Although the solution result is correct, there is some error in the solution process. The mistake is that a relationship of the form a^a=b^b does not necessarily result in a=b because the function f(x)=x^x is not 1-1 For example, the equality a^a=(1/2)^(1/2) is true for both x=1/2 and x=1/4 The equation (x^3)^(x^3)=0.9 has two roots: x1=0.31095 and x2=0.96123 For more details see video 26 at ua-cam.com/video/QtlO6xEARgo/v-deo.html on my channel named L+M=N - Maths For Everyone Also verification is not necessary. The solution procedure ensures that the value x=6^(1/3) is a root of the equation.
dude doesnt know what he is doing, just uses memorized rules step by step, could be ai for all we know x to the x is a highly complicated function that requires a year worth of study and the concept of infimum and supremum
It's worse. If you listen to the audio at 05:00 , "between the lines" he implies that if a^b=c^d then a=c or b=d. He deduces that a=6 or a=6, hence a=6.
The other two imaginary solutions is x=-(³√3/2^(2/3))±(i3^(5/6)/2^(2/3)) All you need is to -6 from both sides and present 6 like (³√6)³ and solve equation x³-(³√6)³=0 x³-y³=(x+y)(x ²-xy+y²)
Why not write them from the polar form r∠θ directly converted to r*(cos(θ) + i*sin(θ)) form? The cube roots are easily found for the polar form, which is converted into Cartesian form via the r*cis(θ) expansion. With x³ = 6, we have: x = ³√6, ³√6*(-1 ± i*√3)/2 That's ³√6 times the 3 cube roots of 1 . But if you go that route, then there are infinite other complex roots for nⁿ = 6⁶. So, you get to find the 3 cube roots of each of those infinite solutions for n. Look up the Lambert function to get an idea of what this means.
Solution: x^(x³) = 36 |()³ ⟹ [x^(x³)]³ = (6²)³ ⟹ (x³)^(x³) = 6^6 |The same operations are done with (x³) on the left side of the equation as with 6 on the right side of the equation, therefore it must be: x³ = 6 |∛() ⟹ x = ∛(6) ≈ 1,8171
With x³ = 6 , we have: x = ³√6, ³√6*(-1 ± i*√3)/2 That's ³√6 times the 3 cube roots of 1 . But if I go that route, then there are infinite other complex roots for nⁿ = 6⁶. So, I get to find the 3 cube roots of each of those infinite solutions for n. Look up the Lambert function to get an idea of what this means.
I’ll try solving for them x^3 = 6 Subtract 6 from both sides x^3 - 6 = 0 It’s in difference of cubes form, so let’s factor it in differnce of cubes (x - cbrt6)(x^2 + xcbrt6 + (cbrt6)^2) = 0 Discard the real case since it was already solved, solve the complex one. x^2 + xcbrt6 + (cbrt6)^2 = 0 Use the quadratic formula x = (-(cbrt6) +- sqrt((cbrt6)^2 - 4(1)((cbrt6)^2)))/2(1) x = (-cbrt6 +- sqrt(cbrt6 - 4((cbrt6)^2)))/2 x = (-cbrt6 +- isqrt(cbrt6 + 4((cbrt6)^2)))/2 We can clean this up a little bit x = (-cbrt6 +- isqrt(cbrt6(1 + 4cbrt6)))/2 My answer looks a little different than Wolfram’s
I see , it`s quite a job to find the other two solutions. Even more if you want to check if they are okay. That`s probably the reason he did not consider them.
Before watching: (a^m)^n = a^(mn). So, we will cube both sides. This gives us: x^(3x^3) = (x^3)^(x^3) = 36^3 = (6^2)^3 = 6^6. So, (x^3)^(x^3) = 6^6. Ergo, x^3 = 6, and thus... x = 6^(1/3).
Thank you so much for your work my only suggestion is that people who are interested in math subjects have family good knowledge of math work so you may want to go faster and skip too much explanation of every item . That way we get to the end of solution much faster . Your work is very much appreciated
Although the solution result is correct, there is some error in the solution process. The mistake is that a relationship of the form a^a=b^b does not necessarily result in a=b because the function f(x)=x^x is not 1-1
For example, the equality a^a=(1/2)^(1/2) is true for both x=1/2 and x=1/4
The equation (x^3)^(x^3)=0.9 has two roots: x1=0.31095 and x2=0.96123
For more details see video 26 at ua-cam.com/video/QtlO6xEARgo/v-deo.html
on my channel named L+M=N - Maths For Everyone
Also verification is not necessary. The solution procedure ensures that the value x=6^(1/3) is a root of the equation.
dude doesnt know what he is doing, just uses memorized rules step by step, could be ai for all we know
x to the x is a highly complicated function that requires a year worth of study and the concept of infimum and supremum
It's worse. If you listen to the audio at 05:00 , "between the lines" he implies that if a^b=c^d then a=c or b=d. He deduces that a=6 or a=6, hence a=6.
So what
Lambert function.
Clear and excellent as usual!
