doubt you will respond but it’s so interesting to see you go to the school i’m attending this fall (class of 2026). Your videos have and will continue to be an inspiration to me thank you so much
The answer to 1st question is pretty simple, you actually don't even need a pen for it. Just take the RHS to left and add 2 both sides and you get (log(x)-1)²+(log(y)-1)²=2 And you will get maximum value for x=y=b² where b is the base.And we get the desired 16. For the second question I assumed f(x)=y f(f(x))=z and f(f(f(x)))=w And also f(x)≥-2 Now f(w)=0 w=(2)^(1/2)-2 (since w≥-2) f(z)=(2)^(1/2)-2 z=(2)^(1/4)-2 (since z≥-2) f(y)=(2)^(1/4)-2 y=(2)^(1/8)-2 (since≥-2) f(x)=(2)^(1/8)-2 And now x=-(2)^(1/16)-2 or (2)^(1/16)-2 And so r=2×(2)^(1/16) = (2)^(17/16) The values we need.2+17+16=35
I’m looking to do math/physics at a similar level school. I’m only beginning my junior year in high school soon. How has it been for you so far (coursework, support, in general)? If you don’t mind.
@@balloonman2262 Princeton is a great environment to do Math/Physics. On the plus side, it is an incredibly collaborative environment. The math department has daily tea and cookies (and these are not usually cheap cookies/desserts either). The level of theory is very high. Math majors are not required to take a course like differential equations for example because the department doesn't believe it is needed for theoretical work. This may be a downside, but the classes are quite hard. Problem sets are too challenging to do alone and exam averages can be quite low (I'm talking 40-50 % sometimes). You work a lot but if you love math/physics it is quite rewardin
@@benjaminbobell4028 Thank you very much. I believe I would definitely enjoy the level of work. I’m currently taking a physics course over the summer and find resources to practice/learn more mathematics. I cannot wait to get to more advanced classes. I just wish I had a better support system. Nobody else in my classes is as enthusiastic about it as me, and my teachers couldn’t be bothered. Doesn’t deter me though. I’m glad to confirm that it is the opposite there. But enough about myself, thank you again for the response.
For 1st question After taking log(x)= a , log(y) = b (a-1)²+(b-1)²=2 Which is question of circle with centre (1,1) and radius r =2½ a=rcosg +1 b=rsing +1 a+b =r (sing +cosg) +2 Max value of sing +cosg =2½ Max value of a+b=4 Log xy = a +b = 4 (Log(xy))²=16
I found the solution to the first problem with the use of lagrange multipliers. For the equality constraint i used g(x,y)= log(x)^2+log(y)^2-log(x^2)-log(y^2) and it worked perfectly. Maybe not exactly an algebraic solution, but it's nice to see different approaches
A = log(x), B = log(y) A² + B² = 2(A+B), want max of (A+B)², equivalently maximizing A+B, since the equation tells us it's positive. Equation gives us the circle of radius sqrt2 centered at (1,1). Maximum achieved when A+B = c is tangent to the circle, which is either at (0,0) or (2,2). Obviously 2,2.
In first question replace 'log x' with h and 'log y' with k and you will get a equation of circle. I drew the circle and it was obvious h=2 and k=2 will give maximum value.
_Second Question:_ Rewrite *f(x) = (x + 2)^2 - 2* . Compute *f(f(x))* and notice the inner *2s* cancel: *f(f(x)) = (x + 2)^4 - 2 =: f_2(x)* Compute the composition of four instances of *f* . Again the inner *2s* cancel: *f(f(f(f(x)))) = f_2(f_2(x)) = (x + 2)^16 - 2 =: f_4(x)* The only real zeros of *f_4* are *⨦2^{ 1/16 }* and we get *r = 2 * 2^{ 1/16 } = 2^{ 17 / 16 }* The answer is *2 + 16 + 17 = 35* . Via induction one can generalize this to compositions of _n_ instances of *f* : *f_n(x) = f(...f(x)...) = (x + 2) ^ { 2^n } - 2*
For the second question - I failed to understand that "square-free" condition. If you approach it directly, f(x) = x²+4x+2 = (x+2)²-2 so f(f(x)) = ((x+2)²-2+2)²-2 = (x+2)⁴-2 and in general, fⁿ(x) = (x+2)ᵐ-2 where m=2ⁿ, so f⁴(x)=f(f(f(f(x)))) = (x+2)¹⁶-2 meaning you can just brute-force from here: xₘᵢₙ = -2-2¹ᐟ¹⁶ and xₘₐₓ = -2+2¹ᐟ¹⁶ so r = 2∙2¹ᐟ¹⁶ = 2¹⁷ᐟ¹⁶ which means we found co-prime p, q and thus the answer is 2+17+16 = 35
It’s like that so there is a unique answer like 5^3 and 125^1 are numerically the same value but when u add the number u get 8 vs 126 So basically the format is there to have a unique answer by simplifying
@@randomjin9392 If the "square-free" statement wasn't there you could rewrite *r* as: *r = a^{ gp / (gq) } = (a^g) ^ { p / (gq) }* If you choose an integer *g > 1* such that *g* is relatively prime to *p* , the new exponents *p, gq* are still relatively prime. But now the new base *a^g* contains the square *a^2* !
