A Nice math OLYMPIAD exponential question, how to solve for 'm'

A Nice math Olympiad exponential question. How to solve for 'm'. #matholympiadproblem #exponential #math #education #mathcompetition

[@mrjnutube]

Three words: beauty with brains... For me, I was constrained for time, therefore I took a direct substitution: let 3^m=x^2; 2^m=y^2 x^2-y^2=65 (x+y)(x-y)=13*5 x+y=13 x-y=5 ------------ 2x=18 x=9 y=4 ---------- Recall: 3^m=x^2 2^m=y^2 3^m=9^2=(3^2)^2 2^m=4^2=(2^2)^2 3^m=3^4 2^m=2^4 m=4 m=4

[@JJONLINEMATHSCLASSchannel]

Nice one

[@bamlaktesfa4198]

I like you and your short, logical and precise method !

[@JJONLINEMATHSCLASSchannel]

Thanks

[@user-kc4kj5cj3b]

Very good teaching

[@JJONLINEMATHSCLASSchannel]

Thanks and welcome

[@jan-willemreens9010]

... A very interesting solution strategy, executed in a pleasant and well-structured way ... thank you JJ for your always instructive presentations and take good care, Jan-W 🎉

[@JJONLINEMATHSCLASSchannel]

My pleasure!

[@user-ic8nx9fy9c]

Waouh Excellente thanks you .

[@sohaib_mer-]

great vid

[@JJONLINEMATHSCLASSchannel]

Thanks

[@minhhainguyen2671]

Beautiful teacher, Beautiful solution.

[@JJONLINEMATHSCLASSchannel]

Thank you! Cheers!

[@cliveanawana5289]

JJ nne, that is a good one. Keep it up. Lovely mode of deriving the solution.

[@JJONLINEMATHSCLASSchannel]

Many thanks!

[@yvesdelombaerde5909]

Thank you for the strategy. My tactics: 3^m must be >65 let’s try m=4. 81-16=65. Stop

[@JJONLINEMATHSCLASSchannel]

Thanks

[@jorgepinonesjauch8023]

Evidentemente el tanteo no siempre es posible, en este caso fue fácil, ahora bien se debería probar, que no existen otras soluciones

[@selvamd8861]

Strategic solutions

[@issacangra5692]

Great job. Keep spreading it.

[@JJONLINEMATHSCLASSchannel]

Thank you, I will

[@elacerda2008]

Perfect.😀

[@JJONLINEMATHSCLASSchannel]

Thanks! 😃

[@duanefisher2709]

Perfect! Thank you.

[@paulolemosmessias9460]

Congratulations. ❤

[@ShimelesTeacher]

Very clear and very helpful .

[@JJONLINEMATHSCLASSchannel]

Glad it was helpful!

[@kenitra.west.6389]

Thank you.

[@youngdkout]

Good job my darling teacher

[@JJONLINEMATHSCLASSchannel]

Thanks

[@lamusat]

Great job. Please keep it up!

[@JJONLINEMATHSCLASSchannel]

Thanks, will do!

[@sureshkar1547]

Super solution MOM, (India)

[@sellamimohamed6492]

Good and well done...

[@JJONLINEMATHSCLASSchannel]

Thank you!

[@olayinkaadeleye3134]

Elegant solution.

[@JJONLINEMATHSCLASSchannel]

Thanks so much

[@sylvanuskouekla8894]

The left side is an increasing function, so only one solution. You try m equals 1, 2, 3, 4

[@856mmedina]

Y que paso con los factores 65 y 1 porqué no los consider

[@alexanderkorotkov1760]

a=33, b=32, a-b=1, a+b=65

[@user-ic8nx9fy9c]

Merci excellente explication grand merci madame ❤❤❤❤👍👋👋👋

[@JJONLINEMATHSCLASSchannel]

U are welcome

[@kisymollel3928]

I used the different approach, I found (m,n) to be (log70/log3; log5/log2) and (log78/log3; log13/log2). Someone can prove it by substituting the values into the given eqn. Thank you very much for watching.

[@gondaimoyana2691]

This lady is just taking chances in Mathematics! 65 has 4 factors which are 1; 5; 13 and 65, therefore, the two possibilities of getting 65 is either 1x65 or 5x13. Why did she choose only 5x13 and not 1x65 too? Was she not supposed to apply both possibilities? Who knows that this would maybe yield two possible values of m? Also, she easoned that a+b is greater than a-b, how does she know that? What if b is negative, will that not make a-b greater than a+b? 😅

[@siyoumabreha3721]

Please remember that b=2^m/2 meaning b can't be -very number. Bcs the base is +ve.