The other two imaginary solutions is
x=-(³√3/2^(2/3))±(i3^(5/6)/2^(2/3))
All you need is to -6 from both sides and present 6 like (³√6)³ and solve equation x³-(³√6)³=0
x³-y³=(x+y)(x ²-xy+y²)
Why not write them from the polar form r∠θ directly converted to r*(cos(θ) + i*sin(θ)) form? The cube roots are easily found for the polar form, which is converted into Cartesian form via the r*cis(θ) expansion.
With x³ = 6, we have:
x = ³√6, ³√6*(-1 ± i*√3)/2
That's ³√6 times the 3 cube roots of 1 .
But if you go that route, then there are infinite other complex roots for nⁿ = 6⁶. So, you get to find the 3 cube roots of each of those infinite solutions for n. Look up the Lambert function to get an idea of what this means.
@@oahuhawaii2141 pretty interesting
Solution:
x^(x³) = 36 |()³ ⟹
[x^(x³)]³ = (6²)³ ⟹
(x³)^(x³) = 6^6 |The same operations are done with (x³) on the left side of the equation as with 6 on the right side of the equation, therefore it must be:
x³ = 6 |∛() ⟹
x = ∛(6) ≈ 1,8171
x^(x³) = 36 = 6²
(x^(x³))³ = (6²)³
(x³)^(x³) = 6⁶
x³ = 6 { skip Lambert func. and complex numbers }
x = ³√6
With x³ = 6 , we have:
x = ³√6, ³√6*(-1 ± i*√3)/2
That's ³√6 times the 3 cube roots of 1 .
But if I go that route, then there are infinite other complex roots for nⁿ = 6⁶. So, I get to find the 3 cube roots of each of those infinite solutions for n. Look up the Lambert function to get an idea of what this means.
36=((6^(1/3))^6
6=((6^(1/3))^3
Therefore x = 6^(1/3)
x³=6 is a cubic equation, so it gives 3 solutions. Why you consider only 1 of them?
He said at the end he wasn't considering the complex roots - though once you know the real root, you can divide through and get a solvable quadratic -
I’ll try solving for them
x^3 = 6
Subtract 6 from both sides
x^3 - 6 = 0
It’s in difference of cubes form, so let’s factor it in differnce of cubes
(x - cbrt6)(x^2 + xcbrt6 + (cbrt6)^2) = 0
Discard the real case since it was already solved, solve the complex one.
x^2 + xcbrt6 + (cbrt6)^2 = 0
Use the quadratic formula
x = (-(cbrt6) +- sqrt((cbrt6)^2 - 4(1)((cbrt6)^2)))/2(1)
x = (-cbrt6 +- sqrt(cbrt6 - 4((cbrt6)^2)))/2
x = (-cbrt6 +- isqrt(cbrt6 + 4((cbrt6)^2)))/2
We can clean this up a little bit
x = (-cbrt6 +- isqrt(cbrt6(1 + 4cbrt6)))/2
My answer looks a little different than Wolfram’s
@@just_another_person_who_li4675 You made a mistake in the 2nd line of "x=...".
Should get
x=(-1±i√3)·6^(⅓)/2
I see , it`s quite a job to find the other two solutions. Even more if you want to check if they are okay. That`s probably the reason he did not consider them.
It was too complex for the purpose of the video
6^6 3^2^3^2 1^13^2 32(x ➖ 3x+2).
Clearly these puzzles are helping - actually got there today :)
Before watching: (a^m)^n = a^(mn).
So, we will cube both sides. This gives us:
x^(3x^3) = (x^3)^(x^3) = 36^3 = (6^2)^3 = 6^6.
So, (x^3)^(x^3) = 6^6.
Ergo, x^3 = 6, and thus...
x = 6^(1/3).
Your "ergo" needs to be proven. Why does a^a=b^b impliy a=b? f(x)=x^x is not 1 to 1.
Lambert function.
Thank you so much for your work my only suggestion is that people who are interested in math subjects have family good knowledge of math work so you may want to go faster and skip too much explanation of every item . That way we get to the end of solution much faster . Your work is very much appreciated
Amazing! Bravo!
X^X^3=36 X=Surd[6,3]
Thanks😊
Thank you very much
Brilliant. Some of the comments below are trying to rubbish this excellent solution. Ignore them. Well done from an ordinary folk😅
Solved with Lambert w function
Nice!
Thé initial formula is ambiguous
X pow x pow 3 could be
Either x pow (x pow 3)
Or (x pow x) pow 3
You should set the () correctly at first
Convention: a^b^c *always* means a^(b^c).
See en.wikipedia.org/wiki/Exponentiation under *Identities and properties*
@@ReasonableForseeability thank you, I didn’t know that
That's because if a^b^c is interpreted as (a^b)^c , that's equivalent to (aᵇ)ᶜ and aᵇᶜ , which is easier to write than a^b^c .
can we also solve his using the super log function?
Please solve 2 power x + x = 5
This is a cubic equation!1Therefore three solutions!!
Or one triple solution, anyway its not polynomial equatiin senso stricte
Ответ - 12.
is this stuff a joke or is it real?
Its real dude
There are infinite other solutions, but they're all complex numbers.
I'm the 100th liked in the video
2015