There is a problem here, when you write (x+y)(x-y)=5x13 you can not deduce that x+y=13 because you do not know if x+y is a natural number. If m is an odd number, then x=3^(m/2) and y=2^(m/2) are not natural numbers. So you have prove that if m is an even number, then m=4. it is very easy to check that there is only 1 solution.

@onlineMathsTV

10 ай бұрын

Noted.

@nicadi2005

4 ай бұрын

@rubikaz "So you have prove that if m is an even number" - Or, you can assume that m is even (thus making the quantities x+y and x-y (positive) integers etc.) and see whether it pays off - which it does, actually. "it is very easy to check that there is only 1 solution." - Indeed. The uniqueness of the solution is a direct consequence of the properties of the exponential function at work here...

@babetesfaye1001

2 ай бұрын

❤❤❤ May I use any number Which have same numerator band denominator ? like 3/3 , 4/4 , 5/5....

@with.love.from.siberia

2 ай бұрын

​@@babetesfaye1001да, потому что оно равно 1

@Quasar900

2 ай бұрын

@@babetesfaye1001 the function f(x) = 3^x - 2^x where x>0 is strictly growing, therefore with x=6 , f(6) > 65 so x must be less than 6, and so on trying integers until finding x=4 or m=4

The comment section is polluted by critics but i learned some tricks from this video.. EXCELLENT

@onlineMathsTV

4 ай бұрын

Wow!!! Thanks and we are glad you gained some values from this video tutorial sir.

@thefireyphoenix

8 күн бұрын

it isnt critism.. in every maths channel you will see people doing that.. they do it for alternate solutions and writing down their own way

@tiehoteele874

2 күн бұрын

@@thefireyphoenix this video wasn't made for them

I got ‘m=4’ by mental arithmetic. Because 3^m > 65 and 2^m The value '65' determines that the value range of m must be less than 5 and greater than 0. When m is a positive integer, test the m=5, 4, 3, 2, 1 and finally get m=4.

@CBSE24

7 ай бұрын

I too

@tuyu6404

6 ай бұрын

Since we don't participate in the Olympics, this short answer is good. But I'm not sure if I would use this method for an Olympics. I think logarithms are simpler than the answer And have a good axiom to condense those complicated answers.❤

@lophocthienuc7345

6 ай бұрын

I dont think so. 2^m 65 why m

@gheorgheneacsu3356

2 ай бұрын

Me too! 😊😊

@chrissyday67

29 күн бұрын

exactly! if students are good enough to do olympiad problems then they'd know powers of 3, 3, 9, 27, 81, 243 at least and also 2 to even higher powrers, 2, 4 , 8, 16, 32 etc so very easy to work out in less than 20 seconds . However I'm used to doing mental arithmatic as I never had a calculater when I was in school

I immediately saw that 65 can be decomposed as 81-16, and conveniently 81 is 3^4 and 16 is 2^4, so matching coefficients suggests m is 4.

@BrightonMutero

5 ай бұрын

You are a genius

@nicadi2005

5 ай бұрын

@mitahaubica6498 "I immediately saw that 65 can be decomposed as 81-16" - That's not decomposition... You can find an infinity of pairs of numbers that have their difference equal to 65. The fact that you selected one such pair that also happens to be powers of the respective bases in the original problem merely indicates you have approached solving this by trial and error... *The question would be whether you can do better than finding the solution that way...*

@counterpoint9260

4 ай бұрын

that is not the right method..may work here but not al the time

@davidmajor1508

2 ай бұрын

You just got lucky. It was pure luck that you used the right numbers to subtract, and that m is an integer in this case.

@danielvazquez6301

29 күн бұрын

Always try some values of m to see the behaviour. m=0 or 1 or 3 or 4... I've found the solution! Obviously it is not an Olympiad problem.

How can one just assume that x+y = 13, and x-y = 5, respectfully, as there's 65 and 1 as well. Furthermore, this wouldn't really work if 65 had many more factors, making it more complicated, opening up to a lot more possibilities.

@onlineMathsTV

11 ай бұрын

Nice question @Santhosh John. There are principles and rules that govern the operations in mathematics as it is for all other sphere of life. Once you are a maths student or tutor you must get urself familiarized with these rules. They become part of you and you know what to do once you have a math challenge/problem before you. Once a mathematician sees a math problem, his head automatically runs through different means of approaching the problem for a better solution.

@thegreathussar9442

11 ай бұрын

When taking difference of squares if smaller part is 1(assuming it is the smaller part and it is) , m=2 and positive part becomes 5, not 65 therefore the result will be 5, not 65. If it has more factors, you just have to make arithmetical inferences and simplify it. That's it.

@naharmath

11 ай бұрын

3^(m/2) is not necesserly an integer!

@thegreathussar9442

11 ай бұрын

@@naharmath but there is no other solution since both parts are exponential and even if m is a rational number the result wouldn't be integer( they are different primes). So this equation requires to be analyzed numerically first

@mariosantangelo9929

11 ай бұрын

Il professore ti ha risposto in maniera adeguata. Però io consiglio al professore di spiegare certe regole anche se ciò richiede qualche minuto in più. I fruitori di you tube non sono tutti matematici, ma persone desiderose di capire ed imparare.

