Which shows that you are NOT a mathematician. This method works if you can split 30 into two numbers x and y, such that x + y = 30 and 3^x = y (ie. 3 +27 = 30 and 3^x = 27), where x must be the solution you are trying to calculate. In other words, you need the solution of the problem in order to split 30 and use this complicated method… to find that number you just used. For example, if you try to solve 3^x + x = 40, you just need to split in your head 40 into 3.279887… and 36.72011..., both with infinite decimals, and since 3.279887… + 36.72011... = 40 and 3^3.279887… = 36.72011..., you can use this method to find the solution, which is 3.279887…, the number you used to find that very number. Brilliant method!
Everyone is acting like what he did in this video uses unnecessary steps, but the goal of this video is not just to teach you how to solve 3^x+x=30, it is to teach you how to use the Lambert function to solve for x in similar situations.
@@Math_Minds So, why choose an example that can easily be solved by inspection? Besides, I had a dual major in math and physics, and have worked in an vocation that uses various math functions for 40+ years - I had never heard of the "lambert W function" and never had a need to use it to solve anything. It is usually very easy to solve by inspection if the result is going to be an integer, and if not, you will have to do a numerical "solution" anyway, so might as well solve it with iteration, which can get you to about 3 significant figures in maybe 3-4 passes, perfectly practical to solve on paper and faster than the solution shown here.
@@brettbuck7362Yeah, the very basic of numerical analysys is plotting and interpolating. If you know its is integer and less than m and greater than n, you wont need even m-n steps to check, just logarithm of it and checking if it is less or more. A dumb computer will halve the possibilities to reduce worst and avarage time complexity, but human intuition will give an better avarage time complexity.
Moving x to the right side, we get 3^x = -x+30. The exponential function is increasing while the linear function is decreasing and there is only one intersection. It's obvious that x=3 is the solution.
In Russia I haven't even been told about the Lambert W Function. Instead I've been taught multiple times about Gauss functions, and many other fundamentals of mathematical framework in many aspects (calculus, vector calculus, complex analysis, tensor analysis)... but W-function is what I've never heard of before...
Define f:R -> R by f(x) = 3^x + x - 30. Note that f is differentiable (and thus, continuous) on (-inf, inf) and f(3) = 0. Note that f '(x) = 3^x ln 3 + 1 > 0 on (-inf, inf) since 3^x > 0 on (-inf, inf). Suppose by contradiction there exists a in R such that f(a) = 0 and a != 3. If a < 3, then by the Mean Value Theorem there exists c in (a, 3) such that f '(c) = [f(3) - f(a)] / [3 - a] = 0, but this is a contradiction since f '(x) > 0 on (-inf, inf). We can use the MVT to show that a > 3 also leads to a contradiction. Therefore, f has one root at x = 3.
It is simpler to solve the problem using modular algebra. Take modulo 3: mod(3^x,3)+mod(x,3)=mod(30,3) [mod(3,3)]^x+mod(x,3)=0 0+mod(x,3)=0 --> mod(x,3)=0 It means that x=3k where k is any integer. Plugging it back to the given equation: 3^(3k)+3k=30 Divide by 3: 3^(3k-1)+k=10 =3²+1 Comparing both sides we get k=1 Thus x=3. As check: LHS=3^x+x=3³+3 =27+3 =30 equals to RHS
@@voicutudor7331: Taking modulo 3: mod[(3^x+x),3]=mod(30,3) As mod[(a+b),k]=mod(a,k)+mod(b,k) and 30 is divisible by 3 then mod(3^x,3)+mod(x,3)=0 As mod(a^n,k)=[mod(a,k)]^n thus [mod(3,3)]^x=mod(x,3) As mod(3,3)=0, [mod(3,3)]^x=0^x and 0^x=0 for any x, not necessarily x is natural number, giving mod(x,3)=0.
@@voicutudor7331: I want to know your comment to the following statements: 3^x+x=30 Take modulo 3 to yield mod[(3^x+x),3]=mod(3^x,3)+mod(x,3) =[mod(3,3)]^x+mod(x,3) =0^x+mod(x,3) =mod(x,3) as 0^x As mod(30,3)=0 then mod(x,3)=0 Is 0^x=0 true only for x a natural number? You'd better consult to someone who you consider good mathematecian.
@@nasrullahhusnan2289 bro your math is fucking trash 3^x mod 3 if x is not a natural number doesn t exist, take 3^1.5 and see wtf is going on. stop with this 5th grade approach
This method works if you can split 30 into two numbers x and y, such that x + y = 30 and 3^x = y (ie. 3 +27 = 30 and 3^x = 27), where x must be the solution you are trying to calculate. In other words, you need the solution of the problem in order to split 30 and use this complicated method… to find that number you just used. For example, if you try to solve 3^x + x = 40, you just need to split in your head 40 into 3.279887… and 36.72011..., both with infinite decimals, and since 3.279887… + 36.72011... = 40 and 3^3.279887… = 36.72011..., you can use this method to find the solution, which is 3.279887…, the number you used to find that very number. Brilliant method!
Because of the sum (30), x should be either 2 or 3. As 3^2+2=11 and 3^3+3=30, then by inspection x = 3.. It won't take more than 1' to get the solution. You can get there easily just by trial..
3 to the power of 3 = 27 + 3. Pretty easy X = 3. Dont need a big ass page explaining it all re writing it multiple times. Just work through to the powers 3x0, 3x3, 3x3x3 = Remainder left 3.
As a generalization, the method presented always works if in the equation a^x + x = b one has b = a^a + a, with a solution x = a. Try it out with some numbers or repeat the derivation as in the video.
