A Nice Exponential Equation

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[aurochrok]

x=0 is *not* a solution since 0^0 is undefined

[FrancisZerbib]

0^0 is undetermined in my country, boss

[TedHopp]

While lim_{x→0+} x^(x^2-2x)=0, technically the function is not defined there. The domain of definition is the *open* interval (0, infinity). It can be continuously extended to include 0, but that has to be done explicitly. (Note also that the left limit at 0 does not exist.)

[mystychief]

5:17 must be 2. b=0 while a NOT 0, so the solution x=0 is not valid (giving 0^0).

[jackperl]

Great video as always -- I really like the graph of this equation: elegant!

A third method would be convert subtraction in the exponent to a fraction, giving:
x^(x²) / x^(2x) = 1

multiply through by x^(2x):
x^(x²) = x^(2x)

then either x = 1 (1st solution) or:
x² = 2x

then either x = 0 (2nd solution) or divide through by x:
x = 2 (3rd solution)

[SidneiMV]

(xˣ)ˣ⁻² = 1

x ≠ 0 because 0⁰ is undefined

xˣ ≠ -1 because (1 - 2) is odd
=> x ≠ -1

xˣ = 1 => *x = 1*

x - 2 = 0 => *x = 2*

[KNOCK_OUT_]

As a mathematician I do appreciate your work.

[MathsScienceandHinduism]

Its easy.
We need to check by taking x= 1 and -1. X=1 satisfies the equation but x=-1 does not. So 1 is a solution. X cannot be 0 since 0^0 is undefined. Now another possibility is that the power should be 0, considering which we get x= 2 since we already know x cannot be 0. Hence the solutions are 1, 2

BTW I am uploading integrals on my channel.

[Monokumaaaaaaaaa]

The solution depends on what field we are actually solving this problem.

If x is an integer, then yes, we should analyze 2 possible options from which the expressions a^b can give us 1 as a result:
1) the base is equal to 1 => x = 1
2) the power is equal to zero => x = 0, x = 2; but the base shouldn't be zero at the same time => we left only with x = 2.

If x is a real number (which, I guess, is your interpretation of the problem, since you logged left side), then we're dealing with f(x)^g(x) function, where f(x) > 0 is a compulsory condition. The rest is the same.
But if you change expression in the exponent, you might find some negative "roots" which don't actually make sense (they're simply outta D(f)), so it's always better to avoid the ambiguities by providing strict restrictions on "x" :)

[taniacsibi]

In țara mea (RO) 0 ^0 e nedeterminate , x=1, x=2 Ok punctul (0, 0) nu aparține graficului, met.2. M-a placut

[IshanGupta-oh]

x^(x^2-2x)=1
x^(x^2-2x)=x^0
Thus, x^2-2x=0
So, x(x-2)=0
Therefore, x=2 and x cannot be 0

[i_am_a_gugugu]

x²-2x=0
x(x-2)=0
x=0 or (x-2)=0
x-2=0, x=2
2^4-4=2^0 => 1

[zancle]

Le soluzioni immediate che ho trovato sono x=1 x=2, se x=0 viene 0^0, ma non ricordo in questo momento se viene 1 oppure è una forma che richiede analisi matematica superiore

[kirillbarkov]

Hello mr can you solve these equations? xe^(1/x)=C // (1/x)e^x=C Thanks

[taniacsibi]

La met. 2, a ^b =1, b =0, a diferit de 0, in rest totul e ok

[neuralwarp]

Mathematicians will say 0⁰ is undefined. But Physicisrs will say of course it has a value, maybe many values, just depends on your direction of travel. Mathos are just lazy.

[asheredude]

x ∈ {1, 2} I was thinking -1 but it doesn't make the power even so I think two solutions