I worked it out a bit different. My solution was simply determine what of 3 exponent would get me a number greater than 65 that would be an odd number (3^4). I then subtracted that 65 from that number (81) and I got 16 which is 2^4. In other words you can rewrite the equation in this instance as 3^m - 2^m=65 3^m - 65 = 2^m The first exponent of 3 which results in a number greater than 65 is 4 so 3^4 = 81 81-65 = 2^m 16 = 2^m 16 can be written as 2^4 m=4

@Xhopp3r

7 ай бұрын

That's exactly what I did.

@Smith_14

7 ай бұрын

m can be a negative number?

@leishajuneja2994

7 ай бұрын

You cant take any value of 'm'.consider the question is same but with a very large value(instead of 65),probably in crores,it wout take an eternity to reach that number

@leoosu

7 ай бұрын

I did the same😂

@DownhillAllTheWay

7 ай бұрын

You're assuming that m is an integer?

65=13•5=(9+4)×(9-4)=9^2-4^2=3^4-2^4 4 is the m; My high school math teacher used to tell me,the easiest way to understand an equation is to make them look the same,that is to say,we should make the brief side more complicated other than simplifing the complicated side for the most of the time.

I was impressed by the analytic demonstration used to figure out that m = 4. Sometimes procedures can be more interesting to follow along than just knowing the result.

@italixgaming915

6 ай бұрын

Well, you'll be impressed to see that the proof is not valid. If you suppose that m is an even number then x+y and x-y are integers but if it's an odd number then they are IRRATIONALS and you can't use 65=5*13. Plus the fact that even in the case where x+y and x-y are integers, 5*13 is not the only way to get 65, you must also look at 1*65...

@jarikosonen4079

3 ай бұрын

​​​​​@@italixgaming915This looks like diophantine method used here, which would work only with the integers. In case the solution works like in above case it could prove that no other integer solutions exist... But it seems 65x1 was not checked. This could work for 3^m-2^n cases also. Variables 'm' and also 'n' are often used for integers, but not necessarily always.

@claudiohagra

2 ай бұрын

You are the top!😊 You are a great teacher!!!!

@elmehdiazzouz7888

Ай бұрын

the proof is not valid, we need to proceed much more cautiously with analytic ways, i d say : some more conditions/discussions needed to be added to the video .... i gree with italixgaming the arithmetic way stays safer ..

Similarly you can as well re-write 65 as 81-16....From there you make the bases of the two numbers to be similar with what you have on the left hand side..From there you take one of the corresponding bases and equate them together,when bases are the same powers will also be the same.

@usmanmusa8028

8 ай бұрын

This is what I actually expected from him

@TomJones-tx7pb

8 ай бұрын

If you do not notice that 16 = 2**4, you are not a computer geek.

@lgmoses3876

8 ай бұрын

I did it,in my braine.

@sfqamd

8 ай бұрын

kzread.info/dash/bejne/pX2gtKqgZ5aWc6Q.htmlsi=ujk2KD_xjoMGqiP5

@shivaprasadmallikarjunaiah3751

8 ай бұрын

you are not mathematically solving the problem, but doing so by trial and error. These were smaller numbers so it is easy for anyone to come to that conclusion ( becausethe solution is "visible" in the numbers in front of you). In other words, how would you solve the same problem with entirely different and larger numbers involved? ...say 2059 for instance.

Hi there! Here is an alternative solution. The difference of the powers on the rhs increases with m, and it is already greater than 65 for m=5. So m must be less than 5. Assuming m is an integer, and noticing that m=3 doesn't work, the only possible solution is 4.

@onlineMathsTV

11 ай бұрын

Ya boss. You are very very correct and I like you approach sir. This is an indication that you are a master in this area. Respect sir...👍👍👍

@ronitmahawar1193

11 ай бұрын

but this only works if m is integer which we dont know it would be

@romanyukvictor

11 ай бұрын

@@ronitmahawar1193 Exactly! To solve this you have to prove that the equation has single solution. Just shift 2^m to the right side and divide the entire equation on 2^m. As a result you will get the increasing function of m on the left (3/2)^m and decreasing function on the right (1+65/2^m) which can intersect only once. So there is only one root. The only way to find the root is to enumerate the integers and find out m=4.

@ifomichev

11 ай бұрын

@@ronitmahawar1193 the solution proposed by the author of the video also works only for integers, because it relies on factorization of 65

@jccamargo99

11 ай бұрын

For this reason many people don't like math.

You can also factor 65 into 65 and 1. This gives values of a and b as 33 and 32 hence b=2^(m/2)=32, m= 10 but m will have different value for 3^(m/2)= 33. Using logs(or ln) m=(log 33/log 3)=6.365

@eliasgitau7353

6 ай бұрын

This makes sense and it is mathematically correct

@nicadi2005

5 ай бұрын

@samuelmayna "You can also factor 65 into 65 and 1." - Yes, you can, but these won't be proper factors for 65, in the sense that ANY NUMBER could be "factored" as itself and one... Also, as you've seen yourself, this breaks the consistency of the original equation by forcing the unknown to take different values simultaneously - which is obviously not possible.

