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

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PLAYLISTS 🎵 :

[XJWill]

The reason that guess and check is not the best solution method is because it rarely works outside of contrived problems. In many applications the problems to be solved are not so carefully contrived and guess and check will not work.

[WolfgangKais]

„Guess and check“ dos not solve problems, it helps to answer questions that someone has created from first knowing the answer.

[XJWill]

Consider a generalized form of the equation in the video: x^sqrt(x) = a
where a > exp(-2/e) and we are looking for real values of x


This can be solved by employing the product-log aka Lambert-W. This is best achieved by manipulating the equation to the form u * exp(u)

I will use the notation Wp for the prinicipal branch and Wm for the lower branch
of the product log, and simply W for the generalized product log

x^sqrt(x) = a
ln(x) * x^(1/2) = ln(a)
(ln(x)/2) * x^(1/2) = ln(a) / 2
(ln(x)/2) * exp(ln(x)/2) = ln(a) / 2
ln(x)/2 = W(ln(a) / 2)
x = exp(2 * W(ln(a) / 2))

Note that there are no real solutions for a < exp(-2/e) since
ln(exp(-2/e)) / 2 = -1/e and W(x < -1/e) is complex valued

Note also that for exp(-2/e) < a < 1 there are two real solutions since
Wp and Wm have overlapping domains in -1/e < x < 0

Now we can solve for the specific case of a = 1/2, and there are two
solutions for x since exp(-2/e) < 1/2 < 1

xp = exp(2 * Wp(ln(1/2)/2))
= exp(2 * ln(1/2)) since Wp(u*ln(u)) = ln(u) for u > 1/e
= exp(ln(1/4))
= 1/4

xm = exp(2 * Wm(ln(1/2)/2))
= exp(2 * Wm(ln(1/4)/4))
= exp(2 * ln(1/4)) since Wm(u*ln(u)) = ln(u) for 0 < u < 1/e
= exp(ln(1/16))
= 1/16

It is worth noting that that solutions will not be rational for most values of a. The solutions were only rational in the video because of a judicious choice for a.

[mike.]

I agree that you successfully found all the solutions to the equation, but this method was more like get lucky and find two solutions and then show that they must be the only two solutions. It didn't feel at all like solving an equation.

[kushaldey]

Here's your favorite method, substitution.
Let x=y², then (y²)^y=½
=> y^(2y) = ½
=> y^(y) = (½)^(½)
=> y*ln(y) = -ln(2)/2.
Let y=1/z.
=> ln(z)/z = ln(2)/2.
Note that f(z)=ln(z)/z is interesting because f(1)=f(infinity)=0, take limit using L Hopital rule, and f(z) has maxima at z=e. Thus z=2 is the obvious solution but by a little hit and trial we also have z=4. Thus x=1/z² = 1/4 and 1/16.

[BOBPERIO]

Beautiful! Crystal clear and easy to follow. I really love your teaching style! 👍🏻

[juanmolinas]

greetings!, that was a nice analisys syber!

[SunilKumarMaths.ac.]

Apart of the very nice explanation about the topic, I like how you even reply to comments also.

[carlosayam]

Replace x - > y^2; then u have (y^2)^y = y^(2y) = 1/2; then the easy guess y=1/2 works...

[youssefbettaibi]

Thank you, sir,
What program do you use to write?

[tonyennis]

I don't like G&C as a method to solve the problem, but I do like it to find the bounds of the solution. it gives some intuition.

[Tletna]

Good analysis with the derivative. I wish you would have shown a way to solve without guess and check though. :( I tried and was getting lost.

[walterufsc]

x^x^(1/2) = 1/2

Therefore, the exponent 1/2 at the left side can be written as x^x^(1/2):

x^x^(x^x^(1/2) = 1/2. Doing it over and over again:

x^x^x^x^x...= 1/2 ==> x^(x^x^x^x...) = 1/2 ==> x^(1/2) = 1/2 ==> x = (1/2)² ==> x = 1/4

Question: Why is the solution x = 1/16 not obtained with this method?

[emanuellandeholm]

u=sqrt(x), x = u^2 => (u^2)^u = 1/2 => (u^u)^2 = 1/2 => u^u = 1/2 ^ 1/2 => u = 1/2 => x = u^2 = 1/4

[pedrovargas]

You stick a square root on both sides, then the equation solves itself.

Thanks. ✌️

[gdtargetvn]

You can rewrite the equation to x^(x^1/2)=1/2 and then apply that one method that blackpenredpen or MindYourDecision showed, basically replace 1/2 with x^(x^1/2), repeat this process until you have an infinite tower of power, which in this case is x^(x^(x^(x^...)))=1/2
Again, replace x^(x^(x^...)) with 1/2 and it becomes x^1/2=1/2
Or sqrt(x)=1/2. Therefore, x=1/4.

[paper]

you are a great teacher! thanks for all your fun videos👍

[devondevon]

X=1/4
1/2=1/2 raised to the power of one
Square the base 1/2 (1/2 x 1/2 =1/4) to 1/4 and compensate
by halving the exponent 1 to 1/2. That is, 1/4 raised to the power of 1/2 =1/2. But 1/2 = the square root of 1/4. So 1/4 raised to the power of the square root of 1/4 = 1/2. And x is in this form. So X = 1/4 Answer

[MatheusCarvalhok]

I dont know why (if Its really hard or Its Just "medium" and i just dont understand), but i cant solve equations like "f(x)^g(x) + h(x) = y"

Like "x^x + x = 6", Its 2, and its easy to get this result, but i cant Generalize It, like "x^x + x = y" (do some kinda of inverse function to solve x for Any y)

[carloshuertas]

Another great explanation, SyberMath!