For the equation x^x = y^y, there exists a family of solutions { (n/(n+1))^n, (n/(n+1))^(n+1) } where n is natural. In this case, x and y are (2/3)^2 and (2/3)^3. But (4/5)^4 and (4/5)^5 also work, etc.
4/9 answer x^x = (2^2/3^2) ^8/9 *1/2 square the base , and compensate by halving the exponent. For example n^ p = (n^2) ^ p/2 So 3^4 = 9^2 2^5 = 4^2.5 'x^x = (4/9)^ 8/18 x^x = (4/9)^ 4/9 since 8/18 =4/9 x = 4/9 if n^n = p^p, then n= p Answer 4/9
Sloppy phrasing at 04:04. 0^0 has no agreed value. It is the limit of x^x as x approaches 0 that has the value 1; with other limits that approach 0^0 having different values. This is a subtle but important difference, and getting it right wouldn't have been a problem here.
I keep hearing that 0^0=1 and that's final. I call BS. Mathematicians that are way smarter than me are in dispute with each other about it, so that tells me that the matter is far from settled.
The rule to differenciate x^x is to do it first taking the x of the exponent as a constant and after taking the base as a constant: (x^x)' = x (x^(x-1)) + x^x lnx = x^x (1 + lnx) 🤗
No need to spend so much time convincing us that 8/9 = 8/27 * 3, LOL. In general, I think you can skip more easy steps in your proofs, most people watching can handle it.
For the equation x^x = y^y, there exists a family of solutions { (n/(n+1))^n, (n/(n+1))^(n+1) } where n is natural. In this case, x and y are (2/3)^2 and (2/3)^3. But (4/5)^4 and (4/5)^5 also work, etc.
4/9 answer
x^x = (2^2/3^2) ^8/9 *1/2 square the base , and compensate by halving the exponent.
For example n^ p = (n^2) ^ p/2
So 3^4 = 9^2 2^5 = 4^2.5
'x^x = (4/9)^ 8/18
x^x = (4/9)^ 4/9 since 8/18 =4/9
x = 4/9 if n^n = p^p, then n= p
Answer 4/9
This video was amazing
Allô the solution is x=4/9😊(4/9)^4/9=(2/3)^8/9 easy to find by using xln(x)=(8/9)ln(2/3)=(2/3)^2ln(2/3)^2 and by identification x=(2/3)^2=4/9
Sloppy phrasing at 04:04. 0^0 has no agreed value. It is the limit of x^x as x approaches 0 that has the value 1; with other limits that approach 0^0 having different values. This is a subtle but important difference, and getting it right wouldn't have been a problem here.
I keep hearing that 0^0=1 and that's final. I call BS. Mathematicians that are way smarter than me are in dispute with each other about it, so that tells me that the matter is far from settled.
The rule to differenciate x^x is to do it first taking the x of the exponent as a constant and after taking the base as a constant:
(x^x)' = x (x^(x-1)) + x^x lnx = x^x (1 + lnx) 🤗
xlnx = (8/9)ln(2/3)
xlnx = 2³/3²ln(2/3)
xlnx = (2³/3²)(a/a)ln(2/3)
xlnx = (2³/3²)(/a)ln(2/3)ᵃ
a = 2 [ by inspection ]
xlnx = (2²/3²)ln(2/3)²
*x = (2/3)² = 4/9*
Исчерпывающее объяснение свойств одной из самых интересных показательных функций.
I thought it was x^x=(2/3)^(4/9) 💀💀💀
x^x=((2/3)^2)^(4/9)=(4/9)^(4/9)...x=4/9
The function x^x has a local maximum at x=0 ))
Are you sure?
x^x=(2/3)^(8/9)
=((2/3)^((2/3)^2))^2
=((2/3)^2)^((2/3)^2))
∴x=(2/3)^2=4/9
X=4/9
I got 4/9, but not 8/27.
No need to spend so much time convincing us that 8/9 = 8/27 * 3, LOL. In general, I think you can skip more easy steps in your proofs, most people watching can handle it.
x = 4/9