@user-uy7uu1rm5u

4 ай бұрын

m is an integer

@samuelmayna

4 ай бұрын

​@@nicadi2005 but it is mathematically logical.Mathematics is about thinking all cases.

@samuelmayna

4 ай бұрын

​@@user-uy7uu1rm5u but my approach is sound which shows that 65 and 1 won't work but it can give different answers for some equations.

The way I see it, 3 cubed is 27, less than 65, and 3 raised to 4 is 81. Therefore 65 is between 27 and 81. Upper bound, lower bound or range even. Now 81 would mean m is 4. So 2 should be raised to 4 also which gives us 16. 81 minus ➖ 16 is 65. So m is 4. The other ways are using log or binomial series which is overkill for smaller numbers. - Amy

You tried here tutor Jakes. I have learnt something here. Just keeping on running this channel. More grace, love from Port Harcourt ❤.

@onlineMathsTV

11 ай бұрын

It is our pleasure to serve you sir. Thanks for watching

Love your video! I got lost half way and I did not know that you could just square the exponents and make them equal

@onlineMathsTV

11 ай бұрын

We are glad you love what is happening here. We promise to give more educative contents in the area of mathematics with the help of God. Love you.....💖💖💕💕

@karlm9584

5 ай бұрын

I like this too. I have done it with roots before. Roots are just indices but I wouldn't have thought of doing it.

Maybe it's a bad argument but I would say for a^m - b^m = X, and a, b, m belong to N, a^m> X > a^m-1. Here 3^m>65>3^m-1 81>65>27 => m=4.

Thank you for reminding me that you can't believe everything on the internet. Props to you i almost believed it untill i tested it with calculator. BRAVO YOU SHOULD WIN AN OSCAR

A matemática é uma língua universal como a música. Parabéns, ótima técnica ❤

Parabéns. Não sei falar nada em inglês e mesmo assim consegui aprender com sua aula. Até eu estou surpreso de ter assistido sua aula até o final sem saber se iria entender o seu modo de esnsinar. A matemática pode ser universal, mas o jeito de ensinar é fundamental.

@onlineMathsTV

9 ай бұрын

@Airtonreis, we want to sincerely say you are such a wonderful person and thanks a million for watching our contents despite the language barrier. We all @OnlineMathstv deeply cherish and love you from the depth of our hearts sir. ❤️❤️❤️💖💖💖💕💕💕🙋🙋🙋

@airtonreis2675

9 ай бұрын

@@onlineMathsTV 🤝

@MATHSHEADMASTER

7 ай бұрын

You are welcome to subscribe to our channel for more interesting math problems.

@pedrooo13

3 ай бұрын

Brasil tá em todo lugar não tem jeitoooooo

I like how you used indices to get past that first section. I've never really been good at spotting where to use substitution, the section when you brought in x and y

@onlineMathsTV

5 ай бұрын

Hahahaha....thanks a bunch my good friend and thanks for watching our contents consistently. We all here love deeply....💖💖💕💕😍😍

Before watching: Alright, so, we are looking at exponential functions. 3^m - 2^m = 65. First, we can rule out m=1 and anything below; the difference between those would be smaller than between 3 and 2, and thus much lower than 65. Next, notice that we are dealing with an integer on the right. This heavily implies (but *does not necessarily guarantee* ) that 3^m and 2^m are both integers as well. If both are integers, then m must also be an integer. So we should start with integers. (If we were dealing with a mixed number instead, this would be much more complex; as it stands, we can just plug in integers and see which one works). Let's start with the first x that gives us 3^m >65, namely 4. 3^4 = 9^2 = 81, and 2^4 = 4^2 = 16 So for m= 4, we have 81-16 = 65. Fortunately for us, this checks out, and thus we have our answer of *m = 4*

@onlineMathsTV

11 ай бұрын

Thanks for this thorough explanation. You the best. 👍👍

@RoderickEtheria

11 ай бұрын

Negative m gets fractions.

@Psykolord1989

11 ай бұрын

@@RoderickEtheria Yes, you are correct, and I see I made a typo in the "so for m=4 we have..." section by putting a negative in front of the 4. Fixed now. I don't imagine it was a huge problem since in the section right above it,and at the very end of that section, I used 4, but it still may have confused some people.

Another option: the solution is pretty clear, it's a small number one can calculate with the mind. Then it reduces to show that their solution is unique. One can use calculus to show that the analog continuos function is monotone for x>4 and be done with it, or use induction to show that it grows on the integers for n>4.

@onlineMathsTV

11 ай бұрын

Your approach is superb sir. I find it very fascinating and I will try it out in subsequent math challenges. Thanks for sharing this nice and wonderful procedure with OnlinemathsTV. You the boss and much respect boss...👍👍👍

@GirGir183

7 ай бұрын

I think he used a simplified equation for the demonstration. When it's not so simple and the numbers are much larger, then this method can be used as well.

Couldn't have thought of this approach. Thanks for this.

Solução genial. Parabéns

I did it in a different way. Write 3 as 2+1 and perform binomial expansion, so the 2^m cancels and subtract one from both sides. We have 64 on one side and some series on other side. Notice that the series 2 + 2^2 +.....+ 2^m will definitely be smaller than series on left which is equal to 64. So make this G.P. sum less than 64, you will get that m should be less than 5, and once you have known this, you have proved that you just need to find a solutions less than 5 and those will be the only solutions. Only m=4 works out.

