting closer, three, almost there, four. OHH it's four how convinient.
@@dd-di3mzIt can't be zero then X it's a value between 19,9999999999... 0,000000000...1
Assuming that the examination must be completed in a given time period, it is possible that the point of this question is to see the intuitively obvious solution = 4. I remember (67 years ago) a NY State Science & Engineering Scholarship exam where there was a multiple choice for the product of two huge integers. I noticed that the product ended in 6 and only one set of integers had last digits whose product ended in 6. Solution in a few seconds and onto the next problem!
It’s Stanford university. Think they’d care about the time you’re given?
@@Ron_DeForestNope, they only care how much money your old man gives them, take as many semesters as you like 💵💸💸💸
@@joecaves6235
That may be true of the Admissions office, Joe. But no graduate student TA gives a little brown squishy one about what anybody's old man gives to get his name on a building.
Consider yourself one more data point for "Cynicism is the wisdom of the dolt."
sure, I think the same
This is a really important point you’ve made,@ftgsoftware8119. You cannot be spending this kind of time on a single question. Also note that he’s using Lambert’s W function, which usually students don’t get into until after like numerical analysis or combinatorics or something, not high school math. It would also be really easy to accidentally infer from this video from when he initially mentions it that you could just take any n*(e^n) and just turn it into n. This is not true; you need to have a W function on both sides. Yes he does it properly, but when he mentions W functions initially it’s like he gives two different examples instead of the fact that you need to have a equivalency. Besides, it’s way faster to graph it and you can make a strong educated guess that that’s what they’re looking for because of the simple answer. There are lots of ways to solve problems and prove them, but you don’t want to spend so much valuable test time running through this. That being said, an excellent reminder about Lambert and if you find yourself in the situation where you need it then by all means do so.
I want to thank the author of this video for making this. This was a useful and clear introduction to how repeated use of the Lambert Function W, which I’d never heard of before, can help solve a problem of A^x -x = B.
25 years ago while in my engineering math classes I sounded like a lot of these commenters, that I’d never need all this math wiz-bang hand-waving stuff, that profs were making something simple hard. I certainly heard that from working engineers that they never used the math they learned in school. That may explain a lot of problems we now have in building things that our fathers and grandfather seem to do with much less trouble.
Over the years I’ve learned that the tools in a contrived problem can mask their usefulness in solving a badass problem later down the line. In the coding work I do, I’m using more of what I learned nearly three decades ago than ever before and am actually going back and relearning what I’ve forgotten in multi variable calculus, DiffEq, Fourier and Laplace Transforms.
So, again, thanks!
There are no rewards in life for using the Lambert function to solve a trivial problem of arithmetic. There is absolutely nothing wrong with carefully enumerating possible solutions to find a solution, if that can be done quickly and the uniqueness of the result can be proven.
@@rickdesper I disagree.
U cant iterate easily every similar problem. Imagine there is 21 instead of 20, and 1.5 instead of 2 to the power of x. What then? Ofc we can iterate but its not trivial anymore. Tools provides the answer for every problem regardless the numbers.
@@rickdesper
Until you come across a non-trivial problem and YOUR blade is now rusty from lack of use.
Fourier and Laplace, are essential for DSPs.
Though they never breakdown a problem this way.
This problem only serves to show understanding of math and lovely of grasp of mathematical rules and formulation, much of which mostly is used in a much simpler way when it comes to real world application.
Because the better you understand a problem the better you can derive formulations and use the correct rules for it.
Also, in math, understanding number cycles is essential in knowing how to solve problems meaningfully.
But I agree with you, as I have not used Fourier and Laplace since 2.549 decades ago.
I try 1, then, 2, then 3, then 4 and I find that 4 is the solution. It takes 10 seconds
Same😂😂😂🎉
Didn't need to go this far
I did what you did. Seemed too easy. I was expecting the video was going to say it was a trick question.
Painful to watch. Distortedly complex
Same. The only thing I can think of to make this make sense is that the question must say, "Using the Lambert W function, show that....". This makes no sense otherwise...