No need completely to introduce Lambert's function. If (30-x) · 3^(30-x) = 3^30 from a simple algebraical transformation, as shown on the screen, so (30-x) · 3^(30-x) = 3^3 · 3^27, so (30-x) · 3^(30-x) = 27 · 3^27, so (30-x) = 27, so x = 3, end. The function is very interesting and useful (see even Wikipedia for examples), but not here.
X=3 Bc what's the number that power the number 3 and the addition number It's coming by 3³ and this equal to 27 but 27+3 = 30 So 3x +x =30 is 3³ + 3 =30 So x=3
Solution: 3^x+x = 30 = 3^3+3 |the same operations are done with x on the left side of the equation as with 3 on the right side of the equation, therefore x = 3.
cách giải này không chặt chẽ. Nếu bài toán với x là 1 số thập phân thì sẽ rất khó để đoán được nghiệm của nó. Thậm chí có những bài toán với rất nhiều nghiệm, do đó suy đoán sẽ bỏ xót đáp án. Cách giải này là không có cơ sở để chứng minh nó tuyệt đối. Điều bạn cần là tìm ra nghiệm của bài toán bằng cách chứng minh nó là đúng một cách chặt chẽ.
@@gelbkehlchencó thể dùng đạo hàm, hoặc là áp dụng các công thức toán học khác. Hoặc ít nhất là phương pháp như video. Đó là những phương pháp để chứng minh được nghiệm của bài toán, nó không bỏ xót nghiệm của bài toán. Cách chẩn đoán không thể giải quyết được các nghiệm chặt chẽ, hoặc nếu bài toán đưa ra là số hữu tỉ, chúng ta sẽ khó mà đoán được nghiệm. Và trong toán học là bạn phải chứng minh nó đúng một cách chặt chẽ được công nhận. Toán học không thể sử dụng phương pháp "đoán mò"
"w" is the so-called Lambert W Function (see the title of this video). He did not multiply by "w", he applied the Lambert W Function to the expression on each side of the equation. The Lambert W Function is the inverse of the function f(x) = x*(e^x) . So if y = x*(e^x) , then W(y) = W( x*(e^x) ) = x . [*] Note: the Lambert W Function is usually written with a capital letter W, not with a lowercase w . [*] Actually, the Lambert W Function is the "principal branch" of the equation x*(e^x) = y ; multiple values of x can lead to the same value of y, but the Lambert W Function W(y) only returns the principal root of the equation.
since 3^x and x are both increasing functions the function f(x)=3^x+x is in an increasing function. f(0)=10 for positive x, 3^x+x=30 ln(3^x+x)=ln(30) x*ln(3)*ln(x)=ln(30) x*ln(x)= ln(27) x*ln(x)=3ln(3) x=3
Btw x is congruent to zero mod three, so you can rewrite it as 27^x+3x=30, and immediately we see using the canonical homomorphism between Z[3] and Z/3Z, that the new x must be one, and the original must have been 3.
Giải phương trình :3^x+x=30 +) Tập xác định:D=R +) Phương trình trên tương đương với phương trình sau: 3^x+x-30=0 Xét hàm số f(x) =3^x+x-30, x∈R =>f'(x) =(3^x)ln3+1>0∀x =>f(x) đồng biến trên R =>phương trình f(x) =0 có tối đa 1 nghiệm trên tập số thực +) Mà f(3) =0 =>X=3 là 1 nghiệm duy nhất của phương trình f(x) =0 Kết luận: tập nghiệm của phương trình là S={3}
30 is divisible by 3. Assuming that X is a whole number then 3^X is divisible by 3 which means that X must also be divisible by 3. As 30 is small I tried 3 and it worked.
я придумал два способа немного легче... 1 способ: перенести х вправо и нарисовать две функции у=3^х и у=30-х, найти пересечение и всё 2 способ: снова перенести х и получить: 3^х=30-х, увидеть, что одна функция возрастает, а другая - убывает, следовательно, есть хотя бы одно решение и методом подбора нашёл, что х=3
At 7:00, how did you decide to break the exponent 30 (of 3^30) into 3+27 ? Isn't that insight actually the same as solving the original equation 3^x + x = 30 on-sight: x = 3 ? 3^x + x = 30 I'm looking at the 30 on the righthandside, I want to break that up into two parts, one part that equals 3^x and another part that equals x ; and I immediately see that 27 + 3 will do the trick. No long derivation nor application of Lambert W Function needed! How is what you did at 7:00 - 7:21 any different?
@@bikramdev007 He hasn't answered my question, but as far as I can see, his calculation depends on "guessing" (as you call it) in order to arrive at the (simplified) answer that he eventually got (although granted, he could still have presented a non-simplified answer if he hadn't broken up 30 into 3+27). Well, what stops me then from "guessing" (I'd rather call it "using insight" than "guessing", but have it your way) that in the equation 3^x + x = 30 , I can break up the righthandside into 27 + 3 = 3^3 + 3 , and hence I see that the lefthandside matches if x = 3 ? In the next step, I'll then deliver _proof_ by entering the "guessed" x=3 back into the original equation and see that it works out; that is proof. So we don't actually need to fuddle with Lambert W Function etc., this approach of mine is much shorter and just as good, because _it is based on the same insight_ (oh, pardon, i mean "guess").
@@vitorrj4211 As another commenter already said, in this particular case, it's easily seen that the function f(x) = 3^x + x is a continuous and increasing function, and hence there is only one solution to the equation f(x) = 30 (in the real domain). By the way, how did his method prove that there are no complex solutions? The Lambert W Function W(y) is only the _principal_ branch of roots to the equation y = x^(e^x) . He chose to break up the exponent 30 into 3 and 27 ; who says there is no solution when breaking up 30 into z and (30-z) where z is complex-valued?