@albajasadur2694

7 ай бұрын

a good method to find the range of m and it makes sensible checking easier with limited number of m

@theboss73104

14 күн бұрын

Yeah

If you guess the solution, m=4 and present the equality as 3^m - 81 = 2^m - 16 , then it would be easy to prove that both functions (on the right and on the left) increase and therefore their graphs have only one intersection.

@ca1498

8 ай бұрын

You need more than that. Two increasing functions can intertwine and cross each other all the time. But if they have one intersection, and after that one of them grows faster than the other all the time, then it follows that they won't intersect again. It's like two cars racing. They both increase their distance from the start all the time, but they could swap places many times during the race--unless one of them is always faster than the other after the point in which they were even with each other.

@ca1498

8 ай бұрын

@@reginaldocalvo4361Each of the two sides can be a function e.g. y = 3^x - 81. The solution of the equation 'side 1' = 'side 2' is a number x (or m) where the two functions give the same y for the same x. You first find, by guessing, one x (or m) for which the two functions have the same y, which would mean that the particular m is one solution to the equation of function 1 = function 2 for some x. You then show that each of the functions only grows, and that the difference between the two y-s for each x (which is a function of x as well) also only grows. Therefore there can be only one value of x for which the difference is 0, so only one m (the one we already guessed) is the solution to the equation where one of the functions has the same y as the other for a given x. If I were participating in this Olympiad and solving this problem, I would have guessed 4, and then I would have argued that the first side 3^m... grows much faster than the other side 2^m for each subsequent m. I wouldn't be using derivatives, as I did not know any calculus in high school. But if they ask about integer solutions, I would talk about growth of y with respect to changes in m by +1. And if they did not limit it to whole numbers, I would probably still try to talk about slope of the graphs of the functions and hope to make an acceptable argument, as I don't see how you can analyze these functions without using calculus.

@user-xl3mg3om7s

8 ай бұрын

@@reginaldocalvo4361 3^m-81= f(m) this can be considered as a function depending on m 2^m-16= g(m) also could be considered as function of m.

@user-xl3mg3om7s

8 ай бұрын

@@ca1498 you are right, but When you see this function it is easy to notice that the growing path or direction is already know but partially

@Vitzyk

8 ай бұрын

Consequence is false.Contra-example x and x^3. Both increase but have 2 intersections

Difference of squares into u-substitution. Excellent methodolgy.

Parabéns. Resolução muito boa.

It doesn't work if m is odd. In this case (x+y) and (x-y) aren't integer and can't be assumed as 5x13.

@danielrivera2278

11 ай бұрын

If you do the analysis, and supposing m is an integer, you can conclude m is even. That's because 3^m-2^m must be congruent to 0 mod (5). If m is even you get 1 mod 5 or 4 mod 5. But m being even, you have 0 mod 5 always. Anyways, it's not proven in the video, maybe it would be amazing to have hows and whys in the video

@user-oi3iv7oo4z

11 ай бұрын

@@danielrivera2278 of cause. I mean that "trick" in the video is not universal.

@Change_Verification

11 ай бұрын

@@danielrivera2278 and who even said that m must be an integer ?

@danielrivera2278

11 ай бұрын

@@Change_Verification exactly. I also think like that, that's because assuming integer was the first thing I said

@Eismann1

8 ай бұрын

Yes, he didn't do the preliminary work. But this is an important step to make tricks like this one work.

profe en olimpiadas de las matematicas se aprende mucho. y con usted bastante. felicitaciones

@onlineMathsTV

3 ай бұрын

Thanks a million sir, we appreciate this comment my good friend.

Solução genial. Parabéns. Muy interesante, hoy aprendí un buen método, gracias.

I love how the way you work out the question..❤❤

Сначала поделить обе части на 2^m. Тогда слева будет возрастающая функция, справа убывающая. Тогда уравнение имеет не более одного корня. Подобрать корень не сложно. 9 класс, ничего сложного. За проведенное решение минус: нигде не доказано отсутствие других решений, переход к системе ничего не обосновывает.

Very well solution but the rules are very lengthy or Expensive ☺️ So if we assume the value of "m" here from 1-4 We are easily getting the value of m Such as Let , m=1,2,3,4,... ♾️ and now, 3^1-2^1=1≠65 again 3^2-2^2=5≠65 And now if we let, m=4 then 3^4-2^4=65=65 So we can easily get m=4😊

@onlineMathsTV

10 ай бұрын

Bravo👍👍👍

@WhiNikkatherealone

8 ай бұрын

i mean hit and trial is always the last resort

m≤2 makes it too small and don't work. So, m must be even, as 3^m ≡ 3 (mod 4) for m odd, but 65 ≡ 1 (mod 4). On the other hand, 3^m - 2^m = 3^{m-1} + ... ≥ 3^{m-1} and is strictly increasing. Since, 3⁴ = 81 is already too big, we must have m

3ᵐ-2ᵐ=65 y=3ˣ-2ˣ y=65 если построить оба эти графика, то будет видно, что уравнение имеет единственное решение, поэтому можно попробовать подобрать корень подбором: х=1: 3¹-2¹=1; 1≠65 х=2; 3²-2²=5; 5≠65 х=3; 3³-2³=19; 19≠65 х=4; 3⁴-2⁴=65; 65=65 х=4 корень m=4

In a slightly fancier approach we can make a recursion 2(3^m - 2^m) + 3^m = (2+1)3^m - 2^(m+1) = 3^(m+1) - 2^(m+1) Put a(m) = 3^m - 2^m a1 = 3^1 - 2^1 = 1 a2 = 2 a1 + 3^1 = 5 a3 = 2 a2 + 3^2 = 19 a4 = 2 a3 + 3^3 = 65, done

@onlineMathsTV

10 ай бұрын

Wow!!! This approach is impressive but a bit obscure sir.

