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.
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,
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.
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.
"X isn't real it can't hurt you"
The integers no more real than the Imaginary numbers.
@@inyobill Integers literally are "real numbers" while imaginary ones are not:
en.wikipedia.org/wiki/Real_number
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.
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,
No complex solutions?
@@gamingplus8625 You can the 2 complex solutions from the polynomial x^3 - 4 =0
@@craftinators7107 yes,but it was not in the video
no big deal
😀
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).
The complex solutions are
4^(1/3) * (i*sqrt(3)-1)/2
and 4^(1/3) * (-i*sqrt(3)-1)/2
The square roots of these are again complex.
er moet nog 2 maal een minnetje voor!
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.
You are super sir ❤❤❤
You are great sir ❤❤❤❤
Love from India 🇮🇳
And why discarding the two complex solutions? They are still solutions, and we didn't ask for real solutions only. What do I miss?
You were asked for real positive solutions implicitly, by putting x under the radical sign.
title is Sqrt(3^x^2)) = x^(sqrtx) but thumbnail and video show Sqrt(3^x^2)) = 3^(sqrtx)
5:29 As we know that we are looking for a positive real x , why may we not directly deduce from x³ - 4 = 0 that x = ∛4 ?
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.
@@cret859 Glad I could help!
Thank you!
hey I atleast found one being 4^1/3, yet I didn't know how to prove anything 🤣
Complex solutions:
(-1 (+-) i* sqrt(3)) / ( 2^ (1/3))
🇮🇹🇮🇹🇮🇹
6:30 you can put 4 in the right hand so your life don't get harder
Ahhhh, never realized that exponentiation isn't associative(?). Makes sense, 27**3 would be several orders of magnitude smaller than 3**27
Gotta love still using a blackboard!
yo can u do jee advanced questions
A good try, wish u all sucess.By the way ,if u substitute cube root of 4 in the exp. Equation ,how does it behave.since it is a root of the equation.
Does this require the use of the Product log function?
No dont make it complex man it sucks
Just wanted to comment to say the title is wrong, should be 3^sqrtx, no?
But a cuadratic equation tends to have a positive answer, right?
9:12 we can
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]}.
Obviously he has, the video presumably was only considering real solutions. No need to be so pretentious about it.
@@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.
The radical sign and negative or complex numbers do not coexist.
@@PrimeNewtons sqrt(-1) = i
sure looks like they coexist just fine in that equation.
but maybe you just didn't explain your actual thought.
Sorry to point it out, but the correct title of the video should be: Sqrt(3^x^2) = 3^(sqrtx)
There is no need for the extra work to solve the cubic. You can immediately say that the cube root of 4 is the only solution
Couldn’t it be 1?
Think about it. Is the sqrt(3) equal to 3?
@@joaomane4831 i mean 3^x^2 under sqrt is 3^x?
@@Ankilo-boy No. What you mean would be 3^(sqrt(x^2)) which is NOT equivalent to sqrt(3^(x^2)).
No 3^1/2 is not equal to 3 but x could be 4^1/3
x²/2 = √x
x² = 2√x
x⁴ = 4x
x(x³ - 4) = 0
*x = 0*
*x = ∛4*
You forgot to square the 2 in the 3rd line
@@PrimeNewtons thanks! fixed
3^(0.5x^2) = 3^(x^(0.5))
x^2 = 2x^0.5
x^0.5 (x^1.5 - 2) = 0
x = 0, x = 2^(2/3)