As an engineer, all i did was plot a graph of 20-x and 2^x. Only one intersection shows the existence of only one solution.
After that, a middle school kid can guess the answer is 4.
That is to say, if getting to the answer is your ultimate goal instead of proving it rigorously.
Yes, all those calculations are useless after Microsoft Excel
very god idea
The approach you take could count as a rigorous proof. Or, at least, it could lead to one.
Cómo ingeniero te puede decir que la solución gráfica no es precis
if you wanted to prove it rigorously you just have to show that the derivatives will never change signs, this proves only one intersection if the signs are opposite each other to begin with.
2ˣ = 20 - x
(1/2)⁻ˣ = 20 - x
(1/2)²⁰(1/2)⁻ˣ = (1/2)²⁰(20 - x)
(1/2)²⁰⁻ˣ = (1/2)²⁰(20 - x)
(20 - x)2²⁰⁻ˣ = 2²⁰ = 2⁴2¹⁶ = (16)2¹⁶
20 - x = 16 => *x = 4*
a faster way
2ˣ = 20 - x
2⁻ˣ(20 - x) = 1
(20 - x)2²⁰⁻ˣ = 2²⁰ = (16)2¹⁶
hi...exellent❤
Exactly
Can you justify that A*(2^A) = B*(2^B) implies A=B?
That is why the person in this video clip get paid more , and you don’t get paid.
@@rhodexlx1659 implies A = B as a possible solution
Thanks, never heard of Lambert W function, so had to dig into. Learned something new, always nice!
Well. Without wanting to belittle W for LAMBERT, I did it head on. First: "x" must be even, because the power of 2 will always be even and, consequently, the value of "x" too, since the sum of the two is equal to 20, which is even. So "x " € {0, 2, 4, 6....}. Therefore, when we substitute the options from this set, on the third attempt, we find the answer.
1)knowing the answer is 4, we can always quickly architect an identity to reason the answer (trying to beat the author with time).
2)In actual fact, the quickest method starting from scratch is to a)sketch y =2^x and y=20-x and we can easily explain that there is one and only one intersecting point at x=4
How do you do it when it’s a^x + x = b?
I would say you do it using the Lambert W function. Won’t be quite as clean because you can’t break a or b, but you’ll get an answer better than the xth root of a function of x.
@@eliteteamkiller319 I mean sure, but is that really the point of the question? To see if you know about the lambert W function? The answer would be obvious for those who do, and completely impossible for those who don't. That isn't a good question.
This will make 2 equations. the importance is solving with only with a single equation.
@@denereri Why?
@@Gnowop3 idk
The solution is tremendously overcomplicated. My solution:
x = 2^y
2^2^y + 2^y = 20 --> (2^y)^2 + 2^y = 20
z = 2^y (z > 0)
z^2 + z = 20
Roots of the quadratic equation: z = -5 and z = 4; discard the negative one.
2^y = 4 --> y = 2
x = y^2 = 4
Your solution needs to be pinned so that it stays at the top of the comments!
2^2^y ≠ (2^y)^2
Sorry, it does. When one power is raised to another power, the powers are multiplied (multiplication rule with indices)
2 x y = y x 2
The fact that brackets were used on the right hand side doesn't make any difference. I suspect they were used just to make it clearer what was going on.
@TNK1790 is correct. It turns out that this error isn't caught because
y = 2 (2^2 = 2×2).
Because of PEMDAS,
2^2^y = 2^(2^y).
Since there are no parentheses, exponents must be dealt with first. So one needs to look at the right upper-most exponent, and, unlike × and ÷, concatenated exponents can't be dealt with left to right (which would be the same as adding patentheses). 2^y can't be broken up so that
2^2^y /= (2^2)^y = (2^y)^2.
(/= : does not equal to)
A useful way to look at this problem might be to appreciate the structure that led to the approach. Sure, you can solve this specific problem quickly by inspection, but what if the solution turned out to be something less obvious like √3? Some students may not know that problems with the variable in both polynomial and an exponent form require the Lambert W function. I thought there was secondary value in the way he had to repeatedly manipulate the equation to apply the W() function. That too is a skill worth learning for more difficult questions, IMHO.
