I Solved A Cool Exponential Equation

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[@XJWill1]

You incorrectly state that x = - 2^(1/4) is a solution. The problem is that then you are raising a negative number to a non-integer exponent, which is an problem if you are trying to deal with real numbers. And if you are allowing complex numbers, then you cannot use the solution method you used, since some identities you used are only identical for real numbers.

[@Phantom_Reaper_]

But if the negative sign is outside the root, how is he raising a negative number to a fractional power? Its just the negative of a power of a positive base
Edit: nvm, i see what you mean.

[@SidneiMV]

I totally disagree with you.
Because the final result will be [-2^(1/4)]^[2*sqrt(2)] that is also equal to sqrt(2)^sqrt(2)

[@fadetoblah2883]

@@SidneiMV I thought the same at first, but if you plug (-∜2)^(2√2) into a standard calculator, it will return an error message, illustrating that the result is not a real number. I think the mistake, when doing it on paper, is that we naturally tend to parse (-∜2)^(2√2) as ((-∜2)^2))^√2, but this is arbitrary. If you instead parsed it as ((-∜2)^√2))^2, then it's more apparent that there's a snag.

[@XJWill1]

@@SidneiMV You cannot disagree with the math. And your math is wrong.

[@tryingtomakeanamebelike7245]

I don’t think there’s a problem since the negative is outside of the root

[@yoav613]

I think there is aproblem with the x=-2^0.25 because you get x^x^6= (-2^0.25)^(2^1.5) and it is complex

[@wtspman]

Is there a good reason for not converting the right side from radical notation to exponential notation? You do that very briefly near the end of the first method. My intuition would have been to do it at the start.
Without actually solving the problem, it seems that have both sides of the equation written in a common notation might reveal some obvious solution strategies.

[@MATHUP869]

Thank you for your sharing

[@SyberMath]

Thanks for watching!

[@trojanleo123]

x = 2^(1/4)

[@yoav613]

Nice job👌

[@SyberMath]

Thanks 😁

[@giuseppemalaguti435]

x=e^(W(3√2ln2)/6)

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@broytingaravsol]

easiness

[@neuralwarp]

Square roots again? AGAIN?

[@rakenzarnsworld2]

x = 2^(1/4)