There is a simple solution that giving x an integer from 1 to 3. You can find the answer in three seconds. But if you must prove it, this equation may be solved by your way.
f(x)=3^x g(x)=-x+30 f(x) monotonously increasing, g(x) it decreases monotonously, but knows if there is only one solution. In this case, it is obvious x=3
In equations with obvious roots like this one we can work more simply and faster as I immediately show with 2 ways of solution in R. 1st Way • If x < 0, then 3^x < 3^0 =1, so 3^x + x < 1 + 0 = 1 and the equation has no root < 0 • If x > 3 then 3^x > 27, so 3^x + x > 30 so the equation has no root > 3 • If 0 < x < 3 then 3 ^ x < 27, so 3^x + x < 30 so the equation has no root < 3 • If x = 3 the equation is true, so the unique root of the equation is x =3 2nd Way We will work with the monotonicity of the function f(x) = 3^x + x - 30 If x1 , x2 E R with x1 < x2, then 3^x1 < 3^ x2, so 3^x1 + x1 -30 < 3^x2 + x2 -30 i.e. f(x1) < f(x2) , so f is strictly increasing. Therefore if it has a root, it is unique. Since the equation has the root x = 3, this is also its only root. The monotonicity of f can be easily proven with the help of derivatives. Indeed, f΄(x) = (3^x)ln3 + 1 > 0, so f is strictly increasing, etc. I would like to take this opportunity to announce that I have created a mathematics channel on You Tube called L+M=N My channel is at www.youtube.com/@Nikos_Iosifidis I will be happy if you subscribe to my channel and comment on the solutions of the topics I present.
The answer is three. These small examples are great for learning lambert W, but it’s an unnecessary bit of math baggage when the trial substitution is the best approach.
Wow, what is W function? First time seeing one. (It's been 30 years since highschool graduate, and never heard of it even in college freshman math class. Is W a basic term these days?)
What everyone is failing to understand is the video is not supposed to be about solving the equation easily, but rather be able to solve an exponential-linear function with the help of W Lambert Function. The function is useful for a lot of other problems. Can you not understand that he tried to start off with an easier problem to explain the practical use of the function.
@@Math_Minds "There may be many other roots" The same goes for your method. How did you know at 6:20 that W( (30-x)*ln(3) * e^[(30-x)*ln(3)] ) = (30-x)*ln(3) , unless you already knew the value of x ? Note: the Lambert W Function W(y) is the _principal_ root of the equation y = x*(e^x) , because several different values of x can lead to the same value of y . As a comparison: what you did is equivalent to saying √((x-7)²) = 2x+1 ⇒ (x-7) = 2x+1 ⇒ x = -8 How would we know that the lefthandside would reduce to (x-7) ? This is wrong (as can be seen when plugging the found result back into the original equation), the actual solution is √((x-7)²) = 2x+1 ⇒ (7-x) = 2x+1 ⇒ 6 = 3x ⇒ x = 2
@@Math_Minds I am sorry I did not remember the properties of log correctly and wrote log(30-x) = log30/logx. My solution was wrong. The correct property was log x- log y= log(x/y)
Have you tried using the exhaustive method? The cube of 3 is 27, so X ≤ 3. The answer to the problem is between 0 and 3. Okay, the problem has been resolved.
Yes. When I was studying in college, I felt helpless when I encountered difficulties in math exams. I always wonder why I need to face such complex mathematical problems. I won't be able to use calculus or Newton's XXXX formula in my future life. What is the significance of learning this? However, my cousin is proficient in mathematics. He is an engineer working for China's satellite and rocket systems.@@pascal_0840
Sure, this one was easy to solve because it's using low numbers and x is a whole number. But if x was a rational or imaginary number, we'd need a formula to solve it. That's why the demo
Nice steps I used the following: 3^x + x =30 3^x = 30-x (ln both sides) xln3 = ln (30-x) xln3 = (ln30 / ln x) then xlnx = ln 30 / ln 3 xlnx = ln (30-3) xlnx = ln 27 xlnx = 3 ln 3 x =3 or by using W function ln x e^(ln x) = ln 3 e ^(ln 3) W (ln x e^(ln x)0 = W (ln 3 e ^(ln 3)) then ln x = ln 3 e^ (lnx) = e ^ (ln3) therefore X= 3 Thanks
Is x= 1? No Is x = 2? No Is x=3? YES. Solved in 15 seconds. It took me longer to write this then solve the equation. Then I sat through 10 minutes of this plus ads
to show application of lambert function this is ok. But, why this simple question was chosen. Is it mis use of lambert function. I too use results and adhoc arguments in my channel but why so lengthy??
3^x+x = 30. Just looking at this, it is clear that x cannot be greater than 3, given the constraint of 30. And x cannot be less than 3, because even at 2.999, 3^x + x will be less than 30. So, x has to be 3 👍
@@lmfao3293 Buddy, for any value less then 3 it will never add up to 30. It will always be less than 30. Max value for x = 2.99999, would result in 3^2.99999 < 27. So, when you add them, it can only be less than 30. Glad you made such a bad argument. Makes me look even more smart 🧐🧐🧐🤣🤣🤣
@@lmfao3293 So, when you don't understand the calculation, you call it guess? Well, I don't blame you. With the level of intelligence demonstrated, we cannot expect more..👍
My math professor said that if you use more than 4 lines to solve an equation, you don't understand the music of the universe. You might understand the math, but not what makes numbers sing
3^1 + 3 = 6 3^2 + 3 = 12 3^3 + 3 = 30 done I'm guessing the intent of this video was to show a general method of solution as opposed to a practical way to solve this particular problem.