Yeah I don't really get the people here in comments going “why even bother, it's obviously 4 by inspection”. Like it's there to provide you with something more tangible than just a^x+x=b, to teach a complex method. It's actually really good video showing how to solve the LambertW part without a calculator, I didn't even realize that ofc you could do that by further manipulating the expression inside the W(…).
@@arttu3232 Definitely some big egos in the channel who don't even post their own solutions. 🙄
And yet, there's no explanation for the reasoning that brought in that 2^20. I don't see how a method that relies on an intuitive leap from out of nowhere is a better method.
@@nathanialharris3300he needs to put the 2²⁰ in order to use the W function.
If the question would have been to find the general solution, then this approach is correct. For the specific problem it's obvious.
To the people saying this was a bunch of lines too many and a waste of time, go change 20 to “a” and try again. The point is not to solve one problem. It is to explore a technique.
This method cannot be used for “a” . For example cannot be solved for 21
Something tells me students are not supposed to solve entry tests with answers only representable with the help of special functions. To me this problem is really about observation + proof that the root is unique.
it doesn't work for any number on the right side
It is much faster just to try a few numbers, 1 to 5. The answer is 4.
A binary search gave me the answer.
Since a binary number can be written as a power of 2... You write 20 in binary and then you see the 4 being the only available response.
Ah, numerical analysis guy ;)
20 can be written as 2^4 + 4
So 2^x + x = 2^4 + 4
Therefore x=4
That's when the teacher cannot not give you the points for the question, even if he wanted you to struggle.
I do like that 🙏
This show how unpopular mathematic can be present...
“Don’t ask me why.” This is the problem with math instruction. Maybe the mathematicians don’t know either. “Just do this. Then do that. Don’t ask me why.”
Ranks right up there with PFM (Pure F*ing Magic) and ATAMO (And Then A Miracle Occurs), which is essentially his explanation for why he uses the Lambert W function.
@silverhammer7779 : You can find the proof, but it's either a book or a college course of its own.
I'd never trust anyone who made x's like that
It made the equation more surreal than it should have been. :)
What's unusual about those X?
Look at his e and l. Try not to be confused whenever he writes e^ln2.
Yeah, 2 c's do not equal x.
@@BoundaryElephant That's literally how people write x in cursive. It's the norm.
write 20 into the binary system and compare with sides... you get 2^x + x = 2^4 + 2^2. You can see that 2^2 = 4 so you are finished with 2^x = 2^4. => x = 4.
you actually need 3 seconds. We have an increasing function + increasing function = constant. We know about increasing functions that they are injective, meaning that for an x there exists only one f(x). So when we have 2 increasing functions added, we get one increasing function, which means an injective function => there is only one solution, which you can solve by trial and error. x=4 works obviously, and this is the only solution
This channel loves the Lambert W function.
😊 2⁴ + 4 = 16 + 4 = 20 🎉
x = 4
Looked at a couple of videos. One of the worst math channels on UA-cam. He make several times the same video with every time the same stupid answer. He only changes the title of the video (one is Harvard, the other is Stanford, in both thumbnails there is Hawking who went to neither). And noone learns anything because the bloody Lambert W function is utterly useless.
And yes, you can find x = 4 in 10 seconds.
This channel exists purely for clicks
@@lolilollolilol7773 To be honest, I did feel like I learned something here. Have a minor in math, but that was from back in the 1980s, roughly the bronze age or something. But still, I did feel like I learned something, so I felt like my time spent on this was ok.
@@lolilollolilol7773but you get to see one math trick which may become useful in your next life so I give a thumb up 😊
Elementary x=4
2*2*2*2+4=20
Thank you you are the only person in 52 years to explain a math problem to me in a way I can understand 😊
@@br4524me? Thank you
@@RockBrentwood please could you translate in Italian? I did understand very well
If this was a multi-choice test, then substitution would work, however if essay type test, you would need to show your work. Being Stanford, most likely the later. My math Instructors always said - show your work. If you know how to solve the problem but make a stupid calcultion mistake, you can still get partial credit instead of no credit. I’m 76 and still remember that advise from high school.
