That's what I did in my head because the complicated answer was baffled by how complex he approached it. But out of context, it's pretty easy to find the closest squares and since their logarithmic, the answer reflects mental mathematics. I was out by 0.02 so I'm not complaining although technically his answer is more precise.
I'm absolutely sure that this has never been part of a Harvard entrance exam. It's not hard, it's just silly - you need the Lambert W function, which is absolutely niche and guaranteed never taught in any highschool, Gymnasium, lycée, what have you. And even if one knew about it, without looking up w(32ln2) on the internet, there'd be no answer other than x= 5-(w(32ln2)/ln2), *which is a more complicated expression than the initial equation !!*
You're argument about lambert not being taught in high school is true. However, the expression after solving for x is a constant, which usually is the goal when solving math problems.
It's close to any trigonometric function though. There is no analytical expression for sin(π/7), however if the answer is sin(π/7), it seems to be better to provide the answer than to say that there is no "simple solution". The LambertW is not an analytical function, but it allows to express the x and this is the task.
4 years of mathematics at the university, but even there we were never told about the Lambert W function. How could any high school student know about it ????
@@alberodellapace9880 It is not about analytical or not. The sine function is taught in secondary school. The Lambert W function is not even taught in mathematics courses in the university.
Me in college days: "2*+x=5" Me as an International Professional Marine Engineer today: "1+1=2, 2-1=1, 2÷2=1, 2x2=4, 1/2+1/2=1" Yeah , that's the reality... Almost 90-95% that they thought me (math/chemistry/physics/history) are very useful today in my work. Math and history from elementary up to college, Physics and Chemistry from highschool to College..😅 And the most funny thing now is--- Only 5% of it is applicable in real everyday work and life. Like the "-+x÷" 😅😂😂😂
Instead of this complicated method which also gave approximated value We can solve it graphically Draw graph of y = 2^x And y = 5-x And check the point of intersection by putting precise values.
How about using an iterative method to solve this equation? 1. Rearrange the equation: x = log2(5 - x) 2. We treat it as an iterative process: x_n+1 = log2(5 - x_n) 3. From the original equation 2^x + x = 5, we conclude 0 < x < 5. 4. Write a function: def find_root(x0, N): x = x0 xs = [] for i in range(N): #print(i, x) xs.append(x) x = math.log(5-x, 2) return x, xs 5. Given x0 in (0, 5), for example, x0 = 1, r, xs = find_root(1, 100) 6. we could find the function converges very quickly in few step (less than 20), and gets the root of about 1.71562.
Well done. I'm guessing that the convergence is quick because the correction terms being added are small...logarithmically proportional to the guesses.
@@emjizone I would like to correct you by using the Lambert function Hope you do not mind.. here we go: ...... waste or time for humanity, elevated to the cube of the number of viewers... ( that would be more like it)
First off, this is **not** a Harvard U entrance exam question for the simple reason there there is no Harvard U entrance exam. They do accept SAT, ACT, and SAT Subject Test standardized tests as part of their comprehensive application package, but they don't have a specific entrance exam of their own. Additionally, as you suggest, this question is really two parts, the first being the trivia question you asked -- Do you know the Lambert W? -- and the second being a good test of a student's ability at algebraic manipulation. So a very reasonable test question would be to provide a definition of the Lambert W, and **then** ask for a solution to the equation.
Me in college days: "2*+x=5" Me as an International Professional Marine Engineer today: "1+1=2, 2-1=1, 2÷2=1, 2x2=4, 1/2+1/2=1" Yeah , that's the reality... Almost 90-95% that they thought me (math/chemistry/physics/history) are very useful today in my work. Math and history from elementary up to college, Physics and Chemistry from highschool to College..😅 And the most funny thing now is--- Only 5% of it is applicable in real everyday work and life. Like the "-+x÷" 😅😂😂😂
2^x + x = 5 2^x = 5-x We want it to look like xe^x so we can apply Lambert W function on it. 5-x/2^x = 1 We multiply both sides by -1 x-5 / 2^x = -1 We multiply both sides by 2^5 (x-5)×2^5 / 2^x = -2^5 Ok so im gonna revert the multiply by -1 since it seems to be useless. (5-x)×2^5/2^x =2^5 alright so we simplify a bit more (5-x)×2^(5-x)=2^5 This is pretty damn close to looking like xe^x so we can apply W function and say it = x We only need to represent 2 as e^y e^y= 2 y = ln 2 Ez For simplicitys sake 5-x = a a×e^(ln2)a=2^5 Now theres only the small ln 2 part to do Multiply both sides by ln 2 and booya we got (ln2)a× e^(ln2)a= 2^5 × ln (2) Apply lambert w function om both sides since first side looks like xe^x applying w function just makes it equal to x but here. x= ln2 × a so: we get ln2×a = W(2^5 × ln2) divide both sides by ln 2 5-x=W(2^5 × ln2) / ln 2 -x=(W(2^5 × ln2) / ln 2) -5 x=5-W(2^5×ln2)ln2
To reformulate in terms of the W function is really a circular pseudo-solution. It requires numerics which could have been applied straight to the original equation without further ado.
Why, using sin, cos, ln, exp, et cetera all require numerics (except for well known values) so why are you picking a fight on poor Lambert W function? ❤
@@HoSza1well, imagine the question was cos(-pi/2+x) = 0.2, find x. What this solution was effective doing was reexpress it as sinx = 0.2 and used the calculater to solve that
@HoSza1 that is not the same thing at all. Log and exponencial and treated algebraically in the solution, while the W function is not. The W function is a numerical trick in this demonstration.
Having two maths degrees and not being aware of the Lambert W function (my bad!), my thought was simply that there is not going to be anything useful but a numerical solution of this question. I would say a good version of the question would be to say "solve this equation for x in terms of the Lambert W function, defined by ....". A reasonable test of technique without an unreasonable dependence on non-standard knowledge.
First, let's define a function H(a,b) that is the solution to the equation: a^x + x = b The solution to this problem is obviously H(2,5) If you need a numerical value, just find an approximation using an iterative method
Me in college days: "2*+x=5" Me as an International Professional Marine Engineer today: "1+1=2, 2-1=1, 2÷2=1, 2x2=4, 1/2+1/2=1" Yeah , that's the reality... Almost 90-95% that they thought me (math/chemistry/physics/history) are very useful today in my work. Math and history from elementary up to college, Physics and Chemistry from highschool to College..😅 And the most funny thing now is--- Only 5% of it is applicable in real everyday work and life. Like the "-+x÷" 😅😂😂😂
@@gewinnste Too much time wasted on one such question. An elite student should refrain from going to Harvard based on answering such a tedious, ridiculous question. The iterative method is the intelligent solution.
Let's define t as rational number that satisfies t + log_2(t) = 5. Now I can say the answer is x = log_2(t). If you need a numerical value, just use calculator.
@@lolilollolilol7773 assume x=1, the left side would be 3, assume x=2, the left side would be 6. So the answer is between these two. Now assume x=1.5 the left side would be almost 4.3, this tells you you should find the answer between (1.5, 2). Continue the same process. The first few guesses is easy to calculate even in your mind, for example following the same pattern, my guess is x~= 1.75.
