Would love to see a everyday real world application of all the math I learned in high school and college. I am in IT and I have never used any of it in 30 years.
@@Luke0193: Compound interest rates use math for geometric series. You probably use a fancy built-in function done by a financial calculator or spreadsheet software.
Спасибо за познавательное видео уровнения такого уровня были когда поступал в технический университет в России в 1999 году. Тренировался считал десятками такие. Сейчас уже конечно подзабылись формулы, но с учебником думаю решил бы и сейчас.
You are a person with a good knowledge of the laws of indices , and logs, loved all this at secondary school 45 years ago bit have forgotton some of the rules so get lost near the end , lol, it pays to know these rules and got me out of many a problem in my engineering career , esp algebraic rules
2^x = u. u^3 + u - 30 = 0. Factor: (u - 3)(u^2 + 3u + 10). Real solution u = 3. Quadratic on the second term for the two complex solutions. Back substitute. Messy problem.
If complex numbers are allowed as solutions, then there are many more solutions, namely x=(log(3)+2*k*π*i)/log(2) and x=(log(10)/2+(i*(±atan(sqrt(31)/3)+(2*k-1)*π)))/log(2) for every integer k.
Complex numbers are always allowed in my world. However, this isn't a very interesting expansion of the solution set, as it just says that running around a circle an integer number of times will return you to your starting point. Exp[(2×pi×k + c)×i] = Exp(c×i) for any integer k.
Close, but no cigar. You mixed up the bases for the log functions, if the base for the log function isn't explicitly e. All the real components have a form of log(r)/log(2), regardless of base {such as 2, e, or 10}, as they all yield log₂(r). However, the imaginary component is log₂(e^(i*θ)), which is simplified as ln(e^(i*θ))/ln(2) = i*θ/ln(2) . You wrote the equivalent of i*θ/log(2), but this log function is traditionally base 10.
17/1/24 ( Wednesday ) 9.15 pm. Sorry, I failed maths in school. I thought it was a primary school algebra. Never thought that in order to find the value of X, you need to create a value of Y with log included, which I could not comprehend. I was wrong to think that the value of X was 3. Instead, after so many steps, the value of X was another equation.....no wonder I failed Maths.
X value will be between 1 and 2. Simple solution. Lets take x value 1 then you will get 10 , let’s take x value as 2 now you will get 64 plus 4 that is 68 . 30 is between 10 and 64 so x value will be between 1 and 2. In India we get such questions as multiple choice answer questions and we get less time to solve.
I don't think that anyone who was worried about the music has the patience for the math. Just my opinion. Thank you so much. That was a lot of work. When you're buried in the math, anything else should fade to oblivion.
I'm amazed at how many people get an answer of 3 while completely misunderstanding the nature of logs or that x is a power. The right answer for the wrong reason is the wrong answer.
This was definitely the long way in my opinion. I definitely forgot the other way tho. Too many things you just have to know like 2^3 is 8 and breaking 30 in to 27 and 3. M.
The thumbnail got me very intrigued, and I accepted the challenge. I solved it on my own, and I managed to find x = 3 in a much simpler way. I still applied substitution, t = 2^(x) , so x = log_2(t). But when I reached t^(3) + t = 30, i scomposed left and right side into t (t^(2) + 1) = 2*3*5 And 2*3*5 = 2 (3*5) = 3 (2*5) = 5 (2*3) , so I had to choose the only combination that would match the left side, and the only one that matches is 3 (2*5). In fact, 3 * (3^(2) + 1) = 30. So t = 3, and x = log_2(3) . But then, when I confronted it to the solution you gave, I realized that I found only the real solution. Was it sheer luck because logarithm is defined only for real numbers, or my solution had some sense?
soooo X = 3 as in 8x3 (24) plus 2x3(6) equals 30???? ....... at least that's how I did it in my head....took about 5 seconds maybe less 🤔 unless I'm reading the problem wrong...
Is this the easiest way? So I prefer to study integral calculus, differential calculus, theoretical physics because they are much easier than that (contains irony).
