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

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

@@thefireyphoenix this video wasn't made for them

You are a genius

@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...*

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

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.

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.

That's exactly what I did.

m can be a negative number?

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

I did the same😂

You're assuming that m is an integer?

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...

​​​​​@@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.

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

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 ..

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.

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.

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

@@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

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 too

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.❤

I dont think so. 2^m 65 why m

Me too! 😊😊

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

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

Noted.

@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...

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

​@@babetesfaye1001да, потому что оно равно 1

@@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

This makes sense and it is mathematically correct

@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.

m is an integer

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

​@@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.

This is what I actually expected from him

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

I did it,in my braine.

ukposts.info/have/v-deo/q3mdhn6ebmlhpHU.htmlsi=ujk2KD_xjoMGqiP5

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.

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

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

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

Brasil tá em todo lugar não tem jeitoooooo

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

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...👍👍👍

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

@@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.

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

For this reason many people don't like math.

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

Yeah

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...👍👍👍

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.

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

Nice, you the best. Kudos 👍👍

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).

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

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

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

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

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

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

​​@@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)

Wow!!! Nice sir.

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

Blunt enumeration of roots is not always the best solution)

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.

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

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@@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

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

Bravo👍👍👍

i mean hit and trial is always the last resort

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.

@@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 !

@@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.

@@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 🙂

@@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

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

can u explain what u mean by obscure pls😂

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) )

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.

@@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.

@@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.

@@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

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

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

Negative m gets fractions.

@@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.

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

Очень, очень. Надо же как можно. Удивительно!

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...💖💖💕💕

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

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 - ?

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

Здравствуйте,Лидий! Не ожидал Вас здесь увидеть)

I think because it was easy question

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

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....💖💖💖💕💕💕❤❤❤

@@onlineMathsTV You are very welcome indeed.

@@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.

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

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....💕💕👍👍

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 🎉

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

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

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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...👍👍👍

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

@@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.

@@ralfimuller8948 Agree. Thank you.

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

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

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....💕💕💖💖❤️❤️

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

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.

Thanks for the info.

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

Bravo 👍👍👍 You the best sir.

Bravo 👍👍👍 You the best sir.

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

Thanks a million sir.

Noted sir.

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....💕💕💕

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

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@@onlineMathsTV i am very happy to hear this from your side and appreciate it ❤

Utter garbage