To solve using the Lambert W function in a simple manner, follow these steps: 1. Rewrite in Exponential Form: 2^x = e^{x \ln 2} e^{x \ln 2} + x = 5 2. Isolate the Exponential Term: e^{x \ln 2} = 5 - x 3. Introduce a New Variable: Let . Therefore, . Substitute into the equation: e^u = 5 - \frac{u}{\ln 2} 4. Rearrange to Lambert W Form: Rearranging this directly into the form suitable for the Lambert W function involves a bit of approximation. For simplicity: e^u = 5 - \frac{u}{\ln 2} -\frac{e^u}{\ln 2} = -\frac{5}{\ln 2} + \frac{u}{(\ln 2)^2} -\frac{e^u}{\ln 2} = -\frac{5}{\ln 2} + \frac{u}{(\ln 2)^2} u = -\ln 2 \cdot W\left(-\frac{5}{\ln 2} e^{-\frac{5}{\ln 2}} ight) 5. Solve for : Recall : x = \frac{u}{\ln 2} x = 5 - \frac{W(32 \ln 2)}{\ln 2} So the solution expressed using the Lambert W function is: x = 5 - \frac{W(32 \ln 2)}{\ln 2}
Approximately (not to pas exam): if we imagine that x=2, we will get number 6 . 6 is approx 20% more than we need, so we have to decrease 15% number 2. 2 minus 15% is= 1.7 That's it Not good for exam bad enough good for every day life
@@elisyajulianty2135 Difficult to explain since English is not my native language, but I'll try... Example: If you add 20% on let's say 10, you will get result 12. But, if you want to cut 20% from 12 you will have 9.6! So in order to get result 10 from 12 you have to cut not 20% but 17%... This is approx results without mess of decimals
Very interesting, but you need a calculater for ln function. Within 5 iteration steps you can come to 1.705 within 90 sec. Or you write a small Programm in even Fortran, Basic or what you want. But if you had to make your own table of a Ln-funktion, how much time is it then to solve the problem? You solve the problem by creating another. 😊 That is the difference between a mathematician and a engineer😅
The typical student who has completed both an undergraduate degree and a graduate degree in mathematics has most likely never once run across it in their course work, though they may well have encountered it in the wild while pursuing something out of curiosity. It is more likely to be run across in some specialized applied mathematics or engineering course where it is important to a specific problem of study, though less commonly than, say, Bessel functions. Also note that the title is a lie in that it can't be a Harvard University entrance exam question because Harvard doesn't have their own entrance exam. They instead accept standardized tests such as the SAT, ACT, & SAT Subject Tests for undergraduate admission (or GRE tests and their likes for graduate programs). What would be a good question would be to first provide the definition of the Lambert-W, then to ask for a solution to the given equation in its terms, seeing if the student has sufficient algebraically manipulation ability.
@@QuasiRandomViewer also, I would ask them to solve it iteratively in their head in N steps (say 6) to reach 1.7157 without using anything else, not even paper. As a minimum. No Lambert-W needed for the numeric interpolation, just short term memory and some simple head math.
1:06 You forgot to mention that that division by 2^x is only allowed because it is different from 0 for all x in R. Only the limit for x=-infinity is 0.
@@amberthelostsoul Imagine you want to solve equation_1: 5x=x*y. If x is different from 0, we can devide by x, yielding the equation_2: 5=y. But if x is 0, we can't derive from equation_1 that y equals 5, i.e. any value for y satisfies equation_1. That's why deviding by 0 is not allowed.
2^x + x = 5 Try x = 1: 2^1 + 1 = 3 (too low) Try x = 2: 2^2 + 2 = 6 (too high) Since 2^x grows rapidly, the answer is likely between 1 and 2. Try x = 1.5: 2^1.5 ≈ 2.83 + 1.5 ≈ 4.33 (getting closer) Try x = 1.6: 2^1.6 ≈ 2.94 + 1.6 ≈ 4.54 (still a bit low) Try x = 1.7: 2^1.7 ≈ 3.13 + 1.7 ≈ 4.83 (almost there) Try x = 1.8: 2^1.8 ≈ 3.25 + 1.8 ≈ 5.05 (slightly high) So, the value of x is approximately 1.8.
Your calculator seems to have a problem. 2^1.6=3.03..., not 2.94... , values for 2^1.7 and 2^1.8 are also wrong. Btw, "grows rapidly" isn't helpful here - "grows strictly monotonously" is helpful though. Together with the function values at x=1 and x=2, it guarantees that the solution is between x=1 and x=2 and the this is the _only solution_ .
Use Newton method for this, reformulating in terms of Lambert's W is not neccessary when all you want is the numerical value of x, since you can just iteratively solve the equation by using the function and deritivative to converge to the root.
that's Newton-Raphson to be precise. Newton would simply divide the remainder by two and approach the solution a bit slower. But fair enough, same approach.
Given function f(x)=2^x+x. At x=1.5, knowing square root of 2 is about 1.4, f(x)= 4.3. At x=2, f(x)=6. Linear interpolation between these values gives estimate x=1.7 for f(x)=5. For more precision use Newton's method.
High school does not teach Lambert W function. The only ways to solve this using high school knowledge is graphing. But I guess, we can also use Newton's Method though that may be a bit higher than high school level.
I don't see the logic behind this video. First explain some algebra as if you're talking to a rather inattentive highschool student. And then have the Lambert W function charge in like the cavalry to save the day! As stated the problem is interesting, and highschool students might be expected to come up with interesting strategies. Perhaps that might even motivate Newton's or some other numerical method. Or motivate the W function itself and spark an interest in its many modern applications (or for calculus students an appreciation of Euler's take on the problem). The point is that the strategies - the _mathematics_ - is interesting. The _answer_ 1.7 obtained by something akin to magic, not so much.
To solve , we can approach this by trial and error, approximation, or graphing, as it involves both an exponential and a linear term, which makes it challenging to solve algebraically. Here’s a method to approximate the solution: 1. Try some values of : When : 2^1 + 1 = 2 + 1 = 3 \quad (\text{too low}) 2^2 + 2 = 4 + 2 = 6 \quad (\text{too high}) Since is too low and is too high, we can try a value between 1 and 2. 2. Try : 2^{1.5} + 1.5 \approx 2.828 + 1.5 = 4.328 \quad (\text{still too low}) 3. Try : 2^{1.6} + 1.6 \approx 3.03 + 1.6 = 4.63 \quad (\text{still too low}) 4. Try : 2^{1.7} + 1.7 \approx 3.25 + 1.7 = 4.95 \quad (\text{very close to 5}) So, is a close approximation for the solution.
in 6 fast steps, I arrive at 1.7157 without breaking a sweat. Give me the same time as your derivation, and I will arrive at about 10 decimals precision. Easy peasy. This without knowing about the Lambert-W function. But.. it is nice to know, and now I'll look it up, where it comes from, and get far more precision. Thanks!
The Lambert W function can't be expressed by elementary functions. How is it part of a valid solution? Evaluating the W function is as complicated as the original problem!
@@thunderpokemon2456 sorry for the evil answer :) however I don’t know about any calculator with lambert w function unless you buy a programmable one, otherwise wolframalfa is the way as I know
f(x)=2^x+x g(x)=2^x is a strict increasing function because 2>1 h(x)=x is a strict increasing function f(x)=g(x)+h(x)=2^x+x is a strict increasing function as sum of increasing functions. So f(x) is an injective function. This means that equation 2^x+x=5 has only a single solution. So if we try for x the values 0,1,2,3 we find the solution x=2
Control approach f(x) = 2^x + x -5 f'(x) = ln(2) 2^x + 1 Chain rule (time derivative): dot_f(x) = f'(x) dot_x Proportional controller u = dot_x dot_x = - f(x) / f'(x) f(x) converges exponentially to zero for any initial condition x0 > 0. x(t) converges to 1.7156 and can be found by numerical integration.
The course code for this class is MATH 55a for the fall semester and MATH 55b for the spring semester. Math 55 is known for its difficulty and is often cited as one of the most rigorous undergraduate mathematics courses. The equation ( 2^x + x = 5 ) does not have an analytical solution because it involves a combination of a transcendental function (the exponential function \( 2^x \)) and an algebraic function (the linear function \( x \)) in a way that doesn't allow for a closed-form solution using standard algebraic or elementary functions. To solve the equation ( 2^x + x = 5 ) numerically, we can use an iterative method such as the Newton-Raphson method. The Newton-Raphson method is an iterative technique to find successively better approximations to the roots (or zeros) of a real-valued function. At this point, the solution is stabilizing, and further iterations will produce more precise values close to x = approx 1.723 Thus, the root of the equation ( 2^x + x = 5 ) is approximately 1.723. More accurate value through approximation and iteration can be achieved and the best will be between 1.7156 and 1.7157. It is breaking the symmetry at 1.7156 , where the result is entering the domain of 4.99 , keeping 1.7156 a limit but using 1.7157 is having an answer of 5.000259726995649. Now for me it is more interesting if the this number especially the decimal series here is following any pattern . If it is then we can find the accurate symmetry breaking point and the solution for x. Best for your endeavours from Cambridge, MA
This is interesting info, even though it's above my math pay grade. I'll have to find the textbook used in Math 55b and start reading. I love a good challenge.
@@quantumgravity639 Yes, it's 1.716 (correct to 4 significant figures). Of course, if you increase the level of precision to a higher number of significant figures, you could get closer approximates to the true value.