For the first question, if you simplify it by making log(x) be X and log(y) be Y, and complete the square, you will notice that it is the equation of a circle with radius sqrt(2) and center 1,1. You then know log(xy) is maximized with a 45 degree triangle from the center and solve it. I got 16 like the question asked.
@@Ducster03 if you imagine the (X+Y)^2 as a 3D function it becomes clear that the farthest point on the circle away from the origin will provide ye highest value
@@Ducster03 I don't know how to show it without calculus. But you can try some other points on this circle and see how they add up. In this case log(xy) is X+Y in terms of the dummy variables
@@Noissimsarm i found you can show it by substituing x and y with their values depending on the radius (√2 in this case) and the angle they form from the origin, like their parametric values. From there X + Y is a function depending on only one variable (their angle) which can be easily be differentiated to find that π/4 is a maximun. Replacing this value you get X=Y=2
First one can be solved in a single step by using RMS>=AM. We get: √(log(x)^2+log(y)^2)/2>=(log(x)+log(y))/2 √log(xy)>=log(xy)/2 Log(xy)^2-4log(xy)=0 Log(xy)[log(xy)-4]=0 Log(xy)=4 Log(xy)^2=16
_First Question:_ (Pure algebra without Lagrange-Multiplier) Substitute *a = ln(x)* and *b = ln(y)* , move all terms of the first equation to the left and complete the square: *(a - 1)^2 + (b - 1)^2 = 2* The term to maximize simplifies to *|a + b|^2* . To make it depend on only one variable, substitute *a - 1 = u + v* *b - 1 = u - v* The circle equation simplifies to *u^2 + v^2 = 1 => |u|
In second question you can simplify f(x). That is f(x)= (x+2)^2 - 2 . And its easy to make f(f(f(f(x))))= (x+2)^16 - 2 = 0 . So the only roots are x= 2^-16 -2 ; and x= -2^-16 -2 , because power is even*. Therefore r= 2^(17/16) . So a +p +q is 2+17+16=35.
June Huh has an absolutely amazing story as well. He dropped out of high school, took six years to finish his bachelors, and got his PhD at 31. He now holds the Fields Medal. It’s never too late to go back to school and pursue a passion!
The general method for the solutions of Q1 type: "maximize f subject to g = 0" is by Lagrange Multipliers. I don't particularly like particular solutions like the one given, even though is not hard at all.
For the 2nd one, I decided to go a less rigorous route: if f^4(x)=0, then for g=f^-1, x=g^4(0). g(x) = +-sqrt(x+2) -2 g(0) = +-sqrt(2) -2. Compose this three more times, eliminate +- inside the square roots so we stay in the real numbers, and we get: x = +-2^(1/16) r = 2^(1/16) - -2^(1/16) r = 2^(17/16) a=2, p=17, q=16.
John Nash's wife was my mother's cousin. They were quite a duprass. I got 16 with a little thought. a=log)x, b=log(y), and a=b=2 is a solution of the equation.
At the link given, click the "Archive" link to get to the PUMaC Problem and Result Archive. Then in 2019: SOLUTIONS under Division A ALGEBRA Needing more instructions on how to get to the page shown at 02:50 is probably one of the reasons why I didn't go to Princeton.
I was at stanford a few months ago (late april) and there was a whiteboard left at the bottom of a tree in the engineering quad. I should've re-enacted a bprp video there lmao. I'm sure it's been moved by now though
@@renatoamadori4125 your proof assumes that {sqrt(x^2+y^2)}^0 =1. So you’ve shown (any non zero complex number)^0 = 1 *provided* (any non zero real number)^0 = 1.
