This exponential equation has two main ideas. The first is to understand the difference between an exponent and a tetration. The second is to understand the domain of the Square-root function.
A better explanation would be: Exponentiation is not associative, but there is a convention that 3^x^2=3^(x^2). When it comes to the equation x^3=a, it's much easier to draw the equilateral triangle in the complex plane and read off the three solutions.
you have obtained x^2=2*sqrt(x); let a=sqrt(x) and divide both sides by 'a' => a^3 = 2. so a= 2^(1/3)*exp(i*2*n*pi/3), n=0,1,2. Hence x = a^2 = 2^(2/3)*exp(i*4*n*pi/3).
When we get the equation (1/2)*(x^2) = √x itself we know that we actually have only 2 solutions, and squaring it will give 2 extra values that aren't solutions. So why go the extra step in factorising x^3 - 4 = 0? We know it should give only one proper solution.
Squaring both sides is much easier I think. 3^[x^2]={3^[x^1/2]}^2 3^[x^2]={3^x^1/2} As base is 3 for Both sides X^2 =2(x)^1/2 X^4=4x X(x^3 -4)=0 X=0,x=4^1/3,
We need three solutions, so we must factor as a difference of 2 cubes. Cube root of 4 is only one answer, we can't get rid of the other two inadmissable roots by doing what you said.
@@Th3OneWhoWaitsyou still don’t have to though. This form of difference of two cubes always produces a linear factor and a non-reducible quadratic. Also, the cube root is one to one in real numbers. So no need for all this
@@boringextrovert6719 Makes sense, I was just wanting show that even though there are three solutions to the factoring, not all of them satisfy the problem.
@@Th3OneWhoWaits Thanks you for your help. I now see where/why I was confused. In fact, polynomials of the third degree only have exactly 3 roots when defined on the complex numbers's domain. For polynomials of the 3rd degree defined on reals, they may be only a maximum of three real roots. But the trick here is to verify that 0 and ∛4 are the only possible roots (real or complex). In fact P(z) = z³-4 has three roots (one real root and two conjugated complex roots) { ∛4 ; ∛4×(-1+i.√3)/2 ; ∛4×(-1-i.√3)/2 }. So that z×P(z) = z⁴-4z has four solutions { 0 ; ∛4 ; ∛4×(-1+i.√3)/2 ; ∛4×(-1-i.√3)/2 } Fortunately, ∛4×(-1+i.√3)/2 and ∛4×(-1-i.√3)/2 are not solutions of the initial equation √(3^(x²)) = 3^(√x). So the only solutions are the two real 0 and ∛4. But, if you deliberately limit yourself at the real solutions, as Primes Newton say it at 08:08 , you can spare the determination of the two extra complex roots.
Wow, did you really just say that you can't put anything other than a positive under a radical sign? Of course you can. Have you never heard of √(-1) = i or √0 = 0? In fact any number, including all complex numbers can be placed under a radical. √{re^(iθ)} = √(r)e^(iθ/2) {r>0, θ∈(-π, π]}.
@@baconboyxy Unfortunately not, he specifically refers to the complex solutions and eliminated them because they 'can't go under the radical'. He could easily have said that he was only looking for real solutions. It seems he actually believes this and that definitely needs pointing out as it could be very dangerous for people trying to learn from his videos.
Can somebody explain why other 2 complex solutions are not considered solution? I thought complex world is all about putting negative under square root.