In what course is the Lambert W function taught. It's not taught in Precalculus and it"s not a calculus problem. Would my high school Algebra teacher know this function?
Probability less than 1% with 95% confidence. (i.e. p
I am a mathematician and never heard of the Lambert W function.
I'm not a fan of math, but correct me if I'm wrong here.
The point of this problem isn't to come up with the right answer (Even I knew the answer at a glance). The point is to show that you understand the mathematical principles needed to arrive at the answer. The reason would be that when you come across a problem you can't solve at a glance, you will have a way to get to the correct answer.
Is that a fair summation?
Disclaimer: I watched the end to verify my answer was correct. I watched the beginning until the 5th line and my eyes started to cross and I had to stop. :)
The hard part, IMO, is that there are non intuitive steps. Possible that someone who watched this, would be able to solve some future problem, using similar methods though.
This is a total hoax. At 7:40 when he split 20 in 16+4, he mentally solved the equation 2^x+x=20 mentally, in the same exact way he told not to solve the starting one 😂. With arbitrary numbers how?
However... I think that the purpose of this question is that you have to show the mathematical way of finding the answer, instead of mindlessly trying random numbers XD
"mindlessly"?? Before I start any challenge, I guess a ballpark. Since 2^4 is a little less than 20, I tried 4 for my first try. It was right, so I stopped. I always resented having to invent steps that were not actually needed to find the solution.
Off to find learn about the Lambert W function. Thanks for the question.
x = 4 is an obvious solution
Let us study the function f(x) = 2^x + x - 20 = e^xln2 + x -20 ---> f'(x) = ln2*e^xln2 + 1
f'(x) > 0 ----> f(x) is strickly increasing with f(4) = 0 ----> x = 4 is the unique solution
The purpose of the question wasn't the answer.
The purpose was the proof: or display of knowledge.
the majority of comments not understanding this is frightening
No, taking your time, stealing your time, currency of god spent for idiocy.
Any steps I give other than "x is obviously 4" would be disingenuous. I could make a joke with preposterous steps, but that's not actually the right way to solve such a trivial problem. It's not a test of mathematics, it is a test of wit and conformity.
By simple iteration we get 2^4+4 = 16+4=20
For x=1, we get 2+1=3
So x>1
Further calculating for values of x>1,
We get 20 for x= 4
Ok now 2^x + x = 21
I'm not sure if the goal of this question is to get you to do a real mathematical demonstration or just spotting the obvious.
Stanford does not have an entrance exam. Stop lying.
Lighten up, Francis!
They used to have in 19th century.
@@abefroman7393Stripes
Still a sexy solution.
Wait, really? Then how do you enter it?
There are also 3 shorter ways you can also use to get the value of X, but atimes some problems involving Complex Numbers & Algebra require that you show steps like this.
Knew Lambert W could be used. But instead I did my own proof. So x>0 for a solution. Rewrite as x= 4*(5-2^(x-2)) we know the paranthesis must be positive so x
But how do you know that the 'parenthesis must be positive'?
Note that this solution is not a general solution for the initial formula. It works just exactly for a few numbers like 20 because this is one of the solutions of the W formula. Or said otherwise: The initial formula is a nearly exponentional graph to base 2 if you would plot it. The W function is another exponential graph. Those graphs only match on a few locations, maybe even only on y=20. That's the reason why a simplification using W can even be done. Without that, no easy solution. Just try it with another number than 20.
He used an obscure trick and reasoning that I, quite frankly could not follow. The video teaches us nothing other than how "clever" its author is. Much easier to find it by examination in less than a minute.
Inspection*, I agree with you
Maybe this problem is not meant to teach, but to titillate. I was certainly titilated.
Dont blame others if you didnt understand something. This is completely understandable and its only you who doesnt want to understand and work hard on this question, atleast try to understand. P.S the W is a function which is used to solve these type of equations although he showd how this function worked you can atleast try to understand it.