(X^1) ÷ (X^1) = X^0 by subtracting the exponentials. However since the above equation is a number divided by itself, it is therefore equal to 1. This logic applies for any number, real or complex, except 0, as it would involve performing 0 ÷ 0 which is an indeterminate form (basically a mathematical no no)
Hey, I was wondering if you or any of the smart people in the comments can help with a math problem. 5^x + 6^x + 7^x = 8^y + 9^y + 10^y There should be an infinite number of solutions that can be found by an equation (y = …) that can be made from this one. The question is what is the equation. There should only x’s on the other side
i just did log(x)^2 = 2log(y) and tried simple values and for 100 for x and y found that both equal the same and ended up getting 16. no idea what i proved or why it worked but i did get the answer so cool ig
May I boast that I solved the fist log question quite quickly just using logarithm rules, but I wouldn't even try the second question. It doesn't grab me, and it looks very difficult.
I figured out the first one by just modeling a 3D graph in my head, and drawing the first function and the second function. Where they equate, are the only inputs allowed for the third equation. The first that gets multiplied by two is always too small compared to the second who gets squared. Except, when squaring something and multiplying it by two are the same thing! Therefore logx =2, and so does logy. I didnt have the time to solve the second one, but turning fx into the form (x+2)^2 -2 shows that fffx= sqrt2 - 2. That would mean (x1+2)^2 = sqrt2 - 2 now, and et cetera; it’s simple enough to keep chaining this, until you have derived the original x. Deriving the other x, I dunno. Its probably just negative x. The difference between them would actually become a sum, since a negative is technically smaller than a positive and so goes behind the minus sign.
@@rishavbagri4211 we know that arithmetic mean >= geometric mean . If use no.s whose mean is being taken as logx)² and logy)² and also use relation given in question we would end up with (logxy)² >= (logx)²(logy)² Now since we know the AM=GM only when no.s taken are equal to each other hence we can solve for logx and logy as 2 each and we get minimum value as 16. Hope you understood 🙂👍🏻
@@suyashkarnad5391 question have asked for minimum value of (logxy)² so it should be 4²=16. And ans written in the solution he shown was also 16. May be you misread it.
🇮🇳In 1969, Indian scientist Nambi Narayanan won a prestigious NASA fellowship that helped him find his way into Princeton University, where he gained expertise in chemical rocket propulsion. Proud to be an Indian.🇮🇳
@@Mohd_Zaid_khan In 1969, as a very much undeveloped country, a scientist from this land was successful to go to Princeton under a NASA scholership. This is a proud of our country, this was a moment of glory for India. Being an Indian, I will say with very much high confidence that I AM PROUD TO BE AN INDIAN. YES I AM PROUD OF BEING A PERSON FROM THAT LAND OF NAMBI NARAYANA,RAMANUJAN AND MORE OTHER LEGENDS. YES I AM PROUD TO BE AN INDIAN. You should be ashamed that feel annoying to take the name of your MOTHERLAND. Shame on you.
doubt you will respond but it’s so interesting to see you go to the school i’m attending this fall (class of 2026). Your videos have and will continue to be an inspiration to me thank you so much
Thank you! Wishing you the best there!!
The answer to 1st question is pretty simple, you actually don't even need a pen for it.
Just take the RHS to left and add 2 both sides and you get (log(x)-1)²+(log(y)-1)²=2
And you will get maximum value for x=y=b² where b is the base.And we get the desired 16.
For the second question I assumed f(x)=y f(f(x))=z and f(f(f(x)))=w
And also f(x)≥-2
Now
f(w)=0
w=(2)^(1/2)-2 (since w≥-2)
f(z)=(2)^(1/2)-2
z=(2)^(1/4)-2 (since z≥-2)
f(y)=(2)^(1/4)-2
y=(2)^(1/8)-2 (since≥-2)
f(x)=(2)^(1/8)-2
And now x=-(2)^(1/16)-2 or (2)^(1/16)-2
And so r=2×(2)^(1/16) = (2)^(17/16)
The values we need.2+17+16=35
I do Math/Physics at Princeton and the first question was written by my friend Matthew. Love you man, glad your problem was featured here.