Dude you could've just applied the remainder theorem when you had the equation y^3 + y = 30 or y^3 + y - 30 could only happen when y=3 and then applied the logs.
Wow, I'm in finance and have used algebra in many situations. I don't think I'd have any use for this one. I wasn't expecting the solution to be this long.
Howzit! Monthly mortgage computations are messy, and often require high-precision, iterative calculations and adjustments due rounding to the nearest cent or penny, and lots of decimal places for simple or compound interest. The calculations are messy because the idea value doesn't fit to a perfect dollars and cents value. This introduces errors that must be dealt with in an empirical manner. I wrote a program to find the monthly mortgage payment. It finds the idea value, then rounds that and recalculates every monthly balance so that the final payment for each year takes into account the accrued errors. The math we use in these videos don't have these empirical aspects, and are "cleaner" to calculate.
@@oahuhawaii2141 hmmm. sorry brah, have to disagree with "high-precision, iterative calculations". Any online loan calculator will get you close enough, +/- $50 or so. Excel's native loan ss will also do the job... entered the basic loan figures from mortgage brokers and I'm always within $10 of their proposed payment.
There are 3 principal solutions for x. One is real, which is shown, and two are complex conjugates, which were discarded. In truth, there are an infinite number of solutions, and all but one are complex numbers. 8^x + 2^x = 30 (2^x)^3 + 2^x - 30 = 0 Let y = 2^x: y^3 + y - 30 = 0 (y - 3)*(y² + 3*y + 10) = 0 y = 3, (-3 ± i*√31)/2 For y = 3: 2^x = 3 x = log₂(3) = log(3)/log(2) ≈ 1.5849625007211561814537389439478 --- Actually, e^(i*π*2*k) = 1, for integer k, so we can generalize: 2^x = 3*e^(i*π*2*k) ln(2^x) = ln(3*e^(i*π*2*k)) x*ln(2) = ln(3) + i*π*2*k x = (ln(3) + i*π*2*k)/ln(2) x = log₂(3) + i*π*2*k/ln(2), k integer For y = (-3 ± i*√31)/2: 2^x = (-3 ± i*√31)/2 = r*e^(i*θ) { polar form } r = √((-3)² + (√31)²)/2 = √10 θ = π*(2*k+1) ± atan(√31/3), k integer ln(2^x) = ln(r*e^(i*θ)) x*ln(2) = ln(r) + i*θ x = ln(r)/ln(2) + i*θ/ln(2) x = ln(√10)/ln(2) + i*[π*(2*k+1) ± atan(√31/3)]/ln(2) x = log₂(10)/2 + i*[π*(2*k+1) ± atan(√31/3)]/ln(2), k integer
Why do y sub instead of take the natural log or log of each piece immediately? xlog8 + xlog2 = log30 and go from there? X(log8+log2)=log30 X=(log30)/(log16)
@@marianneoelund2940@marianneoelund2940 Thanks, I understand that, but no other (y-3) on the left is factored/cancelled. I get the 9+ 1 afterwards but.......
@@marianneoelund2940 Yes, it's been 40 years since doing that, forgot that. And to think I was doing integral calculus now forgetting stupid things like that. Thanks.
i would of been out of paper after that equation. Unless you are a scientist developing theories of flux capacitors majority of people will not need this kind of math.
You would be correct if the question were 8x + 2x = 30 8(3) + 2(3) = 30 24 + 6 = 30 x = 3 That's not the question being asked here though, rather it's 8^x + 2^x = 30 Let's see what happens when we plug-in 3 8^(3) + 2^(3) = 30 512 + 8 = 30 X obviously does not equal 3 It's deceptively a much harder problem than it looks. The only thing I knew going into this was x > 1 8 + 2 = 10 x < 2. 64 + 4 = 68 After having watched this video, that's still pretty much all I know, so don't feel bad if this went over your head because I'm right there with ya.
This is 3x longer as it should be. And I guess the goal was to make it over 10 minutes mark. At one point she wrote exactly the same line as the one before, not a single thing was changed... 4:00 mark So,yeah, it's just milking the algorithm at this point. Despicable!