That's great! A strictly consistent proof by representing the left and right sides of the equation in a form convenient for taking the Lambert function, instead of an intuitive, although obvious solution by representing the number 30 in the form of 3^3 + 3 !!!
@@levskomorovsky1762 I wrote "but what's the purpose of the triple factorial? (jk)" where "(jk)" is a common convention for abbreviating "Just Kidding". It was basically a dad joke.
We can just notice that 3power(3)+3=30. Since the function x------3power(x)+x is increasing. Then 3 is the unique solution.
Yeah that's exactly what I thought
I dont think so, can u explain plz?
27 +3 =30.
3^3 =27
Unique on real numbers
that is not math way
Only mathematicians appreciate this elegant solution! Nice introduction to Lambert function.
Thank you.
Which shows that you are NOT a mathematician.
This method works if you can split 30 into two numbers x and y, such that x + y = 30 and 3^x = y (ie. 3 +27 = 30 and 3^x = 27), where x must be the solution you are trying to calculate. In other words, you need the solution of the problem in order to split 30 and use this complicated method… to find that number you just used.
For example, if you try to solve 3^x + x = 40, you just need to split in your head 40 into 3.279887… and 36.72011..., both with infinite decimals, and since 3.279887… + 36.72011... = 40 and 3^3.279887… = 36.72011..., you can use this method to find the solution, which is 3.279887…, the number you used to find that very number.
Brilliant method!
This is my approach: Plot a graph using y=3^x and y=30-x, the solution can be found where the curves meet. Simple and easy.
your eyes are not a proof, can use them for making a guess, but then need to plug in and verify
3^x + x = 30
3^x + x = 3^3 + 3
X=3
... but you may be asked to solve it analytically
@@wellington2779 Well this is useless. What is wrong with you?
@@wellington2779 this is not a proof
Everyone is acting like what he did in this video uses unnecessary steps, but the goal of this video is not just to teach you how to solve 3^x+x=30, it is to teach you how to use the Lambert function to solve for x in similar situations.
Exactly... You are right.
@@Math_Minds So, why choose an example that can easily be solved by inspection? Besides, I had a dual major in math and physics, and have worked in an vocation that uses various math functions for 40+ years - I had never heard of the "lambert W function" and never had a need to use it to solve anything. It is usually very easy to solve by inspection if the result is going to be an integer, and if not, you will have to do a numerical "solution" anyway, so might as well solve it with iteration, which can get you to about 3 significant figures in maybe 3-4 passes, perfectly practical to solve on paper and faster than the solution shown here.
@@brettbuck7362Yeah, the very basic of numerical analysys is plotting and interpolating. If you know its is integer and less than m and greater than n, you wont need even m-n steps to check, just logarithm of it and checking if it is less or more. A dumb computer will halve the possibilities to reduce worst and avarage time complexity, but human intuition will give an better avarage time complexity.
Moving x to the right side, we get 3^x = -x+30. The exponential function is increasing while the linear function is decreasing and there is only one intersection. It's obvious that x=3 is the solution.
Just simply draw a coordinate graph and the two funtions.
Well there are imaginary solutions
what are they?@@windowsxpmemesandstufflol
@@windowsxpmemesandstufflolaaaaaaand no one cares
@@planomathandscience bastardized drunkard
3^x + x = 3^3 + 3
x=3
30 const ; 3^x + x is increasing func -> hence 1 intersection and 1 solution
selection method 3 root.
I hope my car navigation NEVER gives me the directions like this
In Russia I haven't even been told about the Lambert W Function. Instead I've been taught multiple times about Gauss functions, and many other fundamentals of mathematical framework in many aspects (calculus, vector calculus, complex analysis, tensor analysis)... but W-function is what I've never heard of before...
the same in Vietnam and many other countries =]]
По факту)
По факту)
Same in India and I'm a post graduate
I am korean and korean same ..I had to learn from youtube
Define f:R -> R by f(x) = 3^x + x - 30.
Note that f is differentiable (and thus, continuous) on (-inf, inf) and f(3) = 0.
Note that f '(x) = 3^x ln 3 + 1 > 0 on (-inf, inf) since 3^x > 0 on (-inf, inf).
Suppose by contradiction there exists a in R such that f(a) = 0 and a != 3. If a < 3, then by the Mean Value Theorem there exists c in (a, 3) such that f '(c) = [f(3) - f(a)] / [3 - a] = 0, but this is a contradiction since f '(x) > 0 on (-inf, inf). We can use the MVT to show that a > 3 also leads to a contradiction. Therefore, f has one root at x = 3.
It is simpler to solve the problem using modular algebra.
Take modulo 3:
mod(3^x,3)+mod(x,3)=mod(30,3)
[mod(3,3)]^x+mod(x,3)=0
0+mod(x,3)=0 --> mod(x,3)=0
It means that x=3k where k is any integer. Plugging it back to the given equation:
3^(3k)+3k=30
Divide by 3: 3^(3k-1)+k=10
=3²+1
Comparing both sides we get k=1
Thus x=3. As check:
LHS=3^x+x=3³+3
=27+3
=30 equals to RHS
this only works if x in natural
@@voicutudor7331:
Taking modulo 3:
mod[(3^x+x),3]=mod(30,3)
As mod[(a+b),k]=mod(a,k)+mod(b,k) and 30 is divisible by 3 then
mod(3^x,3)+mod(x,3)=0
As mod(a^n,k)=[mod(a,k)]^n thus
[mod(3,3)]^x=mod(x,3)
As mod(3,3)=0, [mod(3,3)]^x=0^x and 0^x=0 for any x, not necessarily x is natural number, giving mod(x,3)=0.
