Bit of work gives x ^ ( -3) = 8^2 x^3 = 1/( 2 ^6) x = 1/4
@macumbonanga7935Ай бұрын
Impossible de ver o que estás a escrever porque tu não se posiciona bem a frente do quadro.
@user-pu5dd8tf7cАй бұрын
To properly solve an equation we must find, or at least check, all solutions. Cubic equations have up to 3 solutions. Here is how to correctly derive all of them: log₈(x) - 4⋅log₈(x) = 2 log₈(x) - log₈(x⁴) = 2 log₈(x/x⁴) = 2 log₈(1/x³) = 2 1/x³ = 8² = 64 = 4³ x³ = ⟮¼⟯³ x³ - ⟮¼⟯³ = 0 Simple solution: x = ¼ Factor the simple solution out: (x - ¼)(x² + ⟮¼⟯x + ⟮¼⟯²) = 0 x = (-¼ ± √(⟮¼⟯² - 4 ⋅ ⟮¼⟯²)) / 2 x = (-¼ ± √(-3 ⋅ ⟮¼⟯²)) / 2 x = (-¼ ± ¼√⟮-3⟯) / 2 | we can stop here if x ∈ ℝ, because √⟮-3⟯ is not a real number x = (-1 ± i√3) / 8 Answers: x = ¼ x = (-1 + i√3) / 8 x = (-1 - i√3) / 8
@ericasi45623 ай бұрын
Это точно учитель математики? Это ж надо из простейшего уравнения столько наваять. И почему аргумент лагорифма вы пишете как показатель степени логарифма? Как вы будете писать, если логарифм будет в степени ? Куда в таком случае показатель напишете?
@jensraab29026 ай бұрын
After watching two of your videos, I realize that I'm clearly not your target audience, and that your didactic approach is not the one I'm used to. You write down extremely many steps but maybe that helps your audience. That said, it seems to me that you make things overly complicated in this video: 1. Is it really necessary to formally factorize in the first step? We have log₈x - 4 log₈x but if you picture the log₈x as one unit, you simply have a difference in the form of A - 4A. Surely, any student who's expected to handle logarithms will be able to understand that if you have one unit of something and subtract 4 units, you end up with -3. So you could go directly from log₈x - 4 log₈x to -3 log₈x without factorization. 2. At the end you have x⁻¹ = 4 and cross multiply. Again, a student at a level to deal with logarithms should be able to simply take the reciprocal of both terms and go directly to x = ¼, right? 3. The transformation you suggest starting at 3:20 is not the one I'd have gone for. Rather than moving the -3 in the exponent of x, why not just move it to the right side of the equation and continue with smaller numbers: -3 log₈x = 2 log₈x = -⅔ x = 8^(-⅔) You can then rewrite 8^(-⅔) as ∛(8⁻²) or (∛8⁻¹)², whatever you like best and further transform it into (∛⅛)² or ∛(⅛)². I'd suggest to first extract the cube root to keep numbers small: (∛⅛)² = (½)² = ¼ And that's it! Easier than manipulating 64. But if you have 64, I'd suggest to rewrite it as 2⁶ because then you can extract the cube root so much easier by simply dividing 6 by 3 and get directly to: ∛(2⁶) = 2² = 4.
Пікірлер: 10
Nice one
Good one.
Thank you for the video on Exponential Logarithm.
@mechnics-maths
5 ай бұрын
Glad it was helpful!
❤
Bit of work gives x ^ ( -3) = 8^2 x^3 = 1/( 2 ^6) x = 1/4
Impossible de ver o que estás a escrever porque tu não se posiciona bem a frente do quadro.
To properly solve an equation we must find, or at least check, all solutions. Cubic equations have up to 3 solutions. Here is how to correctly derive all of them: log₈(x) - 4⋅log₈(x) = 2 log₈(x) - log₈(x⁴) = 2 log₈(x/x⁴) = 2 log₈(1/x³) = 2 1/x³ = 8² = 64 = 4³ x³ = ⟮¼⟯³ x³ - ⟮¼⟯³ = 0 Simple solution: x = ¼ Factor the simple solution out: (x - ¼)(x² + ⟮¼⟯x + ⟮¼⟯²) = 0 x = (-¼ ± √(⟮¼⟯² - 4 ⋅ ⟮¼⟯²)) / 2 x = (-¼ ± √(-3 ⋅ ⟮¼⟯²)) / 2 x = (-¼ ± ¼√⟮-3⟯) / 2 | we can stop here if x ∈ ℝ, because √⟮-3⟯ is not a real number x = (-1 ± i√3) / 8 Answers: x = ¼ x = (-1 + i√3) / 8 x = (-1 - i√3) / 8
Это точно учитель математики? Это ж надо из простейшего уравнения столько наваять. И почему аргумент лагорифма вы пишете как показатель степени логарифма? Как вы будете писать, если логарифм будет в степени ? Куда в таком случае показатель напишете?
After watching two of your videos, I realize that I'm clearly not your target audience, and that your didactic approach is not the one I'm used to. You write down extremely many steps but maybe that helps your audience. That said, it seems to me that you make things overly complicated in this video: 1. Is it really necessary to formally factorize in the first step? We have log₈x - 4 log₈x but if you picture the log₈x as one unit, you simply have a difference in the form of A - 4A. Surely, any student who's expected to handle logarithms will be able to understand that if you have one unit of something and subtract 4 units, you end up with -3. So you could go directly from log₈x - 4 log₈x to -3 log₈x without factorization. 2. At the end you have x⁻¹ = 4 and cross multiply. Again, a student at a level to deal with logarithms should be able to simply take the reciprocal of both terms and go directly to x = ¼, right? 3. The transformation you suggest starting at 3:20 is not the one I'd have gone for. Rather than moving the -3 in the exponent of x, why not just move it to the right side of the equation and continue with smaller numbers: -3 log₈x = 2 log₈x = -⅔ x = 8^(-⅔) You can then rewrite 8^(-⅔) as ∛(8⁻²) or (∛8⁻¹)², whatever you like best and further transform it into (∛⅛)² or ∛(⅛)². I'd suggest to first extract the cube root to keep numbers small: (∛⅛)² = (½)² = ¼ And that's it! Easier than manipulating 64. But if you have 64, I'd suggest to rewrite it as 2⁶ because then you can extract the cube root so much easier by simply dividing 6 by 3 and get directly to: ∛(2⁶) = 2² = 4.