Solving x^{1/sqrt(x)}=1/16

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[armacham]

I used a different way that gets rid of the radical and gets rid of the fractions

First I determined that x > 0 because you can't have negative in a radical and you can't have 1/0

Then I made a new variable a such that x = a^2, with a > 0

putting that in got rid of the square root

(a^2) ^ (1/a) = 1/16
(a^ ( 1/a) )^2 = 1/16
square both sides

a^(1/a) = +-1/4

knowing that a > 0, we can reject the negative branch
a^(1/a) = 1/4

now make a new variable b = 1/a
which means a = 1/b
substitute that in to get:

(1/b) ^ b = 1/4

(b^-1)^b = 4^-1

b^-b = 4^-1
flip both sides to get:

b^b = 4

so you can see that b > 1. if b = 1, b^b = 1. And if b < 1 then b^b must be less than one. positive number less than one to any positive power will be less than one

ln both sides
blnb = ln4

define a function f(b) = blnb, take the first derivative, f'(b) = lnb + 1
that's only equal to zero once, when lnb = -1 (so b = 1/e)
when b > 1, f'(b) is always positive
a function that is always positive can only be equal to a constant once
therefore there is only one solution

which means you can use guess and check. obviously b= 2 is a solution because 2^2 = 4
then you just go back to the definitions of a and x to get your answer.

[moeberry]

The function passes it’s horizontal asymptote just by seeing that at x= 4 the function has a height of 2 which is greater than 1. But at the end as x gets larger and larger it approaches 1 and therefore never comes back to 1/16.

[zahari]

The substitution x = 1/t^2 works immediately.

[rubensramos]

x = exp(-2W(-0.5log(1/16))), where W(x) is the Lambert function.
The Lambert function has four generalizations: 1) Lambert-Tsallis, 2) Lambert-Kaniadakis, 3)Rqq (generalization of Wq) and 4) Rkk (generalization of Wk).

[BackroomsFan-wdwf]

I found a very interesting solution.

x^(1/sqrt(x)) = 1/16

Raise both sides to the sqrt(x) power.
-> x = (1/16)^sqrt(x) = 1/(16^sqrt(x))

Then take the square root on both sides:

sqrt(x) = sqrt(1/(16^sqrt(x)))

Since sqrt(x^y) can be said as sqrt(x)^y we can rewrite the equation based on this fact such that we have:

sqrt(x) = 1/(4^sqrt(x))
-> sqrt(x) = (1/4)^sqrt(x) since the square root of 1 is just itself.

Notice here that we have sqrt(x) on both sides, this means that we substitute in the value of sqrt(x) to the right hand side such that it becomes:

sqrt(x) = 0.25^0.25^0.25^0.25^0.25…
Thus, x = (0.25^0.25^0.25^0.25^0.25…)^2

Because 0.25 is in between e^-e and e^(1/e), we can safely say that sqrt(x) converges to a certain number meaning that x is also finite.

In order to evaluate the infinite tetration, I will be using the formula W(-ln(z))/(-ln(z)) where W is the lambert W function. Now plugging in the value 0.25 yields W(-ln(0.25))/(-ln(0.25). Evaluating this, we see that the infinite series representing sqrt(x) converges to 0.5.

This means that x = 0.5^2 = 0.25

In order to prove that this is the only solution, let’s go back to the previous equation sqrt(x) = (1/4)^sqrt(x)

If we take the derivative we’ll see that we will be comparing 1/2sqrt(x) with -ln(4)/(2sqrt(x) * 4^sqrt(x))

1/2sqrt(x) never reaches 0 nor can it go negative, meaning that sqrt(x) must constantly be increasing. But if we look at the other derivative, we’ll see that it is constantly decreasing. That must mean that their only intersection point is at 0.25.

So my final answer is that x = 0.25, very nice question!

[pwmiles]

Isn't it simpler to take both sides to the power of sqrt(x)? I.e.
x = (1/16)^sqrt(x)
= 16^(-sqrt(x))

The LHS is increasing, the RHS is decreasing so there can be at most one solution i.e.
x = 1/4

[walterufsc]

By making a variable substitution: x = 1/z² , we obtain: (1/z²)^z = 1/16
Taking the inverse on both sides: (z²)^z = 16 = 2^4
Hence: z^(2*z) = 2^(2*2) => z = 2 => x = 1/2² = 1/4

[SuperPassek]

How about substituting x = (1/4)^t ? It is clear that 0 < x < 1, so t > 0. Then x^(1/sqrt(x)) = (1/4)^(t 2^t) = 1/16, so t 2^t = 2. It is very clear that t = 1 is a solution, and as f(t) = t 2^t is increasing, it has only one solution. Therefore, x = (1/4)^t = 1/4

[andreabaldacci]

I wrote the original equation as x^x^(-(1/2))=1/16. Setting x^(-(1/2))=u, gives x=u^(-2). The original equation becomes now (u^(-2))^u=1/16, or u^(-2u)=1/16. Taking power 1/2 from both sides provides u^(-u)=1/4 (the negative option can be discarded for domain reasons), or u^u=4, which is u=2. From u=2 follows x=1/4

[pietergeerkens]

Wasn't it trivially obvious that x = (cos π/3)^2?

😉😜

That generates ideas for an April 1 math challenge.

[ytlongbeach]

you explain your logic very clearly. thank you !!!

[Schrodinger_]

Raise your hand if you never bothered to remember the quotient rule, and instead, when seeing u/v, just apply the product rule to u(1/v)

[Jha-s-kitchen]

x = 1/4
watching your videos, Today I tried the question myself first, did what you would do, took ln, made the derivative table and concluded that there is only one solution
Thanks

[MartinPerez-oznk]

THANKS PROFESOR!!!!, VERY INTERESTING!!!!

[dudewaldo]

Guess the solution! Function is monotonic so that's the only one.

These equations do not teach anything I do not like them

[rudela]

I think your penmanship could use some help.

[christianthomas]

obviously x must be positive. by squaring the given equation we get: x^(1/x) = 1/16^2. this last equation could be written as: 1/((1/x)^(1/x))=1/(4^4) and by comparison we get : 1/x=4 So that it appears that x=1/'4 Is the solution.

[vladimirkaplun]

x=1/4 - at a glance .The rest is just time filler.

[tbg-brawlstars]

x = 1/4, I solved in head in 20 seconds 😄
O wait there are more solutions lol
Oh I understand why I only have one solution, I took root and it lost a solution

[sebastiensoubiale]

Sorry but going through logarithm is really useless and disproportionate. Just apply conventional exponential rules and transform left and right sides to find x=0.25