A Non-Standard Equation with One Solution (k is positive)

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[adamsulc]

I solved it using Lambert W function. It would be awesome, if you would make a video about this beautiful function. I would be so grateful and happy😂😃😁. And also thanks for this video.

[hhktleq]

Lambert W function: it’s free real estate

[owhycvb]

Calculus is so strong! I called f(x) =e^x-kx, f'(x) =e^x-k, f' has only one solution x=lnk and it's an increasing function. We can see that f has a minimum at lnk f(lnk) =k(1-lnk). If k>e we have 2 solutions, if 0<k<e we have no solutions and if k=e we have exactly 1 solution. So k=e and minimum of f is f(1)=0. x=1 is our only solution!

[davidgillies]

Of course if you relax the criterion that the two curves intersect at a single point then it's easy to get this in terms of the Lambert W-function. A bit of twiddling gives you a solution at x = -W(-1/k).The second intercept is at the first negative branch of W after the principal one, _viz._ -W_-1(-1/k). In Mathematica you would find these with -ProductLog[-1/k] and -ProductLog[-1, -1/k]. Reassuringly, these are the only two real solutions for k > e, although there''s an infinite number of solutions for x complex which correspond to the infinite branches of W. For 0 <= k < e there are no real solutions.

[alessandroricci]

I think I found an easier solution.
If you think about intersecting the curves y=x and y=ln(x)+ln(k), if you consider that a curve and a line only have 1 intersection when the line is tangent to the curve, u can just take the derivative of ln(x) and set it equal to 1 (the slope of the curve y=x), then you solve for x and get x=1. You substitute it into the original equation and get k=e.

Tell me if I made some any mistake, I solved it before watching

[Daniel-efgg]

Taking the log though was genius.
x = ln k + ln x
x - ln x = ln k
Want to minimize the lhs and set that value equal to the rhs.
d/dx (x-ln x) = 0
1 - 1/x = 0
x = 1
1 = ln k + ln 1= ln k
k = e.

[alessandroarmenti]

Didn't watch the video yet, but I think x should be W(-k)

[adandap]

The tangent case is only one of an infinite family of solutions. For k < 0, y = k x + c has a single intersection with e^x. E.g. y = 1-x intersects at x=0 etc.

[uberless]

I think another interesting question might be:
Find the minimum value for "a" such that y=k[x-a] intersects y=e^x at exactly one point for any given "k."

[patricktolosa]

The trick is finding the second equation

e^x = k (only possible point where the slopes are equal).

From this, combine the original equation, and you reach x=1

Use that and find k=e.

Do we really need all the algebra or special heavy cannons for this?

[snejpu]

Guys. The channel keeps attracting new subscribers, who don't know your rule. If you don't want spoilers from people seeing the thumbnail, just disable the comment sections or don't use premieres. To be honest, why do you care so much about them? If someone wants to try solving the problem by himself, he just won't look for the answer in the comments. Even if there are the answers in the comments, they are not necessarily correct. True viewers will wait, watch and verify them. I don't think you will lose the viewers because of spoilers.

[rhynwlf]

Thank you very much
This kind of questions always appear in the mathematics Olympiad
Lazy me always test for x=0 and x=1 for all exponential functions then skip all calculation(which is not required in the exam) lol

[thomasnielsen]

Great video!
It could be nice to see e^x=kx^3 with one solution(if it exists?). I think you have to use the Lambert function (?).

[musicsubicandcebu]

Divide both sides by e^x . . . 1 = k(xe^-x ). . . the max of xe^-x is 1/e at x=1 . . k has to be e

[MathElite]

First and wow I tried but failed solving this in the past
Nice video again Syber!!! :DDD

[Ssilki_V_Profile]

If param k is negative - then it has 1 solution. If ķ>=0 and k<e - also has no solution. K=e - 1 solution. K>e - has 2 solutions.

[LEVRONE_MARYLAND_MUSCLEMACHINE]

I solved it by multiplying both sides times e to the -x power, then after some algebra you can apply the Lambert W function and you get the result x = - W (-1/k)

[izouurj]

let y=e^x/x, y=k

y'=(x-1)e^x/x^2

solve y'=0 then get x=1 y' turn negative to positive

so k=e y=ex is tangent line of y=e^x

just 1 intersect

[iridium]

Doesnt this have an infinite number of solutions?
e^x=kx <=> (-x)e^(-x) = -1/k <=> x= -W(-1/k) depending on the value of the parameter k, we only solved for one value of k which is k=e, or does it have to do with extending the domain to the complex numbers?

[damianbla]

03:35 Solve e^a=k*a for "a" multiplicating both sides by "-e^(-a) / k" and using Lambert W function :)
Then we get a = -W(-1/k).