I’m looking to do math/physics at a similar level school. I’m only beginning my junior year in high school soon. How has it been for you so far (coursework, support, in general)? If you don’t mind.
@@balloonman2262 Princeton is a great environment to do Math/Physics. On the plus side, it is an incredibly collaborative environment. The math department has daily tea and cookies (and these are not usually cheap cookies/desserts either). The level of theory is very high. Math majors are not required to take a course like differential equations for example because the department doesn't believe it is needed for theoretical work. This may be a downside, but the classes are quite hard. Problem sets are too challenging to do alone and exam averages can be quite low (I'm talking 40-50 % sometimes). You work a lot but if you love math/physics it is quite rewardin
@@benjaminbobell4028 Thank you very much. I believe I would definitely enjoy the level of work. I’m currently taking a physics course over the summer and find resources to practice/learn more mathematics. I cannot wait to get to more advanced classes. I just wish I had a better support system. Nobody else in my classes is as enthusiastic about it as me, and my teachers couldn’t be bothered. Doesn’t deter me though. I’m glad to confirm that it is the opposite there. But enough about myself, thank you again for the response.
For 1st question
After taking log(x)= a , log(y) = b
(a-1)²+(b-1)²=2
Which is question of circle with centre (1,1) and radius r =2½
a=rcosg +1
b=rsing +1
a+b =r (sing +cosg) +2
Max value of sing +cosg =2½
Max value of a+b=4
Log xy = a +b = 4
(Log(xy))²=16
is this correct, i got this but i dont know if its correct
I found the solution to the first problem with the use of lagrange multipliers. For the equality constraint i used g(x,y)= log(x)^2+log(y)^2-log(x^2)-log(y^2) and it worked perfectly. Maybe not exactly an algebraic solution, but it's nice to see different approaches
A = log(x), B = log(y)
A² + B² = 2(A+B), want max of (A+B)², equivalently maximizing A+B, since the equation tells us it's positive.
Equation gives us the circle of radius sqrt2 centered at (1,1). Maximum achieved when A+B = c is tangent to the circle, which is either at (0,0) or (2,2). Obviously 2,2.
.
@@f5673-t1h You may be right but your train of thought is impossible to follow without some more explanation.
In first question replace 'log x' with h and 'log y' with k and you will get a equation of circle. I drew the circle and it was obvious h=2 and k=2 will give maximum value.
I got (h+1)^2+(k+1)^2 = 2, how do you go from there?
@@jesroe5842 the equation of circle becomes (h-1)^2 + (k-1)^2 = 2 **
Here h=2 and k= 2 gives max value for the question
@@tejasg6729 ohh, thanks!
Hey, this is a nice environment change. Thanks for taking the time to film there, man. Great vid.
_Second Question:_ Rewrite *f(x) = (x + 2)^2 - 2* . Compute *f(f(x))* and notice the inner *2s* cancel:
*f(f(x)) = (x + 2)^4 - 2 =: f_2(x)*
Compute the composition of four instances of *f* . Again the inner *2s* cancel:
*f(f(f(f(x)))) = f_2(f_2(x)) = (x + 2)^16 - 2 =: f_4(x)*
The only real zeros of *f_4* are *⨦2^{ 1/16 }* and we get
*r = 2 * 2^{ 1/16 } = 2^{ 17 / 16 }*
The answer is *2 + 16 + 17 = 35* . Via induction one can generalize this to compositions of _n_ instances of *f* :
*f_n(x) = f(...f(x)...) = (x + 2) ^ { 2^n } - 2*
For the second question - I failed to understand that "square-free" condition. If you approach it directly, f(x) = x²+4x+2 = (x+2)²-2 so f(f(x)) = ((x+2)²-2+2)²-2 = (x+2)⁴-2 and in general, fⁿ(x) = (x+2)ᵐ-2 where m=2ⁿ, so f⁴(x)=f(f(f(f(x)))) = (x+2)¹⁶-2 meaning you can just brute-force from here: xₘᵢₙ = -2-2¹ᐟ¹⁶ and xₘₐₓ = -2+2¹ᐟ¹⁶ so r = 2∙2¹ᐟ¹⁶ = 2¹⁷ᐟ¹⁶ which means we found co-prime p, q and thus the answer is 2+17+16 = 35
I think the square free part just means that it shouldn't be a perfect square.