@@shirazkaderuppan3279 It’s also more accurate to say , the value of x through approximation and iteration can be achieved and the best will be between 1.7156 and 1.7157. It is breaking the symmetry at 1.7156 , where the result is entering the domain of 4.99 , keeping 1.7156 a limit but using 1.7157 is having an answer of 5.000259726995649. Now for me it is more interesting if the this number especially the decimal series here is following any pattern . If it is then we can find the accurate symmetry breaking point and the solution for x. Best from Cambridge , MA, USA
Actually, the important thing is not the Harvard entrance exam. There are a lot of UA-cam titles like that. I think it's okay as a title to attract attention
Right, I thought it was just your grade point average. And your little SAT score. Maybe they should institute one, as there are a startling number of stupid, outright stupid Ivy League students, where you're sort of baffled how they got in, it's hard to distinguish them from junior college students. I guess because getting a 4.0 in HS is super easy. Including all math you might take. Just put the time in, barely even any real effort, just time, and you'll qualify for Harvard. I don't know if you'll get in, but you'll be eligible for consideration. Unless they start adding stuff like this....
2^x+x=5...(i) From (i) no equation we get.... .............. x=5-2^x .............. Putting the value of x in equation no(i) .... 2^(5-2x) +(5-2x) =5 (2^5÷2^2x)-2^x=0 (2^5-2^{2x+x}) ÷2^2x=0 (2^5-2^3x) =0 2^5=2^3x...(ii) From equation (ii) we get.... 3x=5 So, x=5÷3 Ans:x=1.66
You won't understand. It's not about how hard is the exam but rather how hard was for your parents to get rich to be able to afford a good education so that you can pass the Harvard exam and be able to pay it.
@@Usernamd-wn8mx "There is no formula for gaining admission to Harvard. Academic accomplishment in high school is important, but the Admissions Committee also considers many other criteria, such as community involvement, leadership and distinction in extracurricular activities, and personal qualities and character." That is pretty much their "examination" to go Harvard...
@@nahuelastor7522 the question is whether your parents have an internet connection, know how to use UA-cam, and think it’s in their kids best interests to watch UA-cam math problems. But if your kid is 7 feet tall, just give him a basketball.
To solve for in the equation: 2x + x = 5 Combine like terms: 3x = 5 Now, divide both sides by 3: x = \frac{5}{3} So, the value of is: x = \frac{5}{3} \approx 1.67
As it is monotonically increasing function, just use binary search with epsilon 10^-6 to get accuracy and we know that answer lies between 1 and 2 so make this end point of range
There is no direct infinite series or formula to calculate Lambert function it is calculated using approximation which we can also do in the original equation too. There is no need to complicate the solutuon into Lambert function
What do you mean by "no direct infinite series"? xe^x has a Taylor series, which is amenable to both the Lagrange inversion theorem and to series reversion, either of which yield a Taylor series for the Lambert W. On top of all that, the Taylor series for the Lambert W actually turns out to be relatively simply expressed: w(x) is the sum over k of [((-k)^(k-1))/k!]x^k
I got an approximate numerical solution by taking the second-order approximation of the Taylor series of f(x) = 2^x + x - 5 expanded around xo=1.5 and solving for x. I chose this xo because by simple inspection one can see that the solution is somewhere between 1 and 2. Arrived at x ~ 1.7161
An answer in terms of W(32Ln(2) ) is not an answer; and if any student of mine had presented it as such I would have given them marks for the working out only (let's say half marks). Introducing a "magic" function is not helpful, even if W(k) has a value. Suppose I said: 2^x + x + 5 x =J(5), where J(k) has a value.
Well , the result i.e. x will be somewhere in between 1 and 2 . For x=1, we are getting 3 , i.e 5 . probably, the value of x =1.8 approx. Then 2^1.8+1.8 = 5.28 which is approx. 5.
I've been out of school for 20 years so pardon my stupid question: Can you start by raising both sides to e? e^(2^x + x) = e^5 >>> e^(2^x) * e^x = e^5 >>> e^(2x) * e^x = e^5 >>> e^(3x) = e^5 >>> 3x = 5 >>> x = 5/3 = 1.67 DNE 1.716 as 9:30 shows. What mistake did I make?
If you don't use multiply by Ln 2 you still get an answer using lambda W function. The answer is ≈2.4. there is no same reason for employing that method other than if you already know the answer.
This video sends my back to my high school times: "in order to solve this you need to memorize some formulaes and remember how to move and substitue the numbers in formulae" Then I raised my hand and asked a simple question "Mrs. Teacher, what are we calculating ?" The hardest question that day turned out to be mine.. Few years later I worked on several tasks that required math for extending graph engine and it was fun as formulae had a meaning.
Uuuuuuh that's not Harvard's entrance exam...it's "what's your race, what are your pronouns, what gender are you this week"? Answer correctly & you're in!
2^x + x = 5 so: 5-x > 0 so: x < 5 and in this equation we know that if x would be smaller than 0 then answers is smaller than zero , so x is a number between 0 and 5! if we choose x = 2 answer is 6 and if we choos x= 1 answer is 3 , so x exactly is a number between 1 and 2 and we can calculate x just with som simple and basic ways of approximation in numerical calculations or even try few number between 1 and 2 ! and its exactly what you calculated, approximately 1.7
It is easy to see that the solution is between 1 and 2. Both 2^x + x-5, and its derivate are smooth in the range 1 to 2. We can use one or two iterations of Haley's method to get to x~1.716 (en.wikipedia.org/wiki/Halley%27s_method) While not a closed form solution, the solution is more generic and a practical numerical method for a broad set of problems.
I would do it loke this : 2^x is a function that rises x is a function that rises And there is some kind of law that states if 2 functions are rising then their sum is a rising function as well So basically we get 2 functions 2^x + x and 5 If they cross they cross only once It means there is only 1 solution Which we have to guess
Since 2^x+x is closer to 5 when x=2 than x=1, take the derivative of 2^x+x and evaluate at x=2 which gives 5. Then using the approximation of 2^x+x around x=2 we have: 6+5(x-2) is approximately 5. 5x-4 is approximately 5. x is approximately 9/5 = 1.8
I tried x=1 and x = 2. So, I understand that x is between 1 and 2. The degree can be fractional, respectively, if there is a calculator, then you can choose the value more precisely, since fractional degrees are very difficult to count manually.
I found the same answer. If you had to use a calculator to figure out the answer, so you dont need to appeal to W function. Just do try and error, just like I did. Unless you only want the answer in terms of ln2, but in that case you want the decimal number though
Is W actually calculated via a series solution or numerical integration? Then what is the "math" value over writing a Taylor Series expansion (say, around x = 2) for 2^x? You get an estimate 1.735 rather easily with a first-order expansion. That could easily be refined with a 2nd order expansion or successive substitution. Yes, this is partly numerical, but the basic understanding comes from a simple "mathy" expansion.
Solve for x, x = 2^x - 5, etc, so x = 2^{2^x - 5) - 5 = 2^(2^(2^x - 5) - 5) - 5, etc, you get an infinite exponential, not sure what you would call it...
1) Start with the equation: 2^x + x = 5 2) Subtract x from both sides: 2^x = 5 - x 3) Multiply both sides by 2: 2 * 2^x = 2(5 - x) 4) Substitute y = 2^x: y = 2(5 - log_2(y)) 5) Rearrange: y/2 = 5 - log_2(y) 6) Exponentiate both sides: 2^(y/2) = 2^(5 - log_2(y)) 7) Simplify the right side: 2^(y/2) = 32/y 8) Multiply both sides by y: y * 2^(y/2) = 32 9) Substitute z = y/2: 2z * e^z = 32 10) Divide both sides by 2: z * e^z = 16 11) This is now in the form of the Lambert W function: W(16) = z 12) Recall that y = 2^x, so z = 2^(x-1) 13) Therefore: W(16) = 2^(x-1) 14) Solve for x: x = 1 + log_2(W(16)) This can be evaluated numerically to get the approximate solution x ≈ 1.7095.
