A Homemade Exponential Equation, x^x=2^{1/x}

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[adandap]

Just one minor comment that doesn't change the solution, but x^x is well defined for some values x < 0. E.g. (-1)^(-1) = -1. Any negative odd integer is OK as is -1/3 etc. But -sqrt(2) isn't, so your answer is fine.

[WahranRai]

Plagiat and copy paste not allowed :
Raise to power x the 2 members : (x^x)^x = x^(x*x) =x ^(x^2) = 2 = (√2)^((√2)^2) → *x = √2*

[MrLidless]

Better to look at y = x^x - 2^(1/x) as a whole and show that is strictly increasing for all x>0, hence only one solution.

[rubensramos]

As Marco said, the solution of x^2log(x) = log(2) is x = exp(0.5W(log(4))). A similar problem is: x^2logq(x) = log(2), where logq is the Tsallis generalized logarithm function. For example, for q = 1/2 one has: x^2logq(x) = log(2) => 2(x^2)(sqrt(x)-1) = log(2), whose solution is expq1(0.5*Wq2(log(4))) = 1.3906. q1=1/2 and q2=3/4. expq(x) is Tsallis q-exponential and Wq(x) is the Lambert-Tsallis function. Wq(x)-expq(x)-logq(x) form a team in the same way W(x)-exp(x)-log(x).

[muhammadmahdidacosta]

Another method:
1) Raise both sides by the power x i.e:
(x^x)^x = (2^(1/x))^x
x^(x^2) = 2

2) Then raise both sides by the power 2 i.e:
x^(x^2)^2 = 2^2
By basic exponent rules:
x^2(x^2) = 2^2
(x^2)^(x^2) =2^2

3) From here we have a form
a^a = b^b therefore a=b where a=x^2, b=2
Therefore
x^2 = 2
x=+-sqrt(2) where the negative can be omitted

[stevenlitvintchouk]

I did it similarly, except I used logs to the base 2, which made the calculation a bit easier. Take logs base2 of both sides. x log x = 1/x, so x^2 log x = 1. Let y = x^2, so x = sqrt (y) and log x = 1/2 log y. Substituting, (y log y)/2 = 1, so y log y = 2. But similarly to the video, 1 = log 2, so 2 = 2 log 2.Therefore, y log y = 2 log 2, so y = 2, so x = sqrt (2).

[marcocappiello]

I tried solving it with lambert W function and my result was x=e^(1/2×W(ln4)), where W(ln4) is equal to ln2. Therefore we have 1/2ln2 which is the same as ln(√2). At the end we can simplify e and ln so the answer is just √2 :)

[ManjulaMathew-wbzn]

Actually if you take the cth power on both sides and square both sides just equalizing the bases and exponents the square root of 2 is the answer. Just something you may want to add as the second method.

[mintentha]

x^x = 2^(1/x)
raise each side to the x power, then we get:
x^(x^2) = 2
We then recursively replace the 2 on the LHS with the LHS as 2 is equal to it, as per RHS
then we get x^x^x^... = 2
then using standard recursive tricks we can turn this into simply x^2 = 2
this simplifies into x = +/- sqrt(2)
I don't think the negative actually works, though.

[NonTwinBrothers]

Glad to find a youtuber not afraid to solve homemade equations
None of that factory garbage

[MichaelRothwell]

Here's my solution to the equation xˣ=2^(1/x) for real x.

First, we assume this is a problem in the real world, i.e. x and all values calculated during the evaluation of the expression on each side are real.

Note that we shall only apply the rules of exponents where the base is positive, or the exponents are integers.

First, as the equation contains the expression 1/x, we see that we must have x≠0.

We shall consider two cases: x>0, x<0.

Case x>0.
xˣ=2^(1/x)
<=> (xˣ)ˣ=[2^(1/x)]ˣ (raising each side to power x. Equivalence as for a>0, xᵃ is a strictly increasing function of x and so one-to-one for x>0)
<=> xˣ^²=2 (rules of exponents for positive bases)
<=> (xˣ^²)²=2² (equivalence as both sides of previous line are non-negative)
<=> (x²)ˣ^²=2² (rules of exponents for positive bases)
By inspection, the equation is satisfied if x²=2, so x=√2 is a solution.
On the other hand, (xˣ)'=xˣ(1+ln x)>0 for x>1, so xˣ is strictly increasing for x>1. Also, for x>1 we have xˣ>1 and, for 0<x≤1, xˣ≤1, so for a>1, xˣ=a has a unique solution for x>0.
Hence, (x²)ˣ^²=2² => x²=2, so, for x>0, x=√2 is the unique solution to the equation.

Case x<0.

1. For real numbers x, y, as long as xʸ is defined as a real number, we have |xʸ|=|x|ʸ.

2. If x<0, y is real, and xʸ is defined as a real number, then y = p/q, where p, q integers, fraction in lowest terms, q odd > 0.

Now xˣ=2^(1/x)
<=> |xˣ|=2^(1/x) (since 2^(1/x)>0, as long as we add the proviso, from now on, that xˣ≥0)
<=> |x|ˣ=2^(1/x) (using result 1 above)
<=> (|x|ˣ)ˣ=[2^(1/x)]ˣ (raising each side to the negative power x. Equivalence as for a<0, xᵃ is a strictly decreasing function of x and so one-to-one for x>0)
<=> |x|ˣ^²=2 (rules of exponents for positive bases)
<=> |x|^(|x|²)=2 (as |x|²=x²)
<=> (|x|^(|x|²))²=2² (equivalence as both sides of previous line are non-negative)
<=> (|x|²)^(|x|²)=2² (rules of exponents for positive bases)
By inspection, the equation is satisfied if x²=2 (and by the same argument as in the case x>0, the equation is satisfied only if x²=2), so x=-√2 appears to be a solution.
However (using result 2 above), irrational powers of negative numbers are not defined as real numbers, so xˣ is not defined as a real number for x=-√2, and so there is no solution to the equation for x<0.

Hence the only real solution of the equation is x=√2.

[davidseed]

guess and check method
x^x =2^(1/x)
x=1 rhs is bigger
x=2 lhs is bigger guess a number in between x=sqrt(2)
lhs= sqrt2^sqrt2
rhs=2^(1/sqrt2)
raise both sides to power of sqrt(2)
lhs=sqrt2^2 =2
rhs=2^(sqrt2/sqrt2)=2
hence x=sqrt2 is a solution.

[VeganSemihCyprus]

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[armacham]

lnx = ln2 / x^2
since x^2 is positive, and ln2 is positive, lnx must be positive
which means x > 1

x^2 * lnx = ln2
knowing that x > 1, both parts of LHS are strictly increasing.
this means there can be only at most 1 solution

so you can use guess & check. And if you find a solution, you're done.

[schematism]

Very wonderful equation!
By the way, I would have used another method, i.e.:
1 - given that the base is x is different from 0 and 1, I can use a log with base x;
2 - so, I have: x=(1/x)log_x(2); then
3 - x^2=log_x(2)
4 - by definition of logarithm: which is the number which, elevated to the square of itself, gives 2?

[ericlan]

The graph of y=2^(1/x) at 9:14 is not correct.

[Harshul]

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


whats hard?

[laurentreouven]

Again the function x ln(x) ? No more challenge here.

[clementyip]

???? I’m scratching my head. Why is the solution so complicated?.

Simply take each side by power 2x and it’s done.

(X^2)^(x^2) = 2^2
=> x^2 = 2 => x = root(2)

[abhi]

x= square Root 2.
Not sure about: x= - square root 2.