@vejayashanker

7 ай бұрын

can u explain what u mean by obscure pls😂

@soltanchalkarow905

4 ай бұрын

can you help me sir? if ab+bc+ca=1 prove: sqrt(a + (1/a)) + sqrt(b + (1/b)) + sqrt(c + (1/c)) >= 2( sqrt(a) +sqrt(b) + sqrt(c) )

It can decide differently. The function (1) f(x)=3^x-2^x -is increasing . { for x>0 (2) 3^x>2^x ; x10 . (3) f(x2)-f(x1)= ……..=3^x1*[3^(x2-x1)-1]-2^x1*[2^(x2-x1)-1 ]> 2^x1*[2^(x2-x1) -1 ]-[ “--“ ]=0 ; (3) f(x2)-f(x1)>0 !!!!! So , it takes all its meanings once a time. f(4)=3^4-2^4=65 . So , x=4 - is the only root of equation ! Respectfully , Lidiy

@onlineMathsTV

11 ай бұрын

The elders in our mist are highly learned. Love your detailed explanation sir. Thanks for finding our time to watch our content and commenting even at this age of yours sir. Much respect and we All @onlinemathstv love you dearly...💖💖💕💕

@Lernen-mit-Rudi

10 ай бұрын

There is no X! 😂but however, good job!

@andreasandre4756

10 ай бұрын

Pay attention that m=constant not variable, so M must be grater then 1 otherwise solution will not be true because ln1=0 or 3-2=1 which is not equal 65. So M>1 and could be anything. So if it is not an integer number? So if it is not equal 65 but 63.5 - ?

@ADSemenov_ru

8 ай бұрын

You took the words right out of my mouth. :)

@Marat7973

8 ай бұрын

Здравствуйте,Лидий! Не ожидал Вас здесь увидеть)

I'm impressed with this explicit methodology 😊(12:16am)

Perfect.Thank you sir.

Instead of solving by equations in eloborate lengthy long time , I simply solve it in the beginning itself my applying m=1 m=2 m=3 Wow.. when I apply m = 4, I got the answer *65* # Instead of 65, if the answer @ RHS is 6584 (or any larger number), we should go for equations as described in the above video .. {But when it is small number less than 100, we can approximately conclude range between by directly applying values for *'m'*

the function f(x) = 3^x - 2^x where x>0 is strictly growing, there for if x=6 f(x) > 65 so x must be

@onbored9627

3 ай бұрын

All I did was think whats the first interger power of 3 that goes past 65, 3 is 27 so the answer is 4. Then just test. and it gave right answer. simple.

@Quasar900

3 ай бұрын

@@onbored9627 Where are you from ? Cause how did you know I was still alive after 4 months ? 🙂Here is a secret about me : I 've never studied Mathematics in English !

@onbored9627

3 ай бұрын

@@Quasar900 Ah, I apologize I only speak English. I'm from the USA. I wasn't trying to take away from your explanation I should've been more clear on that, I just thought yours was so good I didn't even need to say. Figuring out it's strictly growing is clever as hell. I didn't even think of that. I meant simple, as in, my answer was simply a guess really and it worked.

@Quasar900

3 ай бұрын

@@onbored9627 Oh please Sir , no need to Apologise , The fact that I've never studied math in English doesn't mean I don't know English 🙂 It's just I'm not that familar with English terms in math ! But Thank God I do speak and read : French, English, Arabic, Spanish + some Japanese ! I did study math in French (after high school) and Arabic (until high school) , but that waaaas 21 years ago ! What class are you in ? I hope you're safe from those ongoing blizzard storms ! Greetings From Morocco and Free Palestine 🙂

@Quasar900

3 ай бұрын

@@onbored9627 I think you do know these techniques involving the continuity of a function, the intermediate values etc.. to solve equations ! For example : solving in IR set : Arctan(x+1)+Arctan(x-1)=Pi/4

exercicio maravilhoso🥰🥰❤❤❤❤❤❤

@onlineMathsTV

11 ай бұрын

Thanks a millions sir, we love you ❤️❤️💖💖💕💕😍😍

I think it's very easy. We let m = 4, it's correct We will prove "with m > 4 , 3^m -2^m > 65" So, only m = 4 will be equation. m>4 m = 4+a (a>0) 3^m - 2^m > 3^4 - 2^4 Because (3^a-1).3^4 always > (2^a-1).2^4 Thanks from Việt Nam 🎉

You worked hard, I found the result by giving the value of m to 4 in 10 seconds, but it is important how the solution is, but I still think it is too long. Greetings from Turkey.

