Entrance examination to Stanford University

What do you think about this question? If you're reading this ❤️. Have a great day!

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@higher_mathematics

#maths #math

[ericmiller]

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.

[Li.Siyuan]

Sixteen lines of maths too many. It was blindingly obvious the answer was x=4 three seconds after looking at the question.

[DukeofEarl]

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!

[JamesHillhouse]

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!

[icila]

I try 1, then, 2, then 3, then 4 and I find that 4 is the solution. It takes 10 seconds

[Seriouslyfunny]

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.

[amirhoseinkazeminia]

You can guess 4, and since 2^x + x is a strictly increasing function, it has at most one solution.

[ftgsoftware]

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!

[SidneiMV]

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

[JohnKromko-vmof]

I'd never trust anyone who made x's like that

[Gnowop]

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

[eliteteamkiller]

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.

[cleiberrocha]

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.

[robertlezama]

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.

[PerpetualAbidance]

“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.”

[michaelkovalenko]

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

[FXK]

Thanks, never heard of Lambert W function, so had to dig into. Learned something new, always nice!

[lawrencenienart]

This channel loves the Lambert W function.

[ivanmelicher]

This show how unpopular mathematic can be present...

[WahranRai]

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