I think complicated methods to show how to apply something like the W function is fine, but it would be simpler (on a test) to just do x=4 by inspection and prove that the derivative of 2^x+x is always positive, hence there is just that one solution.
This demonstration is an oxymoron, as it already requires seeing the solution in the step 2^20=2^4*2^16
The test seems to be about proving that you can BS some entertaining steps when the answer is obvious.
Simply you can do the following
2^x+x=20 then write 20=2^4+4 after that the right hand will equal the left hand if x=4
I never learned about this W lambda function at univerisity. Weird...
I have minor in math, albeit decades ago. And it didn't ring a bell.
I would list the powers of 2 less than 20 (1,2,4,8,16) and see which would be close enough to lead to a solution to this equation. Hint: in problems like this, it's usually the nearest one. 16=2^4, and 2^4 + 4 = 20.
x=4 is obviously the unique solution since 2^x + x is an increasing function.
Mathematics is beautiful. But, in a test, most of the time we don't have this rich and precious time.
My approach was: 4 is clearly a solution. now show that the function f with expression f(x)=2^x + x - 20 admits a single solution on R, so the the only solution is 4.
f(x) is the increasing function on R (because its first derivative is positive on R)=> the equation f(x)=0 has unique solution.
X=4 from sight
20 is a small number. There are only 4 integer numbers that make 2 to the power of that number less than 20. It takes up to 3 iterations to get the right answer, which can easily be done without use of pen and paper (in mind). It takes about a few seconds to come up with x=4. The algebra of the solution presented here is very curious though. I watched with interest and it was entertaining. Thank you!
No log and no W function in my reasoning (herafter):
2^x + x = 20
2^x + x = 16 + 4
2^x + x = 2^4 + 4
knowing that 2^x + x is a monotonically non-decreasing function (then injective) we can conclude:
■ x = 4
🙂
Oh, logical solution
Damn thats one fascinating piece of maths 😮👍👍
Bro really went the longest possible way around solving a 30-second equation.
2^x is a rising exponential function... and 20-x is linear... the point of intersection is one. The answer is 4, which can be guess within 3 seconds.
Or ... We could just notice that 2^x + x is an ever increasing function and clearly 2^x + x = 20 has no solution for x < 0.
Next notice that when x = 0, then 2^x + x = 1. This fact coupled by the fact that 2^x + x is an ever increasing function means that there is one and only one real x, such that 2^x + x = 20.
By observation, we see that 2⁴ + 4 = 20.
So x = 4 is the one and only real solution.
I love this answer!
@@brian554xx
Thanks 😀.
@@RockBrentwood
I guess x = 4 is a complex solution with no I component. Are you asking how many complex solutions there are?
Having found x=4 by inspection and observing that 2^x + x is strictly increasing, there can be no other solutions.
It's so easy sum
2^x + x = 2 × 2 × 5
2^x + x = 2² × ( 2² + 2⁰)
2^x + x = 2^4 + 4
Therefore x = 4
This is genius 👍🏽
Clever and rather simple!
can you walk me through your thought process for this?
@@hahaureadmyname 2^x + x is the same as 2^4 + 4 . I have only made factors of 20 rather than using functions .
🎉
2^x + x=20 -> 2^x = 20-x (tests the values of x) x=4
Ridiculous, I solved the equation in 10 seconds just in my head while the mathematician took 12 minutes!
And way detoured for that
Why is that better? The other way is way hotter. Literally got turned on seeing Lambert W twice.
Seriously though, replace the number 2 by a and the number 20 by b and try to solve it by inspection. Exactly. The point is not to solve one problem. It is to look at a technique to solve a set of problems.
@@eliteteamkiller319 So use more complex numbers as an example.
And if you did that on an entrance exam you’d get a zero because you didn’t actually demonstrate an understanding of mathematical principles
There's this thing called "Trial and Error" and in Singapore they call it "Guess and Check". I was never taught this when i was young. I was happy when my daughter had homeworks about it when she was in primary 4. Not sure about other countries but i think Singapore is right in having this included in the primary/elementary school Maths
Finding no structured way to solve this problem, I used guess and check I found an immediate solution. x=4, then I checked 2^4 + 4 = 20.