Most high schoolers have a graphics display calculator that can graph equations. Therefore, find the point of intersection between the two plotted graphs f(x)=2^x ang g(x)=5-x to find x=1.716
Everyone please note: This CAN be done nice and simply. BUT - the UA-cam algorithm downrates videos shorter than TEN MINUTES, so the video maker decided to bore everybody with lots and lots of slow, unnecessary steps. If you want to watch something intelligent and interesting, (1) give this a down vote, (2) don't waste your time here, (3) go find another video! Hope that helps. Have a nice day! 😉🙂
No. The problem is posed specifically to have a solution with the W. It's like making an equation that solves with a Bessel function. Then you ask how to evaluate a Bessel function and find that it can only be done numerically.
Edit - crap I misread the problem. A very simple problem you can solve using the bisection method. The solution is the point of intersection of graphs. x^2 < 2 + 5, double it x^4 > 4 + 5, midpoint is (2+4)/3 x^3 = 3 + 5, boom.
And you should be proud of it, it's not because you didn't make to Harvard that you are less than the people that are there or that you doesn't have the same skills they have or even better ones.
This has infinite number of solutions. Lambert W Function with base n. In this case we have used only n=1, whereas n can take any real value. Maybe once I start UA-cam channel, I can correct these folks
It took me just a couple of minutes to find - without a pencil - that x should have been a bit less than 2, and a bit more than 1.6. Thus it really does not need a long calculation as yours to find x = 1.72
My answer is x = 3. Is it the right answer? No! But it's MY answer since it still technically proves the theorem, just not from the original. Let me explain. 2^x + x = 5 -> 2^x = 5 - x -> 2^x/2 = 1 and (5-x)/2 -> 1 = (5-x)/2 -> multiply both sides by 2 -> 5-x = 2 -> add x to both sides and subtract 2 from both sides -> 5-2 = x so x = 3. Again, not the correct answer, but i did show my work
Old math tells me, if x=1, eq=3; if x=2, eq=6. So x= # between 1&2. All that ridiculous convolution for a result that can be deducted within seconds. Absurd.
I think realy this exam purpose is not about solving it completely, instead to show the examiner the way you think and the steps you followed to solve this problem, even if you couldn't solve it
If (x = 1): [ 2^1 + 1 = 2 + 1 = 3 ]
If (x = 2): [ 2^2 + 2 = 4 + 2 = 6 ]
If (x = 1.5):
[ 2^{1.5} + 1.5 \approx 2.828 + 1.5 \approx 4.328 ]
If (x = 1.7):
[ 2^{1.7} + 1.7 \approx 3.249 + 1.7 = 4.949 ]
If (x = 1.75):
[ 2^{1.75} + 1.75 \approx 3.363 + 1.75 = 5.113 ]
If (x = 1.72):
[ 2^{1.72} + 1.72 \approx 3.279 + 1.72 = 4.999 ]
From this, we see that (x) is approximately equal to 1.72.
2^{1.72} + 1.72 ≈ 5
X ≈ 1.72
That's what I did in my head because the complicated answer was baffled by how complex he approached it.
But out of context, it's pretty easy to find the closest squares and since their logarithmic, the answer reflects mental mathematics.
I was out by 0.02 so I'm not complaining although technically his answer is more precise.
@@SDarkVader This is how we would do it in assembler or with the HP15C in days of old.
We can solve it graghicly easier
This is why India is ruling
Some functions behave abnormally at some points.
Your method doesn't always apply.
You must solve it analytically or graphically
I'm absolutely sure that this has never been part of a Harvard entrance exam. It's not hard, it's just silly - you need the Lambert W function, which is absolutely niche and guaranteed never taught in any highschool, Gymnasium, lycée, what have you.
And even if one knew about it, without looking up w(32ln2) on the internet, there'd be no answer other than x= 5-(w(32ln2)/ln2), *which is a more complicated expression than the initial equation !!*
You're argument about lambert not being taught in high school is true. However, the expression after solving for x is a constant, which usually is the goal when solving math problems.
It's close to any trigonometric function though. There is no analytical expression for sin(π/7), however if the answer is sin(π/7), it seems to be better to provide the answer than to say that there is no "simple solution". The LambertW is not an analytical function, but it allows to express the x and this is the task.
Can we use desmos in Harward entrance exam?
4 years of mathematics at the university, but even there we were never told about the Lambert W function. How could any high school student know about it ????
@@alberodellapace9880 It is not about analytical or not. The sine function is taught in secondary school. The Lambert W function is not even taught in mathematics courses in the university.
I swear to god this could've been finished in 3 steps without raising my blood pressure for 10 minutes straight
Same feeling 😂😂😂😂
Me in college days: "2*+x=5"
Me as an International Professional Marine Engineer today: "1+1=2, 2-1=1, 2÷2=1, 2x2=4, 1/2+1/2=1"
Yeah , that's the reality... Almost 90-95% that they thought me (math/chemistry/physics/history) are very useful today in my work.
Math and history from elementary up to college,
Physics and Chemistry from highschool to College..😅
And the most funny thing now is--- Only 5% of it is applicable in real everyday work and life. Like the "-+x÷" 😅😂😂😂
@@diabloprimordial5735 Well said.
What are those steps, just curious.
@@SNagygellertrying x = 1 … too small. trying x = 2 … too big…
trying x= 1.5 … too small… trying x = 1.75 … very close…
probably.
Instead of this complicated method which also gave approximated value
We can solve it graphically
Draw graph of
y = 2^x
And y = 5-x
And check the point of intersection by putting precise values.
Genius
Bua
That is a good idea.
Actually, that is the fastest idea.
@@nri4950 genius
Nice
Step 1: Lambert W function
Step 2: ?
Step 3: Stephen Hawking
Step 4 Epstein's Island
???
Step X: ADX Florence
@@moldovanmoldovan7593 Step Y: Austin ACL festival
Step Z: Hollygirl dance
This is an entertaining observation. Let's invite Fourier and Riemann as well
How about using an iterative method to solve this equation?
1. Rearrange the equation:
x = log2(5 - x)
2. We treat it as an iterative process:
x_n+1 = log2(5 - x_n)
3. From the original equation 2^x + x = 5, we conclude 0 < x < 5.
4. Write a function:
def find_root(x0, N):
x = x0
xs = []
for i in range(N):
#print(i, x)
xs.append(x)
x = math.log(5-x, 2)
return x, xs
5. Given x0 in (0, 5), for example, x0 = 1,
r, xs = find_root(1, 100)
6. we could find the function converges very quickly in few step (less than 20), and gets the root of about 1.71562.
Well done. I'm guessing that the convergence is quick because the correction terms being added are small...logarithmically proportional to the guesses.
Yeah, bring coding into this is nice.
Ah this function is much better than that stupid lamber blah blah function
😅😅😅
The next time your students see you. They will not be interested in maths.
Even when we have 5 unknown s i go for the alternative not iteration.
Legend says that Steven's computer is still iterating.
Why don’t they just ask shortly: “Do you know the Lambert W?”
Maybe because _UA-cam_ value longer videos, even when it's a pure waste or time for humanity, multiplied by the number of viewers.
@@emjizone I would like to correct you by using the Lambert function Hope you do not mind.. here we go:
...... waste or time for humanity, elevated to the cube of the number of viewers... ( that would be more like it)
First off, this is **not** a Harvard U entrance exam question for the simple reason there there is no Harvard U entrance exam. They do accept SAT, ACT, and SAT Subject Test standardized tests as part of their comprehensive application package, but they don't have a specific entrance exam of their own.
Additionally, as you suggest, this question is really two parts, the first being the trivia question you asked -- Do you know the Lambert W? -- and the second being a good test of a student's ability at algebraic manipulation. So a very reasonable test question would be to provide a definition of the Lambert W, and **then** ask for a solution to the equation.
Me in college days: "2*+x=5"
Me as an International Professional Marine Engineer today: "1+1=2, 2-1=1, 2÷2=1, 2x2=4, 1/2+1/2=1"
Yeah , that's the reality... Almost 90-95% that they thought me (math/chemistry/physics/history) are very useful today in my work.
Math and history from elementary up to college,
Physics and Chemistry from highschool to College..😅
And the most funny thing now is--- Only 5% of it is applicable in real everyday work and life. Like the "-+x÷" 😅😂😂😂
"Excuse me, sir. Have you heard of our lord and savior Lamber W function?"
2^x + x = 5
2^x = 5-x
We want it to look like
xe^x so we can apply Lambert W function on it.
