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Why did you impose condition in the first place

My approach 1^x=-1 => (-1)^(1/x)=1 Put,x=1/y So,(-1)^y=1 Therefore,y should be of the form (even/odd) i.e.y=[2m/(2n+1)] So,x=[(2n+1)/2m]

*(a+b)^2=a^2+b^2+2ab* *a^2+b^2=(a+b)^2-2ab=1^2-2×1=-1* *a^3+b^3=(a+b)(a^2+b^2-ab)=1[(-1)-1] =-2*

So 1^(real number) = -1 ?

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It is obvious that x = 2. If you want more complexity, take the cube root on both sides to get x = 2.

*(a+1)(b+1) =19=1×19* *(a+1=19và b+1=1) hoặc (a+1=1 và b+1=19)* *(a=18, b=0) hoặc (a=0, b=18)* *(a+1)(b+1) =19=(-1)(-19)* *=>a+b=18* *(a+1=-19và b+1=-1) hoặc (a+1=-1 và b+1=-19)* *(a=-20và b=-2) hoặc (a=-2 và b=-20)* *=>a+b=-22*

a=17/2, b=1

Xis 2

With e^(ix) in Euler's formula x is real number, x cannot be complex number. Your solution is absolutely wrong

Definitely a "creative" approach to the equation. However, 1^x=1 for every x being a complex number, making it a non-solvable equation. If you input the value(s) you obtained for x in the complex function 1^x, you'll always obtain 1 as the result

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This is wrong. 1^z = 1 for all complex numbers z In your solution, you are claiming e^(2iπk*ln2/2iπk) = 2 This is true, but it doesn't follow from 1^x=2 The 1/2iπk has an arbitrary k which is not necessarily the same as the first one. Or rather, you should check by writing e^(2πik*ln2/2πin) where k is not necessarily equal to n It simplifies to 2^(k/n) which is not necessarily always the same. Thus, the original equation has no solutions

great video and extremely helpful, but please fix audio my ears are bleeding

Quelle complication. Si x^3=2^3 alors x=2...ça démo est complètement ubuesque.

X=2

9

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A=3,b=3

Xis 2

Are here many possibilities such as a = 18, b = 0

Confusion: if a +ab + b = 18, how can a + b =18? Does this imply ab=0? a or b = 0? Not one of the choices in the above video????

Is a+b=63 instead of 36 . Please clarify.

x=3

x=0; sqrt2; -sqrt2

x=+-sqrt2

x=9

It's enough to solve this with the second equation only if you go through complex numbers a b = 1 Meaning: a + 1/a = 1. You can do a = a^2 / a: a^2 / a + a/a = 1 (a^2+1)/a = 1 a^2+1 = a a^2 - a + 1 = 0 Quadratic equation, the solution is a = plusminus cubic root of -1,. There are 3 cubic roots to -1 but it doesn't matter which one a and b are because a^3 = b^3 = -1 Solution is -2

*a + b = 1* *ab = 1* *a³ + b³ = ?* *_Answer :_* Let a³ + b³ = c. We know that, (a + b)³ = a³ + b³ + 3ab(a + b) or, 1³ = c + 3 × 1 × 1 or, 1 = c + 3 or, c = 1 − 3 or, c = −2 or, *a³ + b³ = −2* 🎉🎉🎉

Just found your channel, pretty cool stuff, but there is no possible value for A and B

This makes no sense, i feel like there are many other solutions according to this logic

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b = 19 / (a + 1) - 1, because 19 is a prime number => 1. a + 1 = 19 => b = 0 => a + b = 18 2. a + 1 = -19 => b = -2 => a + b = -20

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Why 19 + 1 on RHS simply disappear

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Why a+1 can not be equal to ±9.5, and b+1=±2? And how about a+1=3 and b+1=6⅓? Etc. What a fucking method of solving equations? Are you sure you study algebra in America and not in Zimbabawa?

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Very nice. Thanks so much

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