Can you solve this tricky exponential equation?

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[Nikos_Iosifidis]

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
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.

[rvqx]

x³=6 is a cubic equation, so it gives 3 solutions. Why you consider only 1 of them?

[bernardthedisappointedowl]

Clearly these puzzles are helping - actually got there today :)

[ilikemath-masterfail]

The other two imaginary solutions is



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)

[gelbkehlchen]

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

[Psykolord]

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).

[EvanEscher]

36=((6^(1/3))^6
6=((6^(1/3))^3
Therefore x = 6^(1/3)

[habibsabzezar]

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

[BN-hynd]

Brilliant. Some of the comments below are trying to rubbish this excellent solution. Ignore them. Well done from an ordinary folk😅

[oahuhawaii]

x^(x³) = 36 = 6²
(x^(x³))³ = (6²)³
(x³)^(x³) = 6⁶
x³ = 6 { skip Lambert func. and complex numbers }
x = ³√6

[shadowwalkers-cptd]

can we also solve his using the super log function?

[ChrisSquaredTwo]

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

[prime]

This is a cubic equation!1Therefore three solutions!!