Using w func is a shugar. Why not use form if a*e^a = c*e^c, then a=c ?
Solved less than 30 secs just by trying numbers. Even it was a test, you can still guess the "in between" numbers for x.
This is the most roundabout solution I've ever seen to a fairly simple problem
If you can't immediately see that it has to be 4 , then maybe it's not the path for you right now.
mathematicians can spent a lot of paper when software engineers just know the right answer :)
@@linkedhashmap in this case a little reasoning about the bounds of x provides a big clue. 2**x grows much faster than x, so x cannot be large in order for the result to only be twenty. X is also likely a positive integer... so there are only a handful of likely choices... I mean, it's 4 for goodness sake !!
@@thorick590 2^x = 20 - x => x is even
@@linkedhashmap nice !
Maybe I’m missing something, but when 2^20 was split into 2^16.2^4, it looks like the reason to do that rather than say 2^8.2^12 is that you already know the answer is 4 and are trying to prove it, rather than trying to find the answer, otherwise how would you know to choose 4 and 16?
They should kick you out of Stanford for writing your x's like that
4 is an answer by inspection.
Next do d/dx to see if there are critical points. ln 2 2^x +1 = 0, so none on the Real line. (The function is obviously c^infinity).
Thus the only (real) solution is 4.
Complex solutions get tricky.
The answer is so guess-able
A lovely question, even a 10 year old can spot that x=4 is a solution but showing it is the only solution is a lovely use of the Lambert W function
As written in another comment, it's possible to arrive at answer = 4 considering: a) x is included between 0 and 5 (edges not included); b) x is an integer (to obtain 20, x in exponent must be integer, otherwise sum would be irrational). By substitution, we can arrive soon at answer of x = 4.
It can be resolved by geometric method: functions y1=20-x and y2=2^x have just 1 intersection. So we can obviously say that x=4 and its 1 single root.
Everyone saying it was obviously 4. But what if there was a Complex solution? Maybe that's why the working out?
Complex number doesn't even exist nor represents anything in nature. Math is full of useless problems.
The purpose of this particular tutorial is to illustrate the basic technique of solving such a problem, for which reason the tutor employs a basic equation. Through this he illuminates the solution by which the viewer may learn. Those spouting forth "too easy" and "by inspection not only miss the purpose of the video but also splay forth their excellence and mastery (quite dubious, actually). Plato warns us such individuals who allow their skills to go to their heads and become interminable.
On a first look one tries integers to get a feel for the question, and 4 clearly works. The next step is to think about the function to see if there may be other solutions. But y = 2^x + x is clearly an exponential-like function that rises consistently with values bounded below by 0. So only one real solution. As a uni entrance exam, I think going into the lambda function is a bit extreme... especially as the manipulations to get there are more extreme than the trial and error to find 4.
By the way, one of my classmates in a math class in HS tried the slick-trick method, that is gave the correct answer without the analysis to show that he understood the principles to attain the solution. He and the teacher got into a really heated argument with his "got the right answer" and the teacher's "didn't show the work."
2^x + x = 20
=>2^x + x -20 =0
Well the function f(x) = 2^x + x-20 is always strictly increasing bro. f'(x) =(ln2) x 2^x + 1 >0 for all x that belongs to the real set. So it means that it has 1 solution at most. X=4 is the answer
Puts the democratisation of mathematical teaching back 30 years. Come back NancyPi! We miss you more than ever.
One of two real root was easy (natural number), but try to solve another root (real irrational number).
(i)x=4 is a solve.
(ii)2^x + x is a monotonically increasing function.
(iii)Then, x=4 is only one solve.
Here's my honest opinion. What the f. This method doesn't work for every case only if f(x)= 20. And that 20=16+4 requires just the same intuition as solving the problem by just looking at it. I know your step by step solution also implies that there are no other solutions, but you can also prove that by saying that f(x) = 2^x + x is a strictly monotonically increasing function
Obvious answer is x=4. Then the first derivative of this equation is positive and therefore the solution is unique.