Solved 3^m-2^m=65 by just thinking about the first whole number power of 3 above 65.

@dilphek

10 ай бұрын

It is not about finding it. Olympiad is a school competition teaching kids to solve these problems mathematically

@bleh-zj1hy

6 ай бұрын

​​@@dilphekyou guys got a different Olympiad or something? Here Olympiads (for the kids, totally different type from the subjective ones) are mcqs and the subjective ones are like 3 questions in 3 hrs and if you're able to do even 1 you're qualified (you can imagine the toughness so it's really not for the kids)

Great! Its a fun problem. I did something kind of similar with difference of 2 squares.

@onlineMathsTV

11 ай бұрын

Nice, you the best. Kudos 👍👍

I was able to solve the problem in less than 1 minute... I just thought about the Multiple of 3 more than 65 that was 81 and then i subtracted 65 from it and the resulting answer was 16,now we know that 3^4 is 81 and 2^4 is 16,so easily the value of m came out to be 4

Nice one sir

I have been learning Genetic Algorithms in Python; they are good for problems like this. The value I get for 'm' is 3.97, 3.99, 4.00, etc.; different each time as there is a random element for the convergence. A little intuition is also required; If you plug 4.0 into the equation you get correct 65. Program run time ~5 seconds.

@onlineMathsTV

11 ай бұрын

Wow!!! Nice sir.

@pierrecurie

11 ай бұрын

If you're going the numerical route, bisection search is much faster.

@mistertwister1015

10 ай бұрын

Blunt enumeration of roots is not always the best solution)

@victorfildshtein

10 ай бұрын

Hello. I program in PureBasic. I made this task by binary search method, 30 iterations, result 3.9999999991, precision 0.0000000047. Time is almost instantaneous.

@wilsonuche9389

8 ай бұрын

You need to review the Python solutions cos only 4 is an exsct solution. 3.97 is far from it, 3.99 is just an approximate

It took a fraction of a minute to recognize m = 4, but as a mathematician, I did find this interesting. (Dr. Mike Ecker)

@JaroGoraJ

7 ай бұрын

I think because it was easy question

Excellent explanation. It seemed very complicated, but it turned out to be easier than expected. A hug

65 = 81-16 = 3"4 - 2"4 m = 4

Clever solution! One thing I did not get is how you knew that m was a positive integer - or was this just an assumption that happened to work? For example, if the problem had been 4^m - 3^m = 65, m would be approximately 3.36.

@charleskaruru481

6 ай бұрын

m can never be a negative otherwise we wont have 65but fraction

There is a faster workaround, with the assumption that the number m is integer. The term on the right is less than the term on the left. For simplicity, we can assume that the term on the right is zero. This results in: 3^m = 65. m is greater than 3, because 3^3 = 27. m can be 4 because 3^4 = 81. Let's test m = 4: 3^4 - 2^4 = 65 81 - 16 = 65 65 = 65

@onlineMathsTV

11 ай бұрын

Wow!!! This is fantastic. I love this approach sir. We have gained some values from this procedure sir. Thanks for dropping this sir. Respect.....👍👍👍 Much love....💕💕💖💖❤️❤️

@divonsirlopes5409

9 ай бұрын

There is a faster workaround, with the assumption that the number m is integer. The term on the right is less than the term on the left. For simplicity, we can assume that the term on the right is zero. This results in: 3^m = 65 m = log(65)/log(3) = 3,8 m is greater than 3,8 Let's test m = 4 3^4 - 2^4 = 65 81 - 16 = 65 65 = 65

@MONSTER2013

7 ай бұрын

Base on which level you’re at. This question and video are made for yr 10-11? So he gave the solution at that grade. Above yr 12 can use other tools as log/ ln skilfully to solve it.

@divonsirlopes5409

7 ай бұрын

Thanks for the info.

@foudilbenouci482

6 ай бұрын

^you found one solution doesn t mean you found all solutions

Thanks teacher. Today I've learnt something very new in maths. Am really surprised....

Nice one ...... good teacher ...... stay teaching stay doing more math tricks ...... i liked it

Keep up the great work man happy to see somoen in our beloved African continent devoting a portion of their time into this much love from Morocco 🇲🇦 MA

@onlineMathsTV

8 ай бұрын

Much appreciated sir. Thanks for watching our contents and the encouragement sir. Much love from everyone @OnlinemathsTV to you sir 💖💖❤️❤️💕💕💕

@Quasar900

8 ай бұрын

@@onlineMathsTV the function f(x) = 3^x - 2^x where x>0 is strictly growing, there for if x=6 f(x) > 65 so x must be

@Quasar900

8 ай бұрын

the function f(x) = 3^x - 2^x where x>0 is strictly growing, there for if x=6 f(x) > 65 so x must be

If a.b=65, a and b can have infinite values. So, the tutor has just one answer where multiple answers are possible.

