You should learn this trick. A great exponential equation! What do you think about this problem? If you're reading this ❤️ Hello My Friend ! Have a Great Day:) @higher_mathematics #maths #algebra
If you assume your audience knows what Lamberts W function is without any further explanation, I would think you can safely assume you don't have to spell out the initial divisions by x and 2 in painstaking detail.
WRONG! You have to do 3 cases : 1) If x>0 : ln(x²) = 2*ln(x) and x = exp(ln(x)). Then, you have an equation without any solution since -ln(8)/2 < -1/e, so impossible to apply W on both sides. 2) If x
Excellent work in clarifying! A little side note If I understand the W function correctly: I think in case 1 you can get infinite many complex solutions, since the
At the beginning is the equality ln(x^2)=2ln(x) which is correct only when x>0 and at the end this calculation gives x=--0,559..That does’nt seem rigorous, even if the solution is right.
You are absolutely right . Logarithms only result in real values for positive values . If you graph the two functions , you will see that there is only one solution and it is negative . I actually used Newton's method which is a numerical method to come up with a value to 12 decimal places ---> x = minus .559142092566 .
I agree. Mathematica calculates a complex result, which is not a solution. Lambert W function yields complex results for arguments < 1/e. This result has W(-1.039), an argument less than -0.367. As others say, starting the solution by taking logarithms is to head down the road to complex numbers
I didn't know this function too. So after proving there is only one solution with the functions graphics I'd rather started approximating the answer with bisection and gave an answer ~-0.5 ))
Honestly I would have just solve it graphically from the beginning. Then I would have tried to insert some values in the beginning equation to see what’s the interval in which X lays. For example I would have tried values like 0, 1, -1, -2 and so on. I would have gotten a interval and not an exact value, of course. But at least it would have been much easier and faster
You’re describing numerical methods, which are definitely the best way to solve this kind of thing. For example, bisection starts with an interval around the root and cuts the size of the interval in half at each iteration. I would recommend looking up bisection, Newton’s method, and fixed point iteration if you want to know more, and definitely consider taking numerical analysis in college if you get a chance.
@@louiscarl7629 nah thx I’d like to do chemistry in university. I know a bit of maths because my high school is based on science, and maths is always taught.
Lamberts function is such jenious and all. I always wondered though, if there are real life problems that this function is supposed to solve (something more than Olympiad problems, I mean). Is there an engineering - per instance - problem that deals with such exponential functions?
this is like a textbook trivial application of lambert-w. Why did you even make this video if THAT was the thing you wanted to show? Was the purpose of this to just raise awareness of the lambert-W function? What on earth does this have to do with the IMO?
I think the creator of this video missed the point of this problem: you have to show that there is no closed-form solution for problems like this. Actually, you can only write the solution with Lambert's W function or in another way using functions that can't be expressed with a closed-form expression. This is the important thing you have to learn about problems, where a variable is in the basis and exponent at the same time.
I don't think you've answered the question. If you can just denote any zero of the function ln(x)/x - a by Xa and then tell the solution as a function of Xa, then you have not really answered the problem !
How about… 1) When X represents a number being multiplied by an exponent, then X = 8 (so X^2 = 8^2 = 64) 2) When X represents a number’s exponent, then X = 2 ( so 8^X = 8^2 = 64) Maybe I’m not answering the question, maybe I am… but prove me wrong if I’m not lol
Love the explanations thanks for your video ❤ just one feedback it’s not said “NAYtural” - Although the word Nature is distinctly so, the Na in ‘NAtural’ is pronounced same way as NASA, or Na X.. in russian 😇
My biological science ass would just graph and find where they intersected. I like math and I took all the way to calculus 3 at the college level but I honestly never have to use it anything except algebra in my job.
I thought it would be over at 2:11; lnx=1/x and over x = ln8/2. But the x would disappear and there would be no solution to it. I think this eqaution is not solvable.
Because (-ln8/2) is less than (-1/e), W(-ln8/2) is a complex number = -0.291 + 1.36 i. Can you show how you solved for x? e^(real number) cannot be negative.
Это не красивое уравнение. И не красивый ответ Просто посмотрев на него становится понятно, что х отрицательная величина. И она приблизительно -1/2 . Решил?
That 'ln' kinda looks like 'e sub n' just saying. At a glance it would appear confusing to the uninitiated Math student....but nice methodology all the same!
dawg, what you wrote is 64^2=8^4 when the question is x^2=8^x. tragically it doesnt work because 64!=4. im p sure lambert W function is the only way to answer this.
