7 сағат бұрын
16 сағат бұрын
19 сағат бұрын
21 сағат бұрын
14 күн бұрын
Пікірлер
@nikolayguzman331 55 минут бұрын
a>0 b>0 c>0 and a, b, c are integers. Therefore a<8 b<8 c<8 Let a=7 2^7=128. Difference 148-128=20; b=4; c=2; If a=6; b=6 and there’s no results for 2^c.😊
@kimsanov Сағат бұрын
So a = 1/2 - sqrt(965)/2 and a = 1/2 + sqrt(965)/2 are lost
@hottuamanullang8046 Сағат бұрын
Thank you it makes me know math a little bit🥰🥰🥰
@franciscoerivandelimalima2569 3 сағат бұрын
x + y = 7 + 5 = 12
@kimsanov 6 сағат бұрын
You divided both sides to (b-a). But you didn’t check branch where a=b
@erwinkurniadi1850 8 сағат бұрын
Where did you get this, log a/log b = log a - log b ?
@Stresss70 13 сағат бұрын
Сразу подставил 0
@lolserm 15 сағат бұрын
Dude you could have solve that in just one line. The obvious answer from the beginning was x=0
@prime423 15 сағат бұрын
This is easier to solve with observation,not perspiration!!5 squared minus two squared is 21.10 seconds not 10 minutes.Mathletes should use "Tricks" only when needed !!
@anatolykatyshev9388 16 сағат бұрын
Obvious solution: a=b=c= ln(148/3)/ln(2)
@user-iw2zn2mj3c 18 сағат бұрын
Справа 1или7
@user-iw2zn2mj3c 18 сағат бұрын
Можно не решать. И так очевидно
@user-iw2zn2mj3c 18 сағат бұрын
О
@bnwesley 18 сағат бұрын
Log(a/b)=loga-logb
@vinayseth5899 22 сағат бұрын
05:33 Why are these "perfect squares" though?
@numbers93 17 сағат бұрын
I think they simply meant squares here. Not necessarily perfect squares (as in squares of integers)
@yashtyagi1255 Күн бұрын
Good teaching❤
@onelastmore5924 Күн бұрын
Nicely simplified. And then you get the calculation wrong….
@PJR1988 Күн бұрын
It's wrong, you cannot say (a-b)^2 >=0 (also it's not possible assume that for the others) you are missing the complex space with this assumtion and it only worksbin this particular case, propper way is using logaritms
@bishtkunal10 Күн бұрын
This question has an oral solution, and u r right it's 0
@opschpiglung Күн бұрын
For any number, not just 148 - just convert it to binary and look at the positions of 1s. That's it. Is it really Olympiad question?
@prosenjitdas8269 Күн бұрын
0
@matiasgualco905 Күн бұрын
0 si x=0 todos los terminos dan 1 y habria 3 positivos y 2 negativos 3-2=1
@user-dq3uh6ee5w Күн бұрын
+_V2 (here real values).
@user-dq3uh6ee5w 2 күн бұрын
0.
@vuyyurisatyasrinivas3849 2 күн бұрын
Iit jee advanced problem....good solution
@mks-learning-easy 2 күн бұрын
Thank you 😊
@ritwikgupta3655 2 күн бұрын
0 was obvious but proving it was tricky. Setting up a sum of three squares was smart. Zamir, Imaginary exponent of a positive integer does not give a negative I think?
@irinabaigozina8551 2 күн бұрын
Amazing!
@mks-learning-easy 2 күн бұрын
Thank you 😊
@nxtkyle8412 2 күн бұрын
good video! keep going dont give up
@mks-learning-easy 2 күн бұрын
🙏🙏
@shashvat1879 2 күн бұрын
Me with value putting 🗿
@jamesharmon4994 2 күн бұрын
Just write 148 in binary. Each "1" is an applicable power of 2
@SuperAnangs 2 күн бұрын
x=16, solved in 4 sec
@mijazukant 2 күн бұрын
really? is that the time it took you to slide the cursor directly to the end of the video ? if not, please explain what you could have possibly thought in the short period of 4 sec, after reading that and saying "yep thats 16" 😂😂😂
@SuperAnangs 2 күн бұрын
@@mijazukant first tried by 2 then I got the wrong answer, then 4 But I aware then that it's in root number So, I multiply 4 with itself to get the solution
@user-dq3uh6ee5w 2 күн бұрын
16.
@user-tg2gm1ih9g 2 күн бұрын
assuming a,b,c integers ... 128 + 16 + 4 ... 2^7, 2^4, 2^2 ... takes under 5 seconds
@jerrypaquette5470 2 күн бұрын
That is how I did it.
@user-ow1fs8et9l 2 күн бұрын
Regular exercise in Russian school program
@xgx899 3 күн бұрын
Without stating that a,b,c are integers, and proving uniqueness (apart from permuting numbers 2,4,7) this is not good at all.
@vshanmugamv5868 3 күн бұрын
2, 4 , 7
@melissajenner8068 3 күн бұрын
Only one of all solutions.
@vuyyurisatyasrinivas3849 3 күн бұрын
Good solution. Not come at a time
@monroeclewis1973 3 күн бұрын
Quicker: factor 5^ (x+4) from left side; then 5^(x+4)[1+ 5^2] Expression in brackets =26. Divide both sides by 26. Then 5^(x+4)=1. But 1=5^0. Therefore, exponents are equal; x+4=0; x=-4.
@mgabiel 3 күн бұрын
X=0 me tomó 1/4 de segundo
@kailasamraja8808 3 күн бұрын
Congrats. It is not the question of time required to solve. The author demonstrates the sequential steps in attempting the question. Suppose that the question has 10 Mark's. If you write the answer as x = 0 in one line, you will get only one mark perhaps. I am kidding dear. Take it as a fun.
@AliAhmad-si4fb 3 күн бұрын
Very nice
@mks-learning-easy 3 күн бұрын
Thank you 😊
@michaelyap939 3 күн бұрын
0.. why bother?
@markusbanach-stb5892 4 күн бұрын
That is a nice problem indeed. Thanks for the video!
@mks-learning-easy 4 күн бұрын
Thank you 🙏🙏
@BruceLee-io9by 4 күн бұрын
I did this: I placed 4^x=a and 5^x=b. Then I divided everything by ab. I replace a/b with y. I get y^2+y-1=0. I solve and I get ln(-1+sqrd5/2)/ln4/5. So x=2.157.
@mks-learning-easy 4 күн бұрын
Thank you for your sharing 🙏🙏
@BruceLee-io9by 4 күн бұрын
@mks-learning-easy Your solution is the most elegant and simple. Thank you for your work.
@shannonwalker6944 5 күн бұрын
4:29. Maybe use * instead of x to indicate multiplication as here x 2 - (2)^.5 / 2 - (2)^.5 looks a lot like x^2-(2)^.5/2-(2)^.5. I am a big fan and a subscriber. Keep up the great work!
@mks-learning-easy 4 күн бұрын
Thank you ❤❤❤
@BruceLee-io9by 5 күн бұрын
👍
@davidmcknlght2700 5 күн бұрын
👍
@mohammedmustafa7563 5 күн бұрын
Good❤
@ArmandoGarzaMx 6 күн бұрын
Great solutiuon I start with logaritms righ away Get the 4... but your solution is elegant
@mks-learning-easy 6 күн бұрын
Thank you 😊
@vanngocngoc9972 6 күн бұрын
Giải rườm rà vậy. X = 35/Y . Thay vào rồi giải phương trình bậc 2 thôi