@onlineMathsTV

11 ай бұрын

Yes it has multiple answer but for the sake of this tutorial we restricted ourselves to this solution sir. Thanks for this observation. Much love....💕💕👍👍

@mustaphaolunrebi8100

7 ай бұрын

I think you add that, where m is an integer. It makes it complete. The only integer factors 65 has are 5 and 13. Nice solution 🎉

@ivandonchev474

7 ай бұрын

Sorry but this video is the most useless shit I have seen. You solved it by guessing and overcomplicated massively

@speedsterh

6 ай бұрын

@@mustaphaolunrebi8100 No, 65 has 4 factors: 1, 5, 13, 65. The equation with 2 other factors should be explored for completeness

Thank you sir for solve this type of problem

Gentleman I appreciate your work.

Hello Online Maths TV, I really like your approach with this and how your deliver the proof. It is elegant and uses several skills. Your pace of delivery is really good too. As a suggestion for completeness in your proof, can you please include a determination table of which factors of 65 and their order are valid for consideration for equating to (x+y).(x-y). A table similar to; (x+y) . (x-y) | x | y | m from x=3^(m/2) | m from y=2^(m/2) | 1 . 65 65 . 1 13 . 5 5 . 13 If you calculate x and y and then m from x=3^(m/2) and from y=2^(m/2) for each of these arrangements, only (x+y) . (x-y) = 13.5 provides valid and consistent values for m. So 3 out the 4 arrangements can be mathematically eliminated. This would verify that the only valid arrangement and values of the factors is 13.5. I hope this helps. Cheers, Paul.

@onlineMathsTV

11 ай бұрын

What a wonderful guide @Paul Middleton. Very very helpful sir. Respect boss. ❤️❤️❤️👍👍👍

@onlineMathsTV

11 ай бұрын

Am compared to drop another comment in respond to your first comment sir. Am speechless and overwhelmed by your systematic construction of this comment. It is so encouraging. This is because you appreciated my little effort in the mist of professors and further suggested a wonderful/a great approach to me on solving this same problem or similar problem in my subsequent videos. Sir, with all humility we are glad to meet with you and have you here. Much love sir....💖💖💖💕💕💕❤❤❤

@paulmiddletonphotography4368

11 ай бұрын

@@onlineMathsTV You are very welcome indeed.

@paulmiddletonphotography4368

11 ай бұрын

@@onlineMathsTV Thank you for your very kind words of appreciation. It certainly is my pleasure to help you and I am really pleased that my suggestion will assist you and into the future with different proofs. May you increase your subscribers, students and people interested in learning your very clear, easy to follow and forthright teaching of maths. All the best to you for the future.

Sir, we can also do this by hit and trial method, assuming different values for m=1,2,3,4.... But your way was also very nice 👍👍. Thank you for this video sir🙏

@onlineMathsTV

11 ай бұрын

Yes, but that will only work if the working process is not in the examination but this is needed when the examiner wants you to show your procedure step by step...👍👍👍

@gregfarnham5651

11 ай бұрын

Yes, trial and error could work if we assume m is a positive integer. I don't believe that was a given, however.

@ralfimuller8948

11 ай бұрын

@@gregfarnham5651 The solution in the video also made use of the assumption that m is an integer. Otherwise the factorization of 65 would not be unique. Actually, the hit and trial method should be entirely ok.

@gregfarnham5651

11 ай бұрын

@@ralfimuller8948 Agree. Thank you.

@danielrivera2278

11 ай бұрын

Also, by trial and error you can't prove that's the obly answer

in maths olympiad time is very critical...what is important is the answer...so the best way is to solve maths olympiad is to have very basic maths then the rest is analysing to get the pattern...so what i did is just look at indecies of 3 that have the last number such that if we subract an indecies of 2 and get 65

You are just great! A very good approach.

Остроумно и красиво! Как и вся математика.

@onlineMathsTV

11 ай бұрын

Thanks for this wonderful comment sir. Love you....💕💕

@user-nv9rw7nh5w

10 ай бұрын

Очень, очень. Надо же как можно. Удивительно!

Слева возрастающая функция при m>0 ( можно найти производную и убедиться ), справа постоянная функция, значит у них может существовать только одна точка пересечения, методом оценки m=4

@onlineMathsTV

10 ай бұрын

Bravo 👍👍👍 You the best sir.

@onlineMathsTV

10 ай бұрын

Bravo 👍👍👍 You the best sir.

@PlumbuM871

5 ай бұрын

Чисто случайно подставил вместо m 4, и всё сошлось! Везёт мне

Excelente. Thank you.

Extraordinario , thanks so much.

Очень интересное решение! Спасибо!

@onlineMathsTV

10 ай бұрын

You are welcome always even as we look forward to seeking better ways of serving you better in our services to you on this channel. We at Onlinemathstv love you without reservation sir....💕💕💕

First you must show that m is even positive integer

@onlineMathsTV

11 ай бұрын

Noted sir.

Очень грамотное изложение, и очень удобно следить на доске. Однозначно плюс!🎉

Wonderfully explained, very informative. But it’s sometimes more convenient to use the easy method

This is so overly complicated.. Just check low values of m and find that m = 4 works.. Then use a simple analysis tool to show uniqueness of the solution e.g. by showing that function f(m) = 3^m - 2^m is strictly increasing for m>=1. Additionally your solution contains errors and missing steps: 1. If you do the trick with 3^(m/2)^2 - 2^(m/2)^2 = 65 then you presuppose that m is even, because if m is odd then 3^(m/2) is not natural anymore, so you cannot use the natural divisors of 65 in the following steps anymore. 2. Even if that approach worked (because you somehow proved that m must be even): After you rewrite the equation as x^2 - y^2=65, then you have to consider *two* pairs of solutions 65 = 65 * 1 and 65 = 5 * 13. Long story short, lots of mistakes in your video unfortunately.