@@nasrullahhusnan2289 divisibility and mods are only well defined over integers
@@voicutudor7331:
I want to know your comment to the following statements:
3^x+x=30
Take modulo 3 to yield
mod[(3^x+x),3]=mod(3^x,3)+mod(x,3)
=[mod(3,3)]^x+mod(x,3)
=0^x+mod(x,3)
=mod(x,3) as 0^x
As mod(30,3)=0 then mod(x,3)=0
Is 0^x=0 true only for x a natural number?
You'd better consult to someone who you consider good mathematecian.
@@nasrullahhusnan2289 bro your math is fucking trash 3^x mod 3 if x is not a natural number doesn t exist, take 3^1.5 and see wtf is going on. stop with this 5th grade approach
This method works if you can split 30 into two numbers x and y, such that x + y = 30 and 3^x = y (ie. 3 +27 = 30 and 3^x = 27), where x must be the solution you are trying to calculate. In other words, you need the solution of the problem in order to split 30 and use this complicated method… to find that number you just used.
For example, if you try to solve 3^x + x = 40, you just need to split in your head 40 into 3.279887… and 36.72011..., both with infinite decimals, and since 3.279887… + 36.72011... = 40 and 3^3.279887… = 36.72011..., you can use this method to find the solution, which is 3.279887…, the number you used to find that very number.
Brilliant method!
Because of the sum (30), x should be either 2 or 3. As 3^2+2=11 and 3^3+3=30, then by inspection x = 3.. It won't take more than 1' to get the solution.
You can get there easily just by trial..
3 to the power of 3 = 27 + 3. Pretty easy X = 3. Dont need a big ass page explaining it all re writing it multiple times. Just work through to the powers 3x0, 3x3, 3x3x3 = Remainder left 3.
As a generalization, the method presented always works if in the equation a^x + x = b one has b = a^a + a, with a solution x = a. Try it out with some numbers or repeat the derivation as in the video.
No need completely to introduce Lambert's function. If (30-x) · 3^(30-x) = 3^30 from a simple algebraical transformation, as shown on the screen, so (30-x) · 3^(30-x) = 3^3 · 3^27, so (30-x) · 3^(30-x) = 27 · 3^27, so (30-x) = 27, so x = 3, end. The function is very interesting and useful (see even Wikipedia for examples), but not here.
X=3
Bc what's the number that power the number 3 and the addition number
It's coming by 3³ and this equal to 27 but
27+3 = 30
So 3x +x =30 is
3³ + 3 =30
So x=3
3 to the power x +3=30
3 to the power x +3=27+3
3 to the power x= 27+3-3
3 to the power x = 27
3 to the power x = 3 to the power 3
Then X= 3......
Solution:
3^x+x = 30 = 3^3+3 |the same operations are done with x on the left side of the equation as with 3 on the right side of the equation, therefore x = 3.
Cách này tương tự mà hay hơn, song đều phải chứng minh rằng đây là nghiệm duy nhất mới chặt chẽ
cách giải này không chặt chẽ. Nếu bài toán với x là 1 số thập phân thì sẽ rất khó để đoán được nghiệm của nó. Thậm chí có những bài toán với rất nhiều nghiệm, do đó suy đoán sẽ bỏ xót đáp án. Cách giải này là không có cơ sở để chứng minh nó tuyệt đối. Điều bạn cần là tìm ra nghiệm của bài toán bằng cách chứng minh nó là đúng một cách chặt chẽ.
@@errornotfound_1972 Wie muss man es dann lösen?
@@gelbkehlchencó thể dùng đạo hàm, hoặc là áp dụng các công thức toán học khác. Hoặc ít nhất là phương pháp như video. Đó là những phương pháp để chứng minh được nghiệm của bài toán, nó không bỏ xót nghiệm của bài toán. Cách chẩn đoán không thể giải quyết được các nghiệm chặt chẽ, hoặc nếu bài toán đưa ra là số hữu tỉ, chúng ta sẽ khó mà đoán được nghiệm. Và trong toán học là bạn phải chứng minh nó đúng một cách chặt chẽ được công nhận. Toán học không thể sử dụng phương pháp "đoán mò"
@@errornotfound_1972 Was sind Derivate?
Someone please explain what the "w" is all about. What is "w" and why did he multiple both sides by it?
"w" is the so-called Lambert W Function (see the title of this video). He did not multiply by "w", he applied the Lambert W Function to the expression on each side of the equation.
The Lambert W Function is the inverse of the function f(x) = x*(e^x) . So if y = x*(e^x) , then W(y) = W( x*(e^x) ) = x . [*]
Note: the Lambert W Function is usually written with a capital letter W, not with a lowercase w .
[*] Actually, the Lambert W Function is the "principal branch" of the equation x*(e^x) = y ; multiple values of x can lead to the same value of y, but the Lambert W Function W(y) only returns the principal root of the equation.
You can graph the function f(x)= exp(x ln3) and g(x) =x-30 and see where f(x)=g(x)
3^x = 30 - x Since 30 - x must be > 0 -> x < 30
x = 3 verifies the equation. 3^x increases while 30 - x decreases, so 3 is unique solution.
Or you just eyeball it and instantly see that the (an?) answer is three. The only tricky part is testing whether there are additional solutions.
Well, I have to admit that you got me there. I never look for other solutions..
@@QUABLEDISTOCFICKLEPOLuckily, is easy to prove this is an increasing function, therefore any solution would be unique.
