Solving x^(1/x)=sqrt(2), a Homemade Exponential Equation

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PLAYLISTS 🎵 :

[scottleung]

I got both solutions without having to deal with any logs!

[johnlashlee]

I love to see the concavity argument. This problem reminds me of my precalculus days where we learned to sketch functions and make graphical arguments such as the ways the graphs of lines and natural log can intersect.

[misterdubity]

Without paper: I logged both sides with a base of 2. (1/x)log(x) = log(sqrt(2)) = 1/2. log(x) = x/2 (base 2). Mentally graphed out y = x/2 and y=log(x) (base 2)

[smalin]

1st solution
start with: x^(1/x)=sqrt(2)
square both sides: x^(2/x)=2
rewrite right-hand side: x^(2/x)=2^(2/2)
true if x=2

2nd solution
start with: x^(1/x)=sqrt(2)
square both sides: x^(2/x)=2
square both sides: x^(4/x)=4
rewrite right-hand side: x^(4/x)=4^(4/4)
true if x=4

3rd solution?
start with: x^(1/x)=sqrt(2)
square both sides: x^(2/x)=2
square both sides: x^(4/x)=4
square both sides: x^(8/x)=16
rewrite right-hand side: x^(8/x)=16^(16/16)
not true if x=2, so you can't find more solutions this way

[orcamaster]

I just raised both sides to the power of 2x to get x^2=2^x, which has 3 solutions of 2, 4, and a slightly negative, non-integer number. You can then eliminate the negative one though because you know the base wouldn't work for the original problem, so it is just 2 and 4.

[aoughlissouhil]

Nice 🙂👍
I used to hate the method guess and check but now I think it's a good way for solving some problems
and now I don't hate it anymore 🙂😆.
And I think in this problem guess and check is better than Lambert W function.

Keep up the good work ❣️❣️❣️❣️

[i-venwang]

You could just do

sqrt(2) = 2nd root of 2
x^(1/x) = x th root of x

x = 2

[zthfeug]

ln both sides and bring the power (1/x) to the front, we have :

1/x*lnx = ln(sqrt2) <==> multiply by -1 --> x^(-1) * ln(x^(-1)) = -ln(sqrt2) [the multiplication became the power). rewrite x^(-1) as e^ln(x^(-1)) so LHS is ln(x^(-1)) * e^ln(x^(-1)) = ln((sqrt2)^(-1)) <==> Apply Lambert W function on both sides and you have ln(x^(-1))= ln(1/sqrt2) thus x = e^(-W(ln(1/sqrt2)).

[bscutajar]

One correction: A linear function can have one intersection only without being a tangent

[paulotabim]

Could you please demonstrate, in the future, how to find the 3rd X solution for this. X= -0.7666... is a negative solution that satisfy the equation.
I only could get this X value using numeric methods.

[mohammedbarbouch]

Solution très facile
X a la puissance 1/x = 2 à la puissance 1/2. Donc x=2

[kadirayam]

As soon as I looked, I understood that it was 2, then I thought it would happen at 4, it was happening.

[kobethebeefinmathworld]

4:55 all the "real" solutions

[frentz]

For the "guess an answer" part of the show .. starting with x^(1/x) = 2^(1/2), I'll guess that x = 2 is at least one of the solutions. But then considering y = x^(1/x), it would seem ln y = (1/x) * ln x, so that y' / y = (-1/x^2) * ln x + (1/x^2) = (1 / x^2) * (1 - ln x). The derivative is zero at x = e, which is a max. It means there's one more solution! (Because x^(1/x) is still increasing at x = 2, then a max at x = e, and then decreasing back down to the magic value 2^(1/2) for some later x ... maybe x = 4? ... 4^(1/4) = B where B^4 = 4 and B > 0 so B^2 = 2 which .. sounds a lot like sqrt(2). So 4^(1/4) is in fact 2^(1/2).

[Triple_Trouble]

This problem you have already covered 4 months ago, when you did x^2 = 2^x. You also did a very similar method last time...

[chromerims]

Good vid 👍
4:19 --- "square both sides" . . . in that very moment, squaring seems like arbitrary selection. E.g., why not cube both sides in that very moment? What is the cosmological source that inspires one to not cube and rather square things right then and there?
6:59 --- "multiply the numerator and the denominator by 2" . . . is equally arbitrary. Why not multiply both numerator and denominator by 7i or e or π?

Yours sincerely.

[rakenzarnsworld]

You are just solving this problem in a hard way.

[christianthomas]

rad(2) = 2^(1/2) therefore by identification, x=2

[--ananas--]

this is so easy, i solved this without watching video

[tulio]

why if I use Lambert it gives only 2 ?