@@HershO. I got what it means but I didn' get why there is such a condition (i.e. in my solution I didn't use it)
It’s like that so there is a unique answer like 5^3 and 125^1 are numerically the same value but when u add the number u get 8 vs 126
So basically the format is there to have a unique answer by simplifying
@@randomjin9392 as Kent Ziti said, its to avoid multiple sols
@@randomjin9392 If the "square-free" statement wasn't there you could rewrite *r* as:
*r = a^{ gp / (gq) } = (a^g) ^ { p / (gq) }*
If you choose an integer *g > 1* such that *g* is relatively prime to *p* , the new exponents *p, gq* are still relatively prime. But now the new base *a^g* contains the square *a^2* !
Still wondering how he was able to read and pinpoint every word whilst staring at the back of the paper 🤔
lmao I don't understand either
the paper must be slightly see-through
When there's light through the other side, you can clearly see ink on the opp side.
For the first question, if you simplify it by making log(x) be X and log(y) be Y, and complete the square, you will notice that it is the equation of a circle with radius sqrt(2) and center 1,1. You then know log(xy) is maximized with a 45 degree triangle from the center and solve it. I got 16 like the question asked.
why is log(xy) maximized when you have a 45 degree triangle?
@@Ducster03 if you imagine the (X+Y)^2 as a 3D function it becomes clear that the farthest point on the circle away from the origin will provide ye highest value
@@Ducster03 I don't know how to show it without calculus. But you can try some other points on this circle and see how they add up. In this case log(xy) is X+Y in terms of the dummy variables
@@Noissimsarm i found you can show it by substituing x and y with their values depending on the radius (√2 in this case) and the angle they form from the origin, like their parametric values.
From there X + Y is a function depending on only one variable (their angle) which can be easily be differentiated to find that π/4 is a maximun.
Replacing this value you get X=Y=2
First one can be solved in a single step by using RMS>=AM. We get:
√(log(x)^2+log(y)^2)/2>=(log(x)+log(y))/2
√log(xy)>=log(xy)/2
Log(xy)^2-4log(xy)=0
Log(xy)[log(xy)-4]=0
Log(xy)=4
Log(xy)^2=16
Smart 💯💯
Professor June Huh, Fields Medal Winner 2022
_First Question:_ (Pure algebra without Lagrange-Multiplier)
Substitute *a = ln(x)* and *b = ln(y)* , move all terms of the first equation to the left and complete the square:
*(a - 1)^2 + (b - 1)^2 = 2*
The term to maximize simplifies to *|a + b|^2* . To make it depend on only one variable, substitute
*a - 1 = u + v*
*b - 1 = u - v*
The circle equation simplifies to
*u^2 + v^2 = 1 => |u|
In second question you can simplify f(x).
That is f(x)= (x+2)^2 - 2 . And its easy to make f(f(f(f(x))))= (x+2)^16 - 2 = 0 .
So the only roots are x= 2^-16 -2 ; and x= -2^-16 -2 , because power is even*. Therefore r= 2^(17/16) . So a +p +q is 2+17+16=35.
This is the first time I actually got a question right while watching these math comp videos.
the recent Fields medalist June Huh is from Princeton!
June Huh has an absolutely amazing story as well. He dropped out of high school, took six years to finish his bachelors, and got his PhD at 31. He now holds the Fields Medal. It’s never too late to go back to school and pursue a passion!
I agree with you about solving math competition questions. It feels great to solve a difficult problem that took some actual thinking to solve
The general method for the solutions of Q1 type: "maximize f subject to g = 0" is by Lagrange Multipliers. I don't particularly like particular solutions like the one given, even though is not hard at all.
For the 2nd one, I decided to go a less rigorous route: if f^4(x)=0, then for g=f^-1, x=g^4(0).
g(x) = +-sqrt(x+2) -2
g(0) = +-sqrt(2) -2.
Compose this three more times, eliminate +- inside the square roots so we stay in the real numbers, and we get:
x = +-2^(1/16)
r = 2^(1/16) - -2^(1/16)
r = 2^(17/16)
a=2, p=17, q=16.
Awesome!!! That was a pretty dang cool vid!
John Nash's wife was my mother's cousin. They were quite a duprass.
I got 16 with a little thought. a=log)x, b=log(y), and a=b=2 is a solution of the equation.