I went to the rest room to pee, before I could pee, I cleaned up all the house. accidently peed in my pants.
since 3^x and x are both increasing functions the function f(x)=3^x+x is in an increasing function. f(0)=10
for positive x, 3^x+x=30 ln(3^x+x)=ln(30) x*ln(3)*ln(x)=ln(30) x*ln(x)= ln(27) x*ln(x)=3ln(3) x=3
???????
@@dumitrudraghia5289 Math without guessing. LOG rules FTW.
Pl explain the letter w
You wrote. I did not understand.
Btw x is congruent to zero mod three, so you can rewrite it as 27^x+3x=30, and immediately we see using the canonical homomorphism between Z[3] and Z/3Z, that the new x must be one, and the original must have been 3.
Giải phương trình :3^x+x=30
+) Tập xác định:D=R
+) Phương trình trên tương đương với phương trình sau:
3^x+x-30=0
Xét hàm số f(x) =3^x+x-30, x∈R
=>f'(x) =(3^x)ln3+1>0∀x
=>f(x) đồng biến trên R
=>phương trình f(x) =0 có tối đa 1 nghiệm trên tập số thực
+) Mà f(3) =0
=>X=3 là 1 nghiệm duy nhất của phương trình f(x) =0
Kết luận: tập nghiệm của phương trình là S={3}
đúng là đạo hàm có thể không là công cụ nhanh nhất nhưng chắc chắn là công cụ mạnh mẽ nhất 😂
3が解になって左辺は明らかに単調増加関数だから3しか解になり得ないっていうのが一番楽
30 is divisible by 3. Assuming that X is a whole number then 3^X is divisible by 3 which means that X must also be divisible by 3. As 30 is small I tried 3 and it worked.
я придумал два способа немного легче...
1 способ: перенести х вправо и нарисовать две функции у=3^х и у=30-х, найти пересечение и всё
2 способ: снова перенести х и получить: 3^х=30-х, увидеть, что одна функция возрастает, а другая - убывает, следовательно, есть хотя бы одно решение и методом подбора нашёл, что х=3
解题步骤如下：因为3power(x)+x=30，所以，3power(x)
At 7:00, how did you decide to break the exponent 30 (of 3^30) into 3+27 ? Isn't that insight actually the same as solving the original equation 3^x + x = 30 on-sight: x = 3 ?
3^x + x = 30
I'm looking at the 30 on the righthandside, I want to break that up into two parts, one part that equals 3^x and another part that equals x ; and I immediately see that 27 + 3 will do the trick. No long derivation nor application of Lambert W Function needed!
How is what you did at 7:00 - 7:21 any different?
When you guess a solution, you get only one solution. What if you have more (maybe complex) solutions?
Maybe he is doing this because math doesn't only depends on the guessing. You have to show the proof besides guessing.
@@bikramdev007 He hasn't answered my question, but as far as I can see, his calculation depends on "guessing" (as you call it) in order to arrive at the (simplified) answer that he eventually got (although granted, he could still have presented a non-simplified answer if he hadn't broken up 30 into 3+27).
Well, what stops me then from "guessing" (I'd rather call it "using insight" than "guessing", but have it your way) that in the equation 3^x + x = 30 , I can break up the righthandside into 27 + 3 = 3^3 + 3 , and hence I see that the lefthandside matches if x = 3 ? In the next step, I'll then deliver _proof_ by entering the "guessed" x=3 back into the original equation and see that it works out; that is proof. So we don't actually need to fuddle with Lambert W Function etc., this approach of mine is much shorter and just as good, because _it is based on the same insight_ (oh, pardon, i mean "guess").
@@vitorrj4211 As another commenter already said, in this particular case, it's easily seen that the function f(x) = 3^x + x is a continuous and increasing function, and hence there is only one solution to the equation f(x) = 30 (in the real domain).
By the way, how did his method prove that there are no complex solutions? The Lambert W Function W(y) is only the _principal_ branch of roots to the equation y = x^(e^x) . He chose to break up the exponent 30 into 3 and 27 ; who says there is no solution when breaking up 30 into z and (30-z) where z is complex-valued?
@@yurenchu Looks like the "guessing" term hit you hard. Sorry if I made you sad (of angry, I guess)
There is a simple solution that giving x an integer from 1 to 3. You can find the answer in three seconds. But if you must prove it, this equation may be solved by your way.
f(x)=3^x
g(x)=-x+30
f(x) monotonously increasing, g(x) it decreases monotonously, but knows if there is only one solution. In this case, it is obvious x=3
In equations with obvious roots like this one we can work more simply and faster as I immediately show with 2 ways of solution in R.
1st Way
• If x < 0, then 3^x < 3^0 =1, so 3^x + x < 1 + 0 = 1 and the equation has no root < 0
• If x > 3 then 3^x > 27, so 3^x + x > 30 so the equation has no root > 3
• If 0 < x < 3 then 3 ^ x < 27, so 3^x + x < 30 so the equation has no root < 3
• If x = 3 the equation is true, so the unique root of the equation is x =3
2nd Way
We will work with the monotonicity of the function f(x) = 3^x + x - 30
If x1 , x2 E R with x1 < x2, then 3^x1 < 3^ x2, so 3^x1 + x1 -30 < 3^x2 + x2 -30 i.e. f(x1) < f(x2) , so f is strictly increasing.
Therefore if it has a root, it is unique.
Since the equation has the root x = 3, this is also its only root.
The monotonicity of f can be easily proven with the help of derivatives.
Indeed, f΄(x) = (3^x)ln3 + 1 > 0, so f is strictly increasing, etc.
I would like to take this opportunity to announce that I have created a mathematics channel on You Tube called L+M=N
My channel is at
www.youtube.com/@Nikos_Iosifidis
I will be happy if you subscribe to my channel and comment on the solutions of the topics I present.