@onlineMathsTV

11 ай бұрын

Thanks for this keen observation and I really appreciate this comment sir. Noted. I will do more detailed work on subsequent videos. You the best and much love for this detailed comment....💕💕💕

@ivandonchev474

7 ай бұрын

Yes this video is shit full of errors. The fact that it has so many errors and is hugely overcomplicated at the same time makes it complete and utter shit

This is how I found the solution : 🙂 We know that 3^m have to be greater than 65 so that when we subtract 2^m from it we can get near 65. Therefore 3³=27

Hedduu Galatoomi! Furmaata bareedaadha bu'uuraa isaa nuf ibsiitee ❤🎉

Great stuff! I went the x²-y³=81-16 route, was that cutting corners?

Nice job! I got the answer by trial and error, but this way of getting a deterministic solution is really cool.

@italixgaming915

6 ай бұрын

Actually this proof is wrong. If m is an even number then x+y and x-y are integers but if m is an even number then they are IRRATIONALS and the unicity of the prime factors works only with integers. Euler did the same kind of mistake when he tried to demonstrate the Fermat theorem (that time he used complex numbers).

He still had to compare and guess which is 13 and which is 5 in the factors. Same guess applied directly to the original problem shows the subtraction of 2^m from 3^m so m

A very simple task. 1. I looked for a power of 3 greater than 65. I got 81. The power is 4. 2. I subtracted 65 from 81. I got 16 3. I calculated the power of 2 to be 16. It turned out to be 4 4. It turned out that m = 4

@Quasar900

3 ай бұрын

You can't be sure that's The Only solution unless you realise that : the function f(x) = 3^x - 2^x where x>0 is strictly growing, therefore with x=6 , f(6) > 65 so x must be less than 6, and so on trying integers until finding x=4 or m=4

I did this in my head in less than a minute. 3,9,27,81(bigger than 65) 2,4,8,16( subtract 16 from 81)…. Works.

@daintydawn2508

Ай бұрын

Yeah, it's a bit nd trial method. But if the value of m had been bigger then, it wouldn't work

1. (3^m-2^m) isincreasing function of m. 2.M is integer,because 65 is integer. 3.f(0)=0,f(5)=243-32=211. 4.we only need to test by 1,2,3,4 5.The answer is 4

You are the best. Thanks

Very instructive task

Very good approach

. I have made 3exp1 -2exp1=1 3exp2-2 exp2 =5 3exp3-2exp3=19 3exp4-2 exp4=65 81 -16 = 65 Différence between two perfect squares

Excellent tutorial. Thank you very much

Nice solution. It makes so easy, the way you explained.

3^m>65 use log of both sides to solve the inequality we get m>3.8 so start to check m=4 in the original equation the equation is satisfied and m = 4

thank u very much abaut it...

Nice. You used Algebraic manipulation. I used my laptop to answer the question with Graphical approach, the answer was 4 too. However, the answer was almost 4 when I used Newton-Raphson and linearization methods, why?

We can try 1,2,3,4,... and see if some patterns or a solution appear. We find that only some values of m can lead to 3^m-2^m = ....5, with 5 at the end. 4 is one of them, and fortunately 3^4-2^4=81-16=65. So 4 is a solution, and it is unique because for any M greater than m, 3^M-2^M > 3^m-2^m Here is why: 3^(m+1)-2^(m+1) - [ 3^m-2^m ] =3×3^m - 2×2^m - 3^m + 2^m =2×3^m-2^m = 3^m+(3^m-2^m) > 0. So above m=4, we will never have 3^m-2^m=65. Conclusion: The unique integer solution is 4.

Thank you so much. That's the easy way to solve it❤👍

thank you sir

Before watching this video i want to share my method - By using binomial expansion 3^m - 2^m =65 =>(1+2)^m - (1+1)^m =65 =>(1+2m + (2m(2m+1)/2!) +........) - (1+m+(m(m+1)/2!)+.......)= 65 (by binomial expansion) Bro i got stuck 😮‍💨

@sweetmess7877

Ай бұрын

You are a genius 🎉🎉

@sweetmess7877

Ай бұрын

You are a genius 🎉🎉

This is terrific! Thank you

3^m - 2^m = 65 > 3^m - 2^m = 81 - 16 > 3^m - 2^m = 3^4 - 2^4 comparing both sides, it can be concluded m = 4

Thank you for the video.

Rewrite the equation as 3^x = 65 + 2^x. We can safely divide by 3^x and then we have that 1 = 65 * (1/3)^x + (2/3)^x. Because the function on the R.H.S. is strictly decreasing, being the sum of 2 other strictly decreasing functions, it means that f(x) = 1 has one solution at max. We notice that x=4 checks, so that is our only solution.

As a math teacher, this is a plus to me. You are amazing 👏

@onlineMathsTV

2 ай бұрын

Thanks a million sir.

Could you explain more on how you arrived at using 13 and 5