It is only a maximum value that being asked and not the sum of all possible value. That's why I choose 16.
in the first question i start with log x= a and log y = b
then some simpel algebra with (a+b)2= a2 +b2 +2ab
then i use AM>=GM theorem
i get log(xy)
Managed to get first solution with Lagrange multipliers, kinda stumped with the rest rn, I’ll give them a try
this guy is kinda obsessed and i love it
My most favourite math problem is solving integrals!!!
calculus is very easy once you know calculus
@@rhombicuboctahedron7811 true calculus is very hard for those who don't study it properly, and easy for those who study it properly
At the link given, click the "Archive" link to get to the PUMaC Problem and Result Archive.
Then in 2019: SOLUTIONS under
Division A
ALGEBRA
Needing more instructions on how to get to the page shown at 02:50 is probably one of the reasons why I didn't go to Princeton.
Sir which books you can recommend for a 12 grade student to learn more concepts and theories of calculas and conics?
Minecraft redstone handbook
@@jocabulous I am so glad that more recognition is being given to this great book, it taught me calculus when I was 8. 10/10 would recomend
@@jocabulous Can vouch for that book. Even though I'm only a 9 year old CS Major, the Electrical Engi's are terrified of my redstone prowess.
For logarithm question i have used arithmetic mean >= geometric mean it just becomes so simple
I was at stanford a few months ago (late april) and there was a whiteboard left at the bottom of a tree in the engineering quad. I should've re-enacted a bprp video there lmao. I'm sure it's been moved by now though
The whiteboard is still there occasionally. Currently not up, but it is used for physics meetings and occasionally physics lectures
@@isaacaguilar5642 ah interesting. Seems like an interesting spot to stick a whiteboard. Could definitely be a nice spot have a meeting though
When you spoke of famous mathematicians who worked at Princeton you forgot Kurt Gödel.
While your there eat some good Jersey food. Harold’s deli in Edison is a good spot
Thanks. I had a lobster roll at la la lobster that day. Unfortunately I was only there for a few hours so I couldn’t explore much.
Don’t forget the “famous” physician from Princeton: Dr Leonard Hofsteader !!
Woah I live in Princeton 👀.
I love these problems thanks
Please find the range of the function
f(x)=√(9-x^2)/√(x^2-1)
For all x belongs to real numbers
Very nice problems
I think the solution for the first question is lacking some additional explanatory steps. The second solution is fine however.
Can you plz make a video on
(Any complex number)^0=1 excluding 0 ,,whyyy?
Let a \in \mathbb{C} / {0}.
Let m \in \mathbb{R}.
Then,
a^m/a^m = 1 (trivially)
and
a^m/a^m = a^(m-m) (Laws of indices)
= a^0
Hence a^0 = 1
QED
Apply Euler's Equation: z = x + i*y = [sqrt(x^2 +y^2 )]*[exp(i*theta)] {This is the Polar Form}
theta = arctg(y/x) {angle in radians}
exp(i*theta) = cos(theta) + i*sin(theta) {This is the Euler's Equation}
z^0 = {[sqrt(x^2 +y^2 )]*[exp(i*theta)]}^0 = {[sqrt(x^2 +y^2 )]^0}*{[exp(i*theta)]^0} = 1*exp(i*theta*0) = exp(0) =
= cos(0) + i*sin(0) = 1 + i*0 = 1
@@renatoamadori4125 your proof assumes that {sqrt(x^2+y^2)}^0 =1. So you’ve shown (any non zero complex number)^0 = 1 *provided* (any non zero real number)^0 = 1.
(X^1) ÷ (X^1) = X^0 by subtracting the exponentials. However since the above equation is a number divided by itself, it is therefore equal to 1. This logic applies for any number, real or complex, except 0, as it would involve performing 0 ÷ 0 which is an indeterminate form (basically a mathematical no no)
Hey, I was wondering if you or any of the smart people in the comments can help with a math problem.
5^x + 6^x + 7^x = 8^y + 9^y + 10^y
There should be an infinite number of solutions that can be found by an equation (y = …) that can be made from this one. The question is what is the equation.
There should only x’s on the other side
Also this is for all real values of x and y
i just did log(x)^2 = 2log(y) and tried simple values and for 100 for x and y found that both equal the same and ended up getting 16. no idea what i proved or why it worked but i did get the answer so cool ig
I got dizzy just trying to read the questions
I did them both correctly in my head, so I feel pretty good about that.