3^×+×=30
x
First glance you all can notice X= 3, and its done. Fact 😂
The answer is three. These small examples are great for learning lambert W, but it’s an unnecessary bit of math baggage when the trial substitution is the best approach.
The nearest value of 3^X to 30 is 3^3 i.e. 27 or X=3.
Then value of 3^X+X for X=3 is 3^3+3 =27+3=30.
Note, X must a whole number..
Guess x=3 as solution. f(x)=3^x+x is strictly monoton increasing, so x=3 is the only real solution. That saves you about 9 1/2 minutes.
By inspection x=3 is one solution and it is the only one because 3^x in increasing and 30-x is decreasing. No need to over elaborate!!!
3^x + x = 30
Now 30 can be written as 3+27 and 27 is just 3³ so 30 = 3³ + 3 but also, 30 = 3^x + x so 3³+3=3^x+x giving x = 3
Can you solve x^2+sqrt(x)=84? We can see that one root is 9, but how to calculate that and other roots?
Wow, what is W function? First time seeing one.
(It's been 30 years since highschool graduate, and never heard of it even in college freshman math class. Is W a basic term these days?)
no its still only in university
but now its more common bc well... we find videos and stuff that shows it
What everyone is failing to understand is the video is not supposed to be about solving the equation easily, but rather be able to solve an exponential-linear function with the help of W Lambert Function. The function is useful for a lot of other problems. Can you not understand that he tried to start off with an easier problem to explain the practical use of the function.
Легко решил в уме, подставляя числа
Beautiful solution. Thank you
3^x + x = 30.
30 = 27 + 3
27 = 3^3
then
3^x + x = 3^3 + 3 =>
=> x=3
What is the W function???
3^x+3 = 30 ---> 3^x+3 = 27 + 3 ----> 3^x+3 = 3^3+3 ---> 3^x = 3^3 ---> x = 3
sir we an get quick answer by putting value of x from 1,2,3 ...
But this is not a valid mathematical procedure, even that value won't be integer or there may be many other roots. Then what you will do?
@@Math_Minds then use your method😅
@@Math_Minds "There may be many other roots"
The same goes for your method. How did you know at 6:20 that W( (30-x)*ln(3) * e^[(30-x)*ln(3)] ) = (30-x)*ln(3) , unless you already knew the value of x ? Note: the Lambert W Function W(y) is the _principal_ root of the equation y = x*(e^x) , because several different values of x can lead to the same value of y .
As a comparison: what you did is equivalent to saying
√((x-7)²) = 2x+1
⇒ (x-7) = 2x+1 ⇒ x = -8
How would we know that the lefthandside would reduce to (x-7) ?
This is wrong (as can be seen when plugging the found result back into the original equation), the actual solution is
√((x-7)²) = 2x+1
⇒ (7-x) = 2x+1 ⇒ 6 = 3x ⇒ x = 2
I gave up on contest math years ago being not nearly brilliant enough for it, just got recommended this and it absolutely blew my mind❤
You are brilliant enough. It takes practice.
I have not found the part 3^30 ln(3)= 3^3 ln(3) 3^27, you had to think about it, well done
Wow, this video on Math Olympiad Exam | Lambert W Function is so informative! I appreciate the detailed breakdown and the insights you've shared here.
30=3power(3)+3 ce qui implique que X=3 par identification...
Much faster: 30=3^3+3. Therefore, x=3.
Better solution:
Transfer x to RHS and take log on both sides. Apply log properties and at last compare LHS=RHS to get x=3
x log3 = log(30-x) now what?
@@Math_Minds I am sorry I did not remember the properties of log correctly and wrote log(30-x) = log30/logx. My solution was wrong. The correct property was log x- log y= log(x/y)
logA - logB = log(A/B)
but log(A-B) can be express as a series only....
@@BeastGamer301 what the hell? log(x/y)=logx - logy
I like this solution. But stuck on how to solve x^x = 27
It's vvvvsimple. It's = 30. So u put number x 1, 2 , 3. Then u find when x=3. Both sides equal. It takes not even 1 min
Have you tried using the exhaustive method? The cube of 3 is 27, so X ≤ 3. The answer to the problem is between 0 and 3. Okay, the problem has been resolved.
yeah, but that would be easy and simple though, and we want to make math the hardest possible
Yes. When I was studying in college, I felt helpless when I encountered difficulties in math exams. I always wonder why I need to face such complex mathematical problems. I won't be able to use calculus or Newton's XXXX formula in my future life. What is the significance of learning this? However, my cousin is proficient in mathematics. He is an engineer working for China's satellite and rocket systems.@@pascal_0840
You don't need a method. It's intuitively obvious after you think for literally one second.
Sure, this one was easy to solve because it's using low numbers and x is a whole number. But if x was a rational or imaginary number, we'd need a formula to solve it. That's why the demo
@@headcode 😴💤
W what is that?
Muito linda a solução..o resultado se consegue sem esforço mas a demonstração e soberba
Nice steps
I used the following:
3^x + x =30
3^x = 30-x (ln both sides)
xln3 = ln (30-x)
xln3 = (ln30 / ln x)
then
xlnx = ln 30 / ln 3
xlnx = ln (30-3)
xlnx = ln 27
xlnx = 3 ln 3
x =3
or by using W function
ln x e^(ln x) = ln 3 e ^(ln 3)
W (ln x e^(ln x)0 = W (ln 3 e ^(ln 3))
then
ln x = ln 3
e^ (lnx) = e ^ (ln3)
therefore
X= 3
Thanks
ERRORS.... ln(a-b)=lna/lnb ??!!
