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(-√2)⁴ = 2. Yep. Negative numbers to the integer powers are well defined.

Damn

Your answer is incomplete, there is 2 real solutions.+/- square root of 2

The rule of interchange involves applying the commutative property of multiplication to the exponents.

When you will post the second solution for the 3 ^ x = x ^ 9 ?

OP totally forgot about comex numbers.

+_2^1/2

Force root.

I just saw, that x=sqrt(2) was a soution. sqrt(2)^4=(sqrt(2)^2)^2=2^ 2=4 Since x^x^4=e^(ln(x^x^4))=e^(x^4*ln(x)) is monoton growing, for x>=0, x=sqrt(2) is the onl real sollution.

@italixgaming915

11 күн бұрын

-sqrt(2) is also solution...

@juergenilse3259

10 күн бұрын

@@italixgaming915 I have onl looked for positive solutions, because negativ values raised to powers with exponents that are not whole numbers may be undefined within the real nubers ...

2^2 (x ➖ 2x+2)

Shouldn't it be +/- sqrt(2) ?

@juergenilse3259

12 күн бұрын

No We can state x>=0, because withhin real numbers, a negative vaue to a non integer power ay be udefined .

Completely wrong. First of all, you never proved why a^a=4^4 =>a=4, the study of the function x^x is missing. If x^4 then x²=2 or -2 then (we assume that x is a real number, otherwise studying x^x would be much more complicated) x is equal to sqrt(2) or -sqrt(2). And both solutions work.

These kinds of videos are extremely misleading