WAIT YOU CAME??? I'm an undergraduate I WOULD HAVE LOVED TO MEET YOUU
good video
I'm bad at math and it makes me feel very insecure. Is this normal? Help
Hello sir i am your student from India
Nice
Calc 1?
May I boast that I solved the fist log question quite quickly just using logarithm rules, but I wouldn't even try the second question. It doesn't grab me, and it looks very difficult.
The opposite for me. The second one was really easy by just changing the form of the quadratic but I had no clue how to do the first one.
@@kritical6033 Ohh it was easy. I should have given it a go. Maybe I'll try again. Thanks 🤠
Nice
I liked the video.
I figured out the first one by just modeling a 3D graph in my head, and drawing the first function and the second function. Where they equate, are the only inputs allowed for the third equation. The first that gets multiplied by two is always too small compared to the second who gets squared. Except, when squaring something and multiplying it by two are the same thing!
Therefore logx =2, and so does logy.
I didnt have the time to solve the second one, but turning fx into the form (x+2)^2 -2 shows that fffx= sqrt2 - 2.
That would mean (x1+2)^2 = sqrt2 - 2 now, and et cetera; it’s simple enough to keep chaining this, until you have derived the original x. Deriving the other x, I dunno. Its probably just negative x. The difference between them would actually become a sum, since a negative is technically smaller than a positive and so goes behind the minus sign.
I've solved both of questions
Is brilliant worth
Why are they so ridiculously hard?
Mind your p’s and q’s!
Ans 1 is 16
1st one was quite basic, classic use of AM GM . 2nd one was a nice and lengthy 😃👍🏻
How can we use am gm?
@@rishavbagri4211 we know that arithmetic mean >= geometric mean . If use no.s whose mean is being taken as logx)² and logy)² and also use relation given in question we would end up with
(logxy)² >= (logx)²(logy)²
Now since we know the AM=GM only when no.s taken are equal to each other hence we can solve for logx and logy as 2 each and we get minimum value as 16. Hope you understood 🙂👍🏻
Yes
Your method is right, but the answer is actually 4, not 16
@@suyashkarnad5391 question have asked for minimum value of (logxy)² so it should be 4²=16. And ans written in the solution he shown was also 16. May be you misread it.
Imagen you Rob his house and he pull up a math question that you need to solve
logx=logy=2
Which maths book are u using
Bro i challenge u to integrate
F(X)2 TO 0 = X²-X+1
That’s cool！Which country are you from?
I think answer of 5th question is 0
Well I will ask something diffrent. Give me hard math, but practical math, not something I will use once a year.
THIS IS INDEED HIGH SCHOOL MATHS!.. 😀😀😁👍👍❤️❤️❤️
Stop yelling in all caps.
Sir I think the answer of first question 16
Sir could we do it by symmetry?
This is yes
f of fof fafof😂😂
f of f of f of f of f
WOW!! PRINCETON UNIV. HOPING TO VISIT THERE TOO!!!.. 🙏🙏🙏😀😀😀😁😁YOU CAN STUDY THERE TOO!.. ❤️❤️👍👍✌️🦅🇵🇭
Stop yelling your post in all caps. You are rude and entitled.
👍🏾👍🏾👍🏾
asnwer=1 isit 🤣😗😄😄😄
meow
interesting 🤔
@@chadlerthethird5683 💀
Lol
🇮🇳In 1969, Indian scientist Nambi Narayanan won a prestigious NASA fellowship that helped him find his way into Princeton University, where he gained expertise in chemical rocket propulsion.
Proud to be an Indian.🇮🇳
ok
stop typing "proud to be indian" it's annoying, i am also indian but you're saying as if india is ruling the world lmao
@@Mohd_Zaid_khan In 1969, as a very much undeveloped country, a scientist from this land was successful to go to Princeton under a NASA scholership. This is a proud of our country, this was a moment of glory for India.
Being an Indian, I will say with very much high confidence that I AM PROUD TO BE AN INDIAN. YES I AM PROUD OF BEING A PERSON FROM THAT LAND OF NAMBI NARAYANA,RAMANUJAN AND MORE OTHER LEGENDS.
YES I AM PROUD TO BE AN INDIAN.
You should be ashamed that feel annoying to take the name of your MOTHERLAND. Shame on you.
@@kenechi__ ok
love the enthusiasm for math
Is brilliant worth