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.
Пікірлер: 52
@Windows__200022 күн бұрын
"X isn't real it can't hurt you"
@inyobill22 күн бұрын
The integers no more real than the Imaginary numbers.
@Windows__200022 күн бұрын
@@inyobill Integers literally are "real numbers" while imaginary ones are not: en.wikipedia.org/wiki/Real_number
@Etothe2iPi22 күн бұрын
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.
@lesliesusil471122 күн бұрын
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,
@gamingplus862522 күн бұрын
No complex solutions?
@craftinators710722 күн бұрын
@@gamingplus8625 You can the 2 complex solutions from the polynomial x^3 - 4 =0
@gamingplus862521 күн бұрын
@@craftinators7107 yes,but it was not in the video no big deal 😀
@johnstanley569222 күн бұрын
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).
@ruud976722 күн бұрын
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.
@dirklutz281822 күн бұрын
er moet nog 2 maal een minnetje voor!
@deriklytten22 күн бұрын
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.
@NOTHING-yu3ry22 күн бұрын
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@zpf628822 күн бұрын
And why discarding the two complex solutions? They are still solutions, and we didn't ask for real solutions only. What do I miss?
@kyintegralson965614 күн бұрын
You were asked for real positive solutions implicitly, by putting x under the radical sign.
@snowman239522 күн бұрын
title is Sqrt(3^x^2)) = x^(sqrtx) but thumbnail and video show Sqrt(3^x^2)) = 3^(sqrtx)
@cret85922 күн бұрын
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 ?
@Th3OneWhoWaits22 күн бұрын
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.
@boringextrovert671922 күн бұрын
@@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
@Th3OneWhoWaits22 күн бұрын
@@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.
@cret85922 күн бұрын
@@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.
@Th3OneWhoWaits22 күн бұрын
@@cret859 Glad I could help!
@TerryFerrellmathematics22 күн бұрын
Thank you!
@FinalMiro22 күн бұрын
hey I atleast found one being 4^1/3, yet I didn't know how to prove anything 🤣
6:30 you can put 4 in the right hand so your life don't get harder
@inyobill22 күн бұрын
Ahhhh, never realized that exponentiation isn't associative(?). Makes sense, 27**3 would be several orders of magnitude smaller than 3**27
@mjpottertx22 күн бұрын
Gotta love still using a blackboard!
@the3stumbleteers870Күн бұрын
yo can u do jee advanced questions
@sheikhfarooq12322 күн бұрын
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.
@the_warpig191922 күн бұрын
Does this require the use of the Product log function?
@thunderpokemon245622 күн бұрын
No dont make it complex man it sucks
@RubyPiec22 күн бұрын
Just wanted to comment to say the title is wrong, should be 3^sqrtx, no?
@ChengxiHu-e1u22 күн бұрын
But a cuadratic equation tends to have a positive answer, right?
@user-lu9fg7pc9q17 күн бұрын
9:12 we can
@TheMathManProfundities22 күн бұрын
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]}.
@baconboyxy22 күн бұрын
Obviously he has, the video presumably was only considering real solutions. No need to be so pretentious about it.
@TheMathManProfundities21 күн бұрын
@@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.
@dante963222 күн бұрын
Can somebody explain why other 2 complex solutions are not considered solution? I thought complex world is all about putting negative under square root.
@PrimeNewtons22 күн бұрын
The radical sign and negative or complex numbers do not coexist.
@stephensimpson779422 күн бұрын
@@PrimeNewtons sqrt(-1) = i sure looks like they coexist just fine in that equation. but maybe you just didn't explain your actual thought.
@user-wl4zu2ok1e22 күн бұрын
Sorry to point it out, but the correct title of the video should be: Sqrt(3^x^2) = 3^(sqrtx)
@boringextrovert671922 күн бұрын
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
@Ankilo-boy22 күн бұрын
Couldn’t it be 1?
@joaomane483122 күн бұрын
Think about it. Is the sqrt(3) equal to 3?
@Ankilo-boy22 күн бұрын
@@joaomane4831 i mean 3^x^2 under sqrt is 3^x?
@novidsonmychanneljustcomme575322 күн бұрын
@@Ankilo-boy No. What you mean would be 3^(sqrt(x^2)) which is NOT equivalent to sqrt(3^(x^2)).