This is a nice work but let me offer a few pointers as some parts of this computation are not needed and the solution x=1/2 can be found much easier. However, please say that we are looking for a real solution of this equation when formulating the problem. 1. Let y=3^x then as you have shown, y^6 - 2y^4 + y^2 - 12 = 0. (*) Observe that if y=a is a solution of (*), so is y=-a a solution yet a negative y is not possible since y=3^x > 0 for any real number x. Then, to avoid the factorization part which might be difficult to guess, let z = y^2. Then, this last equation becomes z^3 - 2z^2 + z - 12 = 0 (**) and we are looking only for positive solutions of (**) since z = y^2 > 0 for any real y. Using a synthetic division, we can easily find that z = 3 is a solution of (**) and that (**) factors as (z - 3)(z^2 + z + 4) = 0. Since we are looking for real solutions only, observe that the factor z^2 + z + 4 is always greater than or equal to 4 for any real z. Thus, the only solution is z = 3. Thus, z = y^2 = 3 so y = 3^(1/2) = 3^x hence x = 1/2 is the only real solution of the original equation (since y must be positive). Comment: I am sorry but in your final argument using logarithms there are two false statements: (1) The complex number (-1 +/- sqrt(15) i)/2 is NOT negative (in fact, the field of complex numbers cannot be ordered, or, simpler, there are no negative or positive complex numbers whereas the field of real numbers can be ordered, and every real number is either positive or negative or zero: this trichotomy does not apply to complex numbers) hence it is false to say that a logarithm of that complex number cannot be found because that number is negative and so we must reject this 'solution'. (2) The second incorrect statement is that the logarrithm of that complex number (-1 +/- sqrt(15) i)/2 does not exist: in fact, log z where z is a complex number does exist. However, the middle part where you solve y^4 + y^2 + 4 = 0 (***) is absolutely unncessary once you observe that (***) has no real solutions since the value of y^4 + y^2 + 4 >= 4 > 0. Thus, this last part is also not needed. Good job. Thanks for this video.
You don't need to square both sides, you can write your equation like y(y² - 1) = 2√3 You can easily see that y=√3 is a solution, so (y - √3) is a factor of the equation y³ - y - 2√3 = 0 So, using polynomial division, you get (y - √3)(y² + √3y + 2) = 0 Now you can easily find the two remaining (imaginary) solutions
Great solution
This is a nice work but let me offer a few pointers as some parts of this computation are not needed and the solution x=1/2 can be found much easier. However, please say that we are looking for a real solution of this equation when formulating the problem.
1. Let y=3^x then as you have shown,
y^6 - 2y^4 + y^2 - 12 = 0. (*)
Observe that if y=a is a solution of (*), so is y=-a a solution yet a negative y is not possible since y=3^x > 0 for any real number x. Then, to avoid the factorization part which might be difficult to guess, let z = y^2. Then, this last equation becomes
z^3 - 2z^2 + z - 12 = 0 (**)
and we are looking only for positive solutions of (**) since z = y^2 > 0 for any real y. Using a synthetic division, we can easily find that z = 3 is a solution of (**) and that (**) factors as
(z - 3)(z^2 + z + 4) = 0.
Since we are looking for real solutions only, observe that the factor z^2 + z + 4 is always greater than or equal to 4 for any real z. Thus, the only solution is z = 3. Thus, z = y^2 = 3 so y = 3^(1/2) = 3^x hence x = 1/2 is the only real solution of the original equation (since y must be positive).
Comment: I am sorry but in your final argument using logarithms there are two false statements:
(1) The complex number (-1 +/- sqrt(15) i)/2 is NOT negative (in fact, the field of complex numbers cannot be ordered, or, simpler, there are no negative or positive complex numbers whereas the field of real numbers can be ordered, and every real number is either positive or negative or zero: this trichotomy does not apply to complex numbers) hence it is false to say that a logarithm of that complex number cannot be found because that number is negative and so we must reject this 'solution'.
(2) The second incorrect statement is that the logarrithm of that complex number (-1 +/- sqrt(15) i)/2 does not exist: in fact, log z where z is a complex number does exist.
However, the middle part where you solve
y^4 + y^2 + 4 = 0 (***)
is absolutely unncessary once you observe that (***) has no real solutions since the value of y^4 + y^2 + 4 >= 4 > 0. Thus, this last part is also not needed.
Good job. Thanks for this video.
You don't need to square both sides, you can write your equation like
y(y² - 1) = 2√3
You can easily see that y=√3 is a solution, so (y - √3) is a factor of the equation y³ - y - 2√3 = 0
So, using polynomial division, you get
(y - √3)(y² + √3y + 2) = 0
Now you can easily find the two remaining (imaginary) solutions
Where did you learn Math ??
Watched and learnt
Thank you