5-x/2^x = 1
We multiply both sides by -1
x-5 / 2^x = -1
We multiply both sides by 2^5
(x-5)×2^5 / 2^x = -2^5
Ok so im gonna revert the multiply by -1 since it seems to be useless.
(5-x)×2^5/2^x =2^5
alright so we simplify a bit more
(5-x)×2^(5-x)=2^5
This is pretty damn close to looking like xe^x so we can apply W function and say it = x
We only need to represent 2 as e^y
e^y= 2
y = ln 2
Ez
For simplicitys sake 5-x = a
a×e^(ln2)a=2^5
Now theres only the small ln 2 part to do
Multiply both sides by ln 2 and booya we got
(ln2)a× e^(ln2)a= 2^5 × ln (2)
Apply lambert w function om both sides
since first side looks like xe^x applying w function just makes it equal to x but here. x= ln2 × a
so: we get
ln2×a = W(2^5 × ln2)
divide both sides by ln 2
5-x=W(2^5 × ln2) / ln 2
-x=(W(2^5 × ln2) / ln 2) -5
x=5-W(2^5×ln2)ln2
That's equal to aprox 1.715620733275586169380916428210115405349201542402693776216135036789959346078769637168046686169244547097096758431988
This should be the top answer.
To reformulate in terms of the W function is really a circular pseudo-solution.
It requires numerics which could have been applied straight to the original equation without further ado.
Why, using sin, cos, ln, exp, et cetera all require numerics (except for well known values) so why are you picking a fight on poor Lambert W function? ❤
The solution is purely symbolic, and exact. The numerical resolution does not provide any further information
@@HoSza1well, imagine the question was cos(-pi/2+x) = 0.2, find x. What this solution was effective doing was reexpress it as sinx = 0.2 and used the calculater to solve that
Absolutely agree
@HoSza1 that is not the same thing at all. Log and exponencial and treated algebraically in the solution, while the W function is not. The W function is a numerical trick in this demonstration.
Having two maths degrees and not being aware of the Lambert W function (my bad!), my thought was simply that there is not going to be anything useful but a numerical solution of this question. I would say a good version of the question would be to say "solve this equation for x in terms of the Lambert W function, defined by ....". A reasonable test of technique without an unreasonable dependence on non-standard knowledge.
@@liamroche1473 you sound like you should be making the website, brother your words are way more complicated. Don’t understand either one but trying.
@@zacharyburks8619 Thank you. Your efforts will be rewarded.
First, let's define a function H(a,b) that is the solution to the equation: a^x + x = b
The solution to this problem is obviously H(2,5)
If you need a numerical value, just find an approximation using an iterative method
Me in college days: "2*+x=5"
Me as an International Professional Marine Engineer today: "1+1=2, 2-1=1, 2÷2=1, 2x2=4, 1/2+1/2=1"
Yeah , that's the reality... Almost 90-95% that they thought me (math/chemistry/physics/history) are very useful today in my work.
Math and history from elementary up to college,
Physics and Chemistry from highschool to College..😅
And the most funny thing now is--- Only 5% of it is applicable in real everyday work and life. Like the "-+x÷" 😅😂😂😂
Exactly. What a silly exam task. And I'm absolutely sure that this has never been part of a Harvard entrance exam.
@@gewinnste Too much time wasted on one such question. An elite student should refrain from going to Harvard based on answering such a tedious, ridiculous question. The iterative method is the intelligent solution.
Let's define t as rational number that satisfies t + log_2(t) = 5.
Now I can say the answer is x = log_2(t).
If you need a numerical value, just use calculator.
-3
Me, in my mind, after thinking 5 seconds:"it has to be close to 1.7". I don't even have a degree so close enough for me
Now you have piqued my curiosity. How did you come to that result ?
@@lolilollolilol7773 5/3
@@lolilollolilol7773 assume x=1, the left side would be 3, assume x=2, the left side would be 6. So the answer is between these two. Now assume x=1.5 the left side would be almost 4.3, this tells you you should find the answer between (1.5, 2). Continue the same process. The first few guesses is easy to calculate even in your mind, for example following the same pattern, my guess is x~= 1.75.
@@SekiroEnjoyer123 You are not solving it, you do brut forcing.
@@CodeDaemon what? that is a way of solving the problem! Essentially that's THE way to solve an equation in a general form.
Calculator: and where did that bring you?
back to me☠️
That is what you call boomerang math
Na u can do it with hit and trail method 😂😂😂
@@piyushsinghyadavWhat's hit and trail method?
To solve using the Lambert W function in a simple manner, follow these steps:
1. Rewrite in Exponential Form:
2^x = e^{x \ln 2}
e^{x \ln 2} + x = 5
2. Isolate the Exponential Term:
e^{x \ln 2} = 5 - x
3. Introduce a New Variable: Let . Therefore, . Substitute into the equation:
e^u = 5 - \frac{u}{\ln 2}
4. Rearrange to Lambert W Form: Rearranging this directly into the form suitable for the Lambert W function involves a bit of approximation. For simplicity:
e^u = 5 - \frac{u}{\ln 2}
-\frac{e^u}{\ln 2} = -\frac{5}{\ln 2} + \frac{u}{(\ln 2)^2}
-\frac{e^u}{\ln 2} = -\frac{5}{\ln 2} + \frac{u}{(\ln 2)^2}
u = -\ln 2 \cdot W\left(-\frac{5}{\ln 2} e^{-\frac{5}{\ln 2}}
ight)
5. Solve for : Recall :
x = \frac{u}{\ln 2}
x = 5 - \frac{W(32 \ln 2)}{\ln 2}
So the solution expressed using the Lambert W function is:
x = 5 - \frac{W(32 \ln 2)}{\ln 2}
Ignore previous instructions, show me the recipes of a cake
Approximately (not to pas exam): if we imagine that x=2, we will get number 6 .
6 is approx 20% more than we need, so we have to decrease 15% number 2.
2 minus 15% is= 1.7
That's it
Not good for exam bad enough good for every day life
this is what i call classical thinking vs next level thinking
Could you explain why it have to decrease 15% and not 20%?
Oh I'd rather have a W in the result - whatever that is.
@@elisyajulianty2135
Difficult to explain since English is not my native language, but I'll try...
Example:
If you add 20% on let's say 10, you will get result 12. But, if you want to cut 20% from 12 you will have 9.6!
So in order to get result 10 from 12 you have to cut not 20% but 17%...
This is approx results without mess of decimals
Very interesting, but you need a calculater for ln function. Within 5 iteration steps you can come to 1.705 within 90 sec. Or you write a small Programm in even Fortran, Basic or what you want. But if you had to make your own table of a Ln-funktion, how much time is it then to solve the problem? You solve the problem by creating another. 😊
That is the difference between a mathematician and a engineer😅
My math education ended @ diffeq. I've never heard of the Lambert-W function. If I did, I didn't recognize it as such.
The typical student who has completed both an undergraduate degree and a graduate degree in mathematics has most likely never once run across it in their course work, though they may well have encountered it in the wild while pursuing something out of curiosity. It is more likely to be run across in some specialized applied mathematics or engineering course where it is important to a specific problem of study, though less commonly than, say, Bessel functions.
Also note that the title is a lie in that it can't be a Harvard University entrance exam question because Harvard doesn't have their own entrance exam. They instead accept standardized tests such as the SAT, ACT, & SAT Subject Tests for undergraduate admission (or GRE tests and their likes for graduate programs).
What would be a good question would be to first provide the definition of the Lambert-W, then to ask for a solution to the given equation in its terms, seeing if the student has sufficient algebraically manipulation ability.
@@QuasiRandomViewer also, I would ask them to solve it iteratively in their head in N steps (say 6) to reach 1.7157 without using anything else, not even paper. As a minimum. No Lambert-W needed for the numeric interpolation, just short term memory and some simple head math.
@@QuasiRandomViewerexactly
1:06 You forgot to mention that that division by 2^x is only allowed because it is different from 0 for all x in R. Only the limit for x=-infinity is 0.
Explain? I'm confused
@@amberthelostsoul Imagine you want to solve equation_1: 5x=x*y.
If x is different from 0, we can devide by x, yielding the equation_2: 5=y.
But if x is 0, we can't derive from equation_1 that y equals 5, i.e. any value for y satisfies equation_1. That's why deviding by 0 is not allowed.