Not many people know about the W function.
I for one, as a PhD mechanical engineer, learned about the W function recently.
I imagine there may be other ways to approach this solution but it just goes to show you that there is actually an element of "art" and "creative thinking" in mathematics. I understood all his steps but I was also thinking, "I would've never thought of doing that."
One glance and you can recognize it's 4, but showing work for proof slightly more complex.
Just put in 4 and that's the proof. Increasing function assures the solution is unique. qed.
20 is a small number, 2^5 is already 32, so x should be between 1 and 4. Also only even numbers would give 20, because 2^x is always even. Therefore we only need to check 2 and 4. And only the number 4 fits the equation.
2^x+x=20 --- solve this expression
2^x=20-x --- 2^x is exponentiation function (with base 2) wich means that expression is above 0 then 20-x too is above 0: 20-x>0 / 20>x / x=0 --- for negative x function 2^x is lower then 1 but 20-1=19 that very bigger than 1 and not equal wich means that x not negative
Log2(2^x)=Log2(20-x) --- set logarithm on left and right expression
x=Log2(20-x) --- logarithm function correct only for expression is above 0: (20-x)>0 / 20>x / x
Short description of solution without or short comments:
2^x+x=20
2^x=20-x / x>0
Log2(2^x)=Log2(20-x)
x=Log2(20-x) / x
20=2^4+4. Therefore, 2^x+x=2^4+4 x=4
I haven't watched the video, but the most obvious solution that came to mind is to build a graph y = 2^x + x - 20 and look at the intersection with the x axis (it is 4).
Some say the answer is obvious. To some it may be, but he is showing the process to go through when an answer isn't obvious. That's what he is teaching. Good for him!
To solve the equation \(2^x + x = 20\), we can use trial and error or numerical methods, as there isn’t a straightforward algebraic solution. Here’s a quick way to find an approximate solution:
1. **Try different values for \(x\)**:
- For \(x = 3\): \(2^3 + 3 = 8 + 3 = 11\)
- For \(x = 4\): \(2^4 + 4 = 16 + 4 = 20\)
We see that when \(x = 4\), the equation holds true.
So, \(x = 4\) is the solution to the equation \(2^x + x = 20\).
F(x)=2^x + x - 20
F(4)=2^4 + 4 - 20 =16+4-20=0
dF/dx = 2^xln2 + 1 >0 (for each real number)
Therefore
x = 4
So I have no idea how you got this conclusion but I mean isn’t this basic math? I figured out 4 hurst by looking at it
easiest way by correlation: 2^×+x=16+4, 2^×=16and x=4, 2^4=16, then 16+4=20
x = 4 "by inspection" - that is all that is sufficient and necessary to answer the question.
Είναι συνάρτηση 1-1 και βρίσκεις τη μοναδική λύση. Όλο αυτό μπορεί να λυθεί σε 3 γραμμές
This and all similar problems are created specifically to have a solution only with the W function. They're trying to make W be considered a fundamental function, like logs and trig functions.
In order to do the step when you split 2^20 into 2^4 * 2^16, you needed to be able to see ahead to the next step when you replaced 2^4 with 16 to make the multiple equal the power. Therefore you needed to have solved the original equation in your head to be able to perform the massively complicated alternative... 🤔🤷
Precisely. Those final steps are precisely what you could have done by simple inspection from the original equation.
I see, the problem is not to find the answer, but to write a bunch of lines that logically goes to the answer.
Решение графически :построение экспоненты 2 в степени х и линии 20-х.Пересечение ответ 4.вариант легче метода логар и формул подстановок. Постановки перебор большой. Степень понизить к уравнению.возможно и здесь 2 варианта.
The RHS is a small number and LHS has x as exponent. It is obvious that x has to be a small number less than 10.. Such sums can be solved by inspection .
Переносим х направо, получаем 2^х=20-х.
Слева функция возрастающая, справа убывающая, значит, корень только 1. Методом подбора: 4
Pretty sure Will Hunting was mumbling 'there's a better way' after the first 20 seconds...