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
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) 🤗
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.
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
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) 🤗
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.
Исчерпывающее объяснение свойств одной из самых интересных показательных функций.
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*
The function x^x has a local maximum at x=0 ))
Are you sure?
x^x=((2/3)^2)^(4/9)=(4/9)^(4/9)...x=4/9
x^x=(2/3)^(8/9)
=((2/3)^((2/3)^2))^2
=((2/3)^2)^((2/3)^2))
∴x=(2/3)^2=4/9
I thought it was x^x=(2/3)^(4/9) 💀💀💀
I got 4/9, but not 8/27.
X=4/9
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