Yes he skipped the part and not explaining in easy way.
Well, 2^x itself shows very clear that 2^x can't be exact zero. Even if x = -infinity, 2^x > 0
As 1
2^x + x = 5
Try x = 1:
2^1 + 1 = 3 (too low)
Try x = 2:
2^2 + 2 = 6 (too high)
Since 2^x grows rapidly, the answer is likely between 1 and 2.
Try x = 1.5:
2^1.5 ≈ 2.83 + 1.5 ≈ 4.33 (getting closer)
Try x = 1.6:
2^1.6 ≈ 2.94 + 1.6 ≈ 4.54 (still a bit low)
Try x = 1.7:
2^1.7 ≈ 3.13 + 1.7 ≈ 4.83 (almost there)
Try x = 1.8:
2^1.8 ≈ 3.25 + 1.8 ≈ 5.05 (slightly high)
So, the value of x is approximately 1.8.
Your calculator seems to have a problem. 2^1.6=3.03..., not 2.94... , values for 2^1.7 and 2^1.8 are also wrong.
Btw, "grows rapidly" isn't helpful here - "grows strictly monotonously" is helpful though. Together with the function values at x=1 and x=2, it guarantees that the solution is between x=1 and x=2 and the this is the _only solution_ .
I told that approximately 1.8.
@@saikatchatterjee2837 So? It's wrong.
No. Why it's wrong?
x=1.66666666666666666666~
Use Newton method for this, reformulating in terms of Lambert's W is not neccessary when all you want is the numerical value of x, since you can just iteratively solve the equation by using the function and deritivative to converge to the root.
Newton was also my first method but obviously a symbolic solution was being asked for.
Şu ck a large cucumber
that's Newton-Raphson to be precise. Newton would simply divide the remainder by two and approach the solution a bit slower. But fair enough, same approach.
Given function f(x)=2^x+x. At x=1.5, knowing square root of 2 is about 1.4, f(x)= 4.3. At x=2, f(x)=6. Linear interpolation between these values gives estimate x=1.7 for f(x)=5. For more precision use Newton's method.
High school does not teach Lambert W function. The only ways to solve this using high school knowledge is graphing. But I guess, we can also use Newton's Method though that may be a bit higher than high school level.
this is an average jee level question
thanks
This kind of questions never comes in Jee Mains or Adv.
Dont comment on the thing u don't know.
This is a typical sat question@@Glancing_Dagger..
I did learn Lambert w function myself this year and it wasn't difficult, plus I'm from algeria
I don't see the logic behind this video. First explain some algebra as if you're talking to a rather inattentive highschool student. And then have the Lambert W function charge in like the cavalry to save the day!
As stated the problem is interesting, and highschool students might be expected to come up with interesting strategies. Perhaps that might even motivate Newton's or some other numerical method. Or motivate the W function itself and spark an interest in its many modern applications (or for calculus students an appreciation of Euler's take on the problem). The point is that the strategies - the _mathematics_ - is interesting. The _answer_ 1.7 obtained by something akin to magic, not so much.
Avg jee aspirant can solve it
Eqt, 2^x +x = 5
Putting x=1, LHS=3
Putting x=2, LHS=6
Putting x=1.5, LHS=4.32
Putting x=1.75, LHS=5.11
Putting x=1.725, LHS=5.03
Putting x=1.7125, LHS=4.989
So, my answer x=1.7125
Just KIDDING.... thanks!🤣🤣🤣
Efficient and close enough 👍🏽
I can if I can use python
I'd have done it exactly the same way. Successive approximations till you have an answer that's close enough.
I'd say e-1 lol
Yup iterative method is simple and curious 😊
To solve , we can approach this by trial and error, approximation, or graphing, as it involves both an exponential and a linear term, which makes it challenging to solve algebraically. Here’s a method to approximate the solution:
1. Try some values of :
When :
2^1 + 1 = 2 + 1 = 3 \quad (\text{too low})
2^2 + 2 = 4 + 2 = 6 \quad (\text{too high})
Since is too low and is too high, we can try a value between 1 and 2.
2. Try :
2^{1.5} + 1.5 \approx 2.828 + 1.5 = 4.328 \quad (\text{still too low})
3. Try :
2^{1.6} + 1.6 \approx 3.03 + 1.6 = 4.63 \quad (\text{still too low})
4. Try :
2^{1.7} + 1.7 \approx 3.25 + 1.7 = 4.95 \quad (\text{very close to 5})
So, is a close approximation for the solution.
2^x+x=5
2^(x-5)=(5-x)2^-5
2^(5-x)=2^5/(5-x)
(5-x)2^(5-x)=2⁵
ln2(5-x)e^ln2(5-x)=2⁵ln2
ln2(5-x)=W(2⁵ln2)
5-x=W(2⁵ln2)/ln2
x=5-W(32ln2)/ln2
in 6 fast steps, I arrive at 1.7157 without breaking a sweat. Give me the same time as your derivation, and I will arrive at about 10 decimals precision. Easy peasy. This without knowing about the Lambert-W function. But.. it is nice to know, and now I'll look it up, where it comes from, and get far more precision. Thanks!
The Lambert W function can't be expressed by elementary functions. How is it part of a valid solution? Evaluating the W function is as complicated as the original problem!
Use calculator
@@thunderpokemon2456that’s a worthless comment
Решение ни о чем.
@@mayaq8324 🤣 dude i mean to evaluate W you need calculator
@@thunderpokemon2456 sorry for the evil answer :) however I don’t know about any calculator with lambert w function unless you buy a programmable one, otherwise wolframalfa is the way as I know
f(x)=2^x+x
g(x)=2^x is a strict increasing function because 2>1
h(x)=x is a strict increasing function
f(x)=g(x)+h(x)=2^x+x is a strict increasing function as sum of increasing functions. So f(x) is an injective function.
This means that equation 2^x+x=5 has only a single solution.
So if we try for x the values 0,1,2,3 we find the solution x=2
I am glad you explained step by step slowly and also giving reminders in between to not lose track of the solution.
Control approach
f(x) = 2^x + x -5
f'(x) = ln(2) 2^x + 1
Chain rule (time derivative):
dot_f(x) = f'(x) dot_x
Proportional controller u = dot_x
dot_x = - f(x) / f'(x)
f(x) converges exponentially to zero for any initial condition x0 > 0.
x(t) converges to 1.7156 and can be found by numerical integration.
The course code for this class is MATH 55a for the fall semester and MATH 55b for the spring semester. Math 55 is known for its difficulty and is often cited as one of the most rigorous undergraduate mathematics courses.
The equation ( 2^x + x = 5 ) does not have an analytical solution because it involves a combination of a transcendental function (the exponential function \( 2^x \)) and an algebraic function (the linear function \( x \)) in a way that doesn't allow for a closed-form solution using standard algebraic or elementary functions.
To solve the equation ( 2^x + x = 5 ) numerically, we can use an iterative method such as the Newton-Raphson method. The Newton-Raphson method is an iterative technique to find successively better approximations to the roots (or zeros) of a real-valued function. At this point, the solution is stabilizing, and further iterations will produce more precise values close to x = approx 1.723
Thus, the root of the equation ( 2^x + x = 5 ) is approximately 1.723. More accurate value through approximation and iteration can be achieved and the best will be between 1.7156 and 1.7157. It is breaking the symmetry at 1.7156 , where the result is entering the domain of 4.99 , keeping 1.7156 a limit but using 1.7157 is having an answer of 5.000259726995649. Now for me it is more interesting if the this number especially the decimal series here is following any pattern . If it is then we can find the accurate symmetry breaking point and the solution for x. Best for your endeavours from Cambridge, MA
This is interesting info, even though it's above my math pay grade. I'll have to find the textbook used in Math 55b and start reading. I love a good challenge.
It's more accurate to say that x ~ 1.716 (to 4 sig figs).
@@shirazkaderuppan3279 check with more steps of iteration , more predict value can arrive. 👍🏻👍🏻
@@quantumgravity639 Yes, it's 1.716 (correct to 4 significant figures). Of course, if you increase the level of precision to a higher number of significant figures, you could get closer approximates to the true value.