This is how you write x
Look at it ❌
Two straight lines crossing each other in the middle.
Beside of that good video, nice and CLEAR solution 👍
а че типа сразу не очевиден результат был? Тут надо лишь доказывать, что других корней нет.
Is x= 1? No Is x = 2? No Is x=3? YES. Solved in 15 seconds. It took me longer to write this then solve the equation. Then I sat through 10 minutes of this plus ads
By inspection, x=3. Time taken = 1 second.
i just can not understand from where u got 3 and 27 ? why is not 1 and 29 or 2 and 28 etc
It is very simple just assume X=3 and work out.
Trivial problem. The answer is 3 by simple inspection.
Does this work just because the answer happened to be 3 or can you solve anything similar with this? Lets say 5^x + 2x = 1723 or something like this?
X= 4.628406 .
Use y/x function and adjust exponet.
to show application of lambert function this is ok. But, why this simple question was chosen. Is it mis use of lambert function. I too use results and adhoc arguments in my channel but why so lengthy??
3^x+x = 30.
Just looking at this, it is clear that x cannot be greater than 3, given the constraint of 30.
And x cannot be less than 3, because even at 2.999, 3^x + x will be less than 30.
So, x has to be 3 👍
no wait i was right we can actually do 3^1,2 or 3^1.7 or something :) no one said this is only natural number
@@lmfao3293 Buddy, for any value less then 3 it will never add up to 30. It will always be less than 30.
Max value for x = 2.99999, would result in 3^2.99999 < 27.
So, when you add them, it can only be less than 30.
Glad you made such a bad argument. Makes me look even more smart 🧐🧐🧐🤣🤣🤣
@@gulboyrathesungod xD Ok ok all you can do it's just guess untill you get right answer :)?
@@lmfao3293 So, when you don't understand the calculation, you call it guess? Well, I don't blame you. With the level of intelligence demonstrated, we cannot expect more..👍
@@gulboyrathesungod :) ok try this: 4^x+x=4294967312 xDD You again try to use calculator to get the answer again right :)
x = 3. 3^3+3 = 27+3 =30
X^2+x=4^2+4 => x=4 là chưa chặt chẽ . Tương tự bài toán trên giải cách này thì cần cm thêm rằng đây là kết quả duy nhất
이걸푸는데 10분이다 걸린다는것에 참 대단하다고 생각한다 ... 그냥 보면 답나오는 문제아니냐;;
3^3は27である。
∴30-27=3
乗数と引いて現れた数が一致する。したがってx=3は満たす。
これじゃ不十分かな？この解法なら小学生でも解ける。一般解じゃないけど文系なので良し！
それだと、他の解の可能性を吟味していないので不十分ですね…。
1,2回微分してグラフ描けばいい
3^(x) + x = 30
3³ + 3 = 27 + 3 = 30
x = 3
Best regards
Marcus 😎
3^3+3=3^×+×
X=3. Simple
I tried by putting x = 1,2,3 n got the answer 3, as very easy solution as we r interested in solution. Method mentioned above is very complicated.
Try to solve it : 3^x + x = 20.
That is the general approach to solve this type of equation.
x = 3. Because
If x > 3 => 3^x + x > 3^3 + 3
If x < 3 => 3^x + x < 3^3 + 3
Simple answer.
Can we use the iterative method?
What is the “ w “ at the end of the solution mean?
Lambert W function
Решается устно за две секунды . Представив графики левой и правой частн уравнения. Корень один, легко найти умным подбором! Х=3
Тоже от Эльмира сюда пришёл?
My math professor said that if you use more than 4 lines to solve an equation, you don't understand the music of the universe. You might understand the math, but not what makes numbers sing
Thank you! That was interesting
Glad you enjoyed it!
(3^x+x=30)
3^1+1=4
3^2+2=11
(3^3+3=30)
3^x+x=3^3+3
x=3
Very interesting way to solve the problem. Thanks for sharing. I just tried some numbers and got the answer very quickly.
3^1 + 3 = 6
3^2 + 3 = 12
3^3 + 3 = 30
done
I'm guessing the intent of this video was to show a general method of solution as opposed to a practical way to solve this particular problem.
You can gain time by noting that if x is an integer it is a multiple of 3...
@@mikebresnahan8682if its 3^1 its +1 not +3 😂😂😂😂
hello, do you have another channel Rashel's classroom?
No, That's not mine.
so good
Thank you.
What is w function?
what it just 3*3*3 + 3 = 30???????!!! why bother so many steps
Brother we are talking about all real Solutions
please, show me more long-winded ways to solve that
What does w mean ? What's the value of w?
この式に関しては、見た瞬間に3って分かった。
他に解があるかどうかは分からんかったが。
Very easy 3³+3=30, some problems don't need to work
That's great! A strictly consistent proof by representing the left and right sides of the equation in a form convenient for taking the Lambert function, instead of an intuitive, although obvious solution by representing the number 30 in the form of 3^3 + 3 !!!
Actually if it didn't exist yet you could simply define it.
but what's the purpose of the triple factorial? (jk)
@@davidkeck6183
Three exclamation marks express the positive emotions of the review author. And how do you express your emotions???
@@levskomorovsky1762 I wrote "but what's the purpose of the triple factorial? (jk)" where "(jk)" is a common convention for abbreviating "Just Kidding". It was basically a dad joke.
@@davidkeck6183 Thank you. I did not know. I am not an English language specialist..
Que gran tutoriual tan excelenye de examen de olimpiadas
What does the small ohm mean?
Mesmo substituindo o 3 por X, mentalmente a resposta é X= 3.
普通に3の倍数になにかを足して30になる数字を1つづつ探す方がシンプルだし良い。