A Radical Exponential Equation | Does 0 Work?

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[@moeberry8226]

-1 also works because it is not necessary for the power to be an integer to yield 1. You can have -1 raised to a rational number and still give 1 as an output. And in this case -1 is raised to the -2/3 power which does give 1.

[@albertoalves1158]

Isn’t (-1) also a solution ??

[@lawrencejelsma8118]

Check the original expression. You would then have if x = -1 a √(-1) = i that doesn't equate out an imaginary expression on the left side to the right side of the original expression.

[@lawrencejelsma8118]

Never mind. You are correct I wrote down √x instead of cube root of x error! Good catch!!

[@lawrencejelsma8118]

Plus a cube root in any polynomial has three roots of solutions. He only ended up with x= 1 and 5.19715 ... only two from a cube root solution set! If I had written down and double checked I also would have gotten the (-1)^ results of -1 = -1. I didn't follow his step rewatching his video of a^b where a = -1 of b = 0 implication confusing remarks.

[@tixanthrope]

x^(x^(1/3)) = x^(x/3) then the solutions are 0 (debatable), 1 and possibly -1. all check out. other option is x^(1/3) = x/3, which is easily solvable.

[@lawrencejelsma8118]

We just need to prove the cube root of 0 is 3 to prove 0^0 is 1 by L'Hopital's Rule: x^(x/3) is (x^x)^(1/3) that is 1 if x^x = 1 and is by the limit by L'Hopital's Rule. x^[x^(1/3)] is e^[x^(1/3)lnx] and L'Hopital's Rule on { lnx/x(^-1/3)} is e^[ 1/x / (-1/3x^(-4/3))] = e^0 = 1 like 0^0 = 1 also.

[@Hobbitangle]

Will it be much simpler to logarithm both sides of the equation? x^⅓•lnx =x•ln x/3 (x^⅓ -x/3)•ln x=0 The first and the obvious solution is ln x = 0, x=1 The next is 3•x^⅓ = x 3³•x=x³ x²=27 x=±√27=±3√3

[@Hobbitangle]

Let's check out the last solutions rhs: (x^⅓)^x=((±3√3)^⅓)^(±3√3)= ±(3√3)^(±√3) lhs: x^(x^⅓)= (±3√3)^((±3√3)^⅓)= (±3√3)^(±√3) Exactly the same, except while exponentioning the negative numbers we must use the complex arithmetics ±

[@tixanthrope]

logarithm is only defined for positive numbers.

[@elf835]

Good video but you glossed over 1 as a solution

[@lawrencejelsma8118]

Great catch. Also found in the comments in cube roots there has to be three roots of solutions. He rambled on and confused me with his a^b discussion of a = -1 having to have an "even integer b" confusing tangent leading us away from (-1)^ being -1 like 1 to the power of its reciprocals being 1 always. I'm just forgetting what I just saw on KZbin and accepting -1 and 1 solutions. I guess the 5.19615... one! It is probably easy by: x^(x^3/2) = x^(x/3) and taking logs. I made a mistake so I'm rechecking.

[@elf835]

@@lawrencejelsma8118couldn’t of said it better myself, both concise and addresses my biggest issues with this video

[@armacham]

And x = -1 since it's an integer power you don't have the same problem as -3rt3

[@lawrencejelsma8118]

@@armacham ... Yes! The other root x = 5.19615242270773 is from in his Wolfram Alpha calculations the base x on both sides of the equation having equivalent exponents in x^(1/3) = x/3 that is solved as e^(3/2ln(3))! Then -1 and 1 are both checked out and solutions making three roots other than x=0. There is an additional root of base "0" to the power "0" that solves the problem by L'Hopital's Rule as the limit as x--> 0 he was discussing as a potential solution. From "Limits using L'Hopital's Rule Calculus" there is a fourth root of x=0 along with all three regular roots of x = -1, x = 1 and x = e^(3/2ln(3)) or 5.19615342270663 by exponents of the same base having to match. He needs to teach more on this problem because so much other math went into the solutions in this problem as this thread implies! This is a half answered solution set even though he did mention the 0^0 Limits Calculus using L'Hopital's Rule extended solution. 😂

[@82rah]

- 3 sqrt(3) is not a solution let f(x) = x^(x^(1/3)) - (x^(1/3))^x then f(-3 sqrt(3) ) = -.0210767 - .0761562 i Both Maple and Wolfram Alpha can be used to evaluate f(-3 sqrt(3) )

[@rakenzarnsworld2]

x = 1

[@DonEnsley-yi2ql]

I got x ∈ { -1, 0, 1, 3√3 } L'Hopital's rule doesn't apply to exponential indeterminant forms. We have other proofs that 0 ^ 0 = 1, so 0 should work.