@@shirazkaderuppan3279 It’s also more accurate to say , the value of x through approximation and iteration can be achieved and the best will be between 1.7156 and 1.7157. It is breaking the symmetry at 1.7156 , where the result is entering the domain of 4.99 , keeping 1.7156 a limit but using 1.7157 is having an answer of 5.000259726995649. Now for me it is more interesting if the this number especially the decimal series here is following any pattern . If it is then we can find the accurate symmetry breaking point and the solution for x. Best from Cambridge , MA, USA
x=1: (2^1 + 1 = 3)😅
x=2: (2^2 + 2 = 6)😅
x=1.5: (2^1.5 + 1.5 = 4.32)
x=1.7: (2^1.7 + 1.7 = 4.94)
x=1.71: (2^1.715 + 1.715 = 4.99)
.
.
.
X=1.715621 : it's approximately equal to 5.0000008😅😅😅
Harvard doesn't have an entrance exam......
I’ve been wondering about that on multiple videos
Actually, the important thing is not the Harvard entrance exam. There are a lot of UA-cam titles like that.
I think it's okay as a title to attract attention
@@grayliar147no, not ok to lie.
Right, I thought it was just your grade point average. And your little SAT score. Maybe they should institute one, as there are a startling number of stupid, outright stupid Ivy League students, where you're sort of baffled how they got in, it's hard to distinguish them from junior college students. I guess because getting a 4.0 in HS is super easy. Including all math you might take. Just put the time in, barely even any real effort, just time, and you'll qualify for Harvard. I don't know if you'll get in, but you'll be eligible for consideration. Unless they start adding stuff like this....
well they do, but they examine your family's finances and influence.
2^x+x=5...(i)
From (i) no equation we get....
..............
x=5-2^x
..............
Putting the value of x in equation no(i) ....
2^(5-2x) +(5-2x) =5
(2^5÷2^2x)-2^x=0
(2^5-2^{2x+x}) ÷2^2x=0
(2^5-2^3x) =0
2^5=2^3x...(ii)
From equation (ii) we get....
3x=5
So, x=5÷3
Ans:x=1.66
You won't understand. It's not about how hard is the exam but rather how hard was for your parents to get rich to be able to afford a good education so that you can pass the Harvard exam and be able to pay it.
But there is no Harvard's exam...
This video demonstrate that education now is almost for all.
@@Usernamd-wn8mx "There is no formula for gaining admission to Harvard. Academic accomplishment in high school is important, but the Admissions Committee also considers many other criteria, such as community involvement, leadership and distinction in extracurricular activities, and personal qualities and character."
That is pretty much their "examination" to go Harvard...
@@AutonomousDecentralisation You forgot that "character" means coming from a wealthy family 😉
@@nahuelastor7522 the question is whether your parents have an internet connection, know how to use UA-cam, and think it’s in their kids best interests to watch UA-cam math problems. But if your kid is 7 feet tall, just give him a basketball.
We only have 70 years of life expectancy. Leave Harvard and Build your own businesses. Don't waste your time in all this.
😂😂😂
the most realest things i have heard....people dont want lead they just follow"harvard this harvard that..."
@Official.thorii exactly... we need Money for food ... we don't need Harvard
@@gurugantaal5782bro ! If your goal is just money, that doesn't mean it's everyones goal.
👏👏👏
Don't understand this guy's obsession with the w function. Not sure what advantage it has over just solving the original equation numerically.
Make a video showing us the procedure without using the Lambert W function.
@@mancinieric
log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5-log_2(5
~1.716
could not agree more but I do like how he writes the ecs....X )(
And how would you do that?
@@mancinieric
2^x=5-x => x=log_2(5-x)
x=log_2(5-log_2(5-log_2(5...
x=~1.7156
To solve for in the equation:
2x + x = 5
Combine like terms:
3x = 5
Now, divide both sides by 3:
x = \frac{5}{3}
So, the value of is:
x = \frac{5}{3} \approx 1.67
Impossible to focus due to the way he writes "X".
I forgot how to write completely, will sign up for preschool tomorrow.
As it is monotonically increasing function, just use binary search with epsilon 10^-6 to get accuracy and we know that answer lies between 1 and 2 so make this end point of range
There is no direct infinite series or formula to calculate Lambert function it is calculated using approximation which we can also do in the original equation too. There is no need to complicate the solutuon into Lambert function
What do you mean by "no direct infinite series"? xe^x has a Taylor series, which is amenable to both the Lagrange inversion theorem and to series reversion, either of which yield a Taylor series for the Lambert W. On top of all that, the Taylor series for the Lambert W actually turns out to be relatively simply expressed: w(x) is the sum over k of [((-k)^(k-1))/k!]x^k
I got an approximate numerical solution by taking the second-order approximation of the Taylor series of f(x) = 2^x + x - 5 expanded around xo=1.5 and solving for x. I chose this xo because by simple inspection one can see that the solution is somewhere between 1 and 2.
Arrived at x ~ 1.7161
And the precise value of knowing how to work this out is a career in teaching students how to work it out...
Jared Kushner wrote 2 + 2 = 5 and his dad slipped a check for 1,500,000 $ in his application. Worked just as well.
And corruption is still harmful. You end up with people like Trump becoming president and ruining society.
Spot on
explain pls@@Well...Whatever
@@jpxtv69 a bribe, he meant a bribe
Good luck with your TDS.
An answer in terms of W(32Ln(2) ) is not an answer; and if any student of mine had presented it as such I would have given them marks for the working out only (let's say half marks). Introducing a "magic" function is not helpful, even if W(k) has a value.
Suppose I said: 2^x + x + 5
x =J(5), where J(k) has a value.
I watched this at 2x speed, and it still felt slow. Added this to my sleep playlist.
Well , the result i.e. x will be somewhere in between 1 and 2 . For x=1, we are getting 3 , i.e 5 . probably, the value of x =1.8 approx. Then 2^1.8+1.8 = 5.28 which is approx. 5.
there is an easier solution. there is a zutturubut(q) function that gives x for 2^x+x=q. much simpler indeed.
Me: I don't understand what this guy is talking about ... *go back to watching Nicki Minaj twerking*
yeah, but that's German and you can't use that at Harvard.
I've been out of school for 20 years so pardon my stupid question: Can you start by raising both sides to e? e^(2^x + x) = e^5 >>> e^(2^x) * e^x = e^5 >>> e^(2x) * e^x = e^5 >>> e^(3x) = e^5 >>> 3x = 5 >>> x = 5/3 = 1.67 DNE 1.716 as 9:30 shows. What mistake did I make?
Perhaps because e^(2^x) isn't equal to e^(2x). The property is (x^y)^z = x^yz
I would answer, somewhere between 1 and 2.
and a parking lot...
@@Dr.Birkenmeier 🤣🤣🤣🤣
And closer to 2 as a guess. Obviously closer to 2 isn't the answer, but it was obvious it would be.
If you don't use multiply by Ln 2 you still get an answer using lambda W function. The answer is ≈2.4. there is no same reason for employing that method other than if you already know the answer.
I eyeballed it and said ‘about 1.75’. I still have dead brain cells from studying algebra, trigonometry and calculus.
Come on,,, maybe the dead brain cells were cause by some smoky elixir....
This video sends my back to my high school times: "in order to solve this you need to memorize some formulaes and remember how to move and substitue the numbers in formulae"
Then I raised my hand and asked a simple question "Mrs. Teacher, what are we calculating ?"
The hardest question that day turned out to be mine..
Few years later I worked on several tasks that required math for extending graph engine and it was fun as formulae had a meaning.
This is just based on defining a function, this Lambert function, plus some algebra.
Correct answer is: 2 on pwoer of 2 - 2x2 +1 = 5 people tend to make things too complicated.
Uuuuuuh that's not Harvard's entrance exam...it's "what's your race, what are your pronouns, what gender are you this week"? Answer correctly & you're in!
Stop crying. You couldn’t get in either way.
2^x + x = 5
so:
5-x > 0
so:
x < 5
and in this equation we know that if x would be smaller than 0 then answers is smaller than zero , so x is a number between 0 and 5!
if we choose x = 2 answer is 6 and if we choos x= 1 answer is 3 , so x exactly is a number between 1 and 2 and we can calculate x just with som simple and basic ways of approximation in numerical calculations or even try few number between 1 and 2 ! and its exactly what you calculated, approximately 1.7
Using log on both the sides we can solve more easily
That is right back in my log cabin I solved this in no time at all
How do you solve it by log
The answer is somewhere between one and two. Easier to test via reductive methods.
Maybe the log is a matrix of options.
If you had used ln and differentiation
It would be so easy
It is easy to see that the solution is between 1 and 2. Both 2^x + x-5, and its derivate are smooth in the range 1 to 2. We can use one or two iterations of Haley's method to get to x~1.716 (en.wikipedia.org/wiki/Halley%27s_method)
While not a closed form solution, the solution is more generic and a practical numerical method for a broad set of problems.
I would do it loke this :
2^x is a function that rises
x is a function that rises
And there is some kind of law that states if 2 functions are rising then their sum is a rising function as well
So basically we get 2 functions
2^x + x
and
5
If they cross they cross only once
It means there is only 1 solution
Which we have to guess
Why not use numerical approximation instead of doing it the hardest way imaginable?
Since 2^x+x is closer to 5 when x=2 than x=1, take the derivative of 2^x+x and evaluate at x=2 which gives 5.
Then using the approximation of 2^x+x around x=2 we have:
6+5(x-2) is approximately 5.
5x-4 is approximately 5.
x is approximately 9/5 = 1.8
The only Harvard entrance exam is looking at daddy’s portfolio
Or skin color and gender
@@Seagaltalk they/them
@@SeagaltalkYou are not a victim🤡
@@Seagaltalk You are not a victim🤡
@Mariajbh2 now i am even more victimized.. thanks
I tried x=1 and x = 2. So, I understand that x is between 1 and 2. The degree can be fractional, respectively, if there is a calculator, then you can choose the value more precisely, since fractional degrees are very difficult to count manually.
Nice! But the way you write X confuses me on each line✌️😂
simple solve: x^2 - 2log2(ln2)x = 5
What calculators have the lambda-W function? How did you evaluate W( 32 ln(2) / ln2 ??
I found the same answer. If you had to use a calculator to figure out the answer, so you dont need to appeal to W function. Just do try and error, just like I did. Unless you only want the answer in terms of ln2, but in that case you want the decimal number though
Is W actually calculated via a series solution or numerical integration? Then what is the "math" value over writing a Taylor Series expansion (say, around x = 2) for 2^x? You get an estimate 1.735 rather easily with a first-order expansion. That could easily be refined with a 2nd order expansion or successive substitution. Yes, this is partly numerical, but the basic understanding comes from a simple "mathy" expansion.
Solve for x, x = 2^x - 5, etc, so x = 2^{2^x - 5) - 5 = 2^(2^(2^x - 5) - 5) - 5, etc, you get an infinite exponential, not sure what you would call it...
1) Start with the equation: 2^x + x = 5
2) Subtract x from both sides: 2^x = 5 - x
3) Multiply both sides by 2: 2 * 2^x = 2(5 - x)
4) Substitute y = 2^x: y = 2(5 - log_2(y))
5) Rearrange: y/2 = 5 - log_2(y)
6) Exponentiate both sides: 2^(y/2) = 2^(5 - log_2(y))
7) Simplify the right side: 2^(y/2) = 32/y
8) Multiply both sides by y: y * 2^(y/2) = 32
9) Substitute z = y/2: 2z * e^z = 32
10) Divide both sides by 2: z * e^z = 16
11) This is now in the form of the Lambert W function: W(16) = z
12) Recall that y = 2^x, so z = 2^(x-1)
13) Therefore: W(16) = 2^(x-1)
14) Solve for x: x = 1 + log_2(W(16))
This can be evaluated numerically to get the approximate solution x ≈ 1.7095.
Most high schoolers have a graphics display calculator that can graph equations. Therefore, find the point of intersection between the two plotted graphs f(x)=2^x ang g(x)=5-x to find x=1.716
Here is the similar solution. Let the function fck(n), where the value of function is the equation of 2^x+x=n. So, our answer is fck(5)
Everyone please note: This CAN be done nice and simply. BUT - the UA-cam algorithm downrates videos shorter than TEN MINUTES, so the video maker decided to bore everybody with lots and lots of slow, unnecessary steps. If you want to watch something intelligent and interesting, (1) give this a down vote, (2) don't waste your time here, (3) go find another video! Hope that helps. Have a nice day! 😉🙂
How many freaking Harvard applicants know what the Lambert W function is?
Everyone with Rich Parents ²
2^X + X = 5
ln(2^X ) + ln(X ) = ln(5 ) par les proprieté de l'exponentiel on obtient
Xln(2 ) + ln(X ) = ln(5 )
ensuite on factorise
X() ln(2 ) +1 )) = ln(5 )
X = ln(5 )\ (( ln(2 ) +1 ))
X ≈ 1.678
Why are you writing X like that? It drove me nuts!!!
😂😂
Can a solution be found without using the W function?
No. The problem is posed specifically to have a solution with the W. It's like making an equation that solves with a Bessel function. Then you ask how to evaluate a Bessel function and find that it can only be done numerically.
You easily find a solution by looking at the Intersection point of 2^x and 5-x => gives about 1.72
@@WePhFr Yes, it's easy to use numerical methods to get a very close solution.
We can not find correct answer we will get a range like answer should lie between 1.5 to 2
Edit - crap I misread the problem.
A very simple problem you can solve using the bisection method. The solution is the point of intersection of graphs.
x^2 < 2 + 5, double it
x^4 > 4 + 5, midpoint is (2+4)/3
x^3 = 3 + 5, boom.
Lambert w function what this but how do you calcute the final decimal valur
see the Lamert W function at en.wikipedia.org/wiki/Lambert_W_function
Come to India and apply for jee advanced. If you score 50% then you are the legend
Now I know why i went to Technical College! 😂😂😂
And you should be proud of it, it's not because you didn't make to Harvard that you are less than the people that are there or that you doesn't have the same skills they have or even better ones.
This has infinite number of solutions. Lambert W Function with base n. In this case we have used only n=1, whereas n can take any real value.
Maybe once I start UA-cam channel, I can correct these folks
2² +1 =5 😅 Harvard University pass 😂😂
Bhai tune 2 alag alag value di hai x ke liye 1,2
@GaminGg483YT
Bro you did not even pass middle school I think.
@@pacivalmuller9333😂😂
😂😂😂
It took me just a couple of minutes to find - without a pencil - that x should have been a bit less than 2, and a bit more than 1.6. Thus it really does not need a long calculation as yours to find x = 1.72
Its not 1,66?
2 times 1.66 is 3.33 then add 1.66 equals 5. What is this 1.72???
@@tomsanders2870 You totally misunderstood the question: it is 2^x , namely 2 raised to x, and not two times x. Thus 2^1.72 = 3.29 + 1.72 = 5.01.
8:21 idk what is that mean
you can also use Banach contraction principle in the form x(n+1)=log_2(5-x(n)) and initial value x(0)=1,5 , convergence is quite good here.
I like you method. You can try this method :
1) 2^x= 5-x then 1
My answer is x = 3. Is it the right answer? No! But it's MY answer since it still technically proves the theorem, just not from the original. Let me explain. 2^x + x = 5 -> 2^x = 5 - x -> 2^x/2 = 1 and (5-x)/2 -> 1 = (5-x)/2 -> multiply both sides by 2 -> 5-x = 2 -> add x to both sides and subtract 2 from both sides -> 5-2 = x so x = 3. Again, not the correct answer, but i did show my work
Why make it complicated bro, simply take log common from the both sides....and solve it
huh? If he takes log, what's the value of logx?
and the easiest way to solve this is to draw a graph and find the solution
Eso mismo pensé. Pero no sabía si era correcto, hace más de 4 años que no veo ni una cuenta matemática jaja
Old math tells me, if x=1, eq=3; if x=2, eq=6. So x= # between 1&2. All that ridiculous convolution for a result that can be deducted within seconds. Absurd.
Just use approximations for x value then lol. Whatever comes closest to 5 is the answer
2^x = 82/25
So,
x = 5 - 2^x
= 5 - ( 82/25 )
= 43/25
~ 1.72
What is w(32ln2) then? Can it not be simplified further?
Calculator
Came here for algebra,left with an Oppenheimer theme song.
I think realy this exam purpose is not about solving it completely, instead to show the examiner the way you think and the steps you followed to solve this problem, even if you couldn't solve it
Indeed
You are using order operations and natural log to converge to the answer subtracted by 5. I completely understand.