[@mdzamiruddin9560]

Many many thanks mam.

[@JJONLINEMATHSCLASSchannel]

Most welcome 😊

[@success7707]

Add Equation 1 to equation 2 You get 2a=18 a=9. Substraction will complicate the solution compare with adding specially for beginers.although yr solution by subtracting is correct

[@JJONLINEMATHSCLASSchannel]

Thanks

[@19sherif]

You should have mentioned that (m) belongs to natural numbers! Can you solve the same problem if you replace 65 by 60?

[@jagadiswarchakraborty295]

Happened to see the video through out. However the beginning of the equation is the key.

[@JJONLINEMATHSCLASSchannel]

🤝🤝

[@naldayaknanmiri]

Ma can you please make a video on calcus , from the scratch. We are suffering to understand it in this 100 level as Nigeria students in university.

[@JJONLINEMATHSCLASSchannel]

Ok noted

[@naldayaknanmiri]

Thank you ma . That will help us a lot.

[@bentleybogle27]

Aren't there other solutions that you could get by factorising 65 as (65)(1) , (1)(65), (13)(5) and even 2 negative factors?

[@bentleybogle27]

I like how you made it manageable by expressing it as the difference of 2 squares.

[@oyindeyefagodspower992]

​@@bentleybogle27 she made huge assumptions.

[@aaoivashchenko6958]

You did not check for possible roots when m below 0. Possibly they are none, but you should have proved it.

[@rui2565]

It is a clever solution. Thank you!

[@JJONLINEMATHSCLASSchannel]

U are welcome

[@sudaannafar9085]

Thank you so much.

[@JJONLINEMATHSCLASSchannel]

You're welcome!

[@paulorodriguesbarros7418]

Quase cheguei fazendo uma tabelinha com potências de 3 e 2 , elevando-os ao mesmo expoente. Só não cheguei ao resultado por um cochilo meu na hora de fazer 3^4.

[@JJONLINEMATHSCLASSchannel]

🤝🤝

[@professormaurice9311]

It would be easier by trial and error.

[@Combinedmathswithpraveen]

How do you there aren't other solutions of m, also you did not classify m to be natural numbers.

[@medben931]

Merci ...

[@RachidOUBRAHIM-z9p]

Vous avez oublié le cas a=65 et b=1 et vice versa

[@paulorodriguesbarros7418]

Tentei por erros e acertos por óbvio aparentemente, mas sem sucesso.

[@GuyDedje]

I would get everything if you were my teacher....

[@JJONLINEMATHSCLASSchannel]

Thanks

[@chukschukwudi9877]

Thanks so much

[@JJONLINEMATHSCLASSchannel]

You're welcome!

[@nikolayguzman331]

This exponential question you can find at least of 10 different videos, And where is Olympiad? 3< m= 65 3^5>>65 m=4

[@carlosfrei7318]

Observen que 65 tiene sólo dos factores posibles 5 y 13 ... esa circunstancia da la posibilidad de aplicar el método propuesto. GOOD

[@JJONLINEMATHSCLASSchannel]

Thanks

[@mikeyu3243]

let m=4, wow, m=4, end of solution

[@youngdkout]

Good afternoon my darling maths teacher

[@JJONLINEMATHSCLASSchannel]

Good afternoon

[@restablex]

Sorry I dont buy your method. You are guessing the factors. So why not guessing a power of 3 higher than 65? The lowest one is 81 = 3^4 Then 81 - 65 = 16 = 2^4. This is also guessing but faster, easier and direct.

[@JJONLINEMATHSCLASSchannel]

Thanks so much. I will appreciate it if you drop your solution as well

[@restablex]

I just did

[@harrymatabal8448]

Mr Restanlex , you are also correct

[@lalbabuprasadmanjhi5464]

Yes can be done by observing

[@JJONLINEMATHSCLASSchannel]

You are guessing as well

[@abhijitganguly264]

65=1×65,65=65×1,you must justify why are you not considering these factors .😊

[@archanabansal9271]

Mam you can also solve it by hit and trial method. Love from India Edit-Mam you make very brilliant videos. i want to ask that from which country does you belong.

[@JJONLINEMATHSCLASSchannel]

Thanks for your contribution. Nigeria

[@jansmotlacha1077]

Your solution is correct, but you must first prove that m is even. Otherwise, you have to consider odd m as well, then m = n+1 and the decomposition has the form (3^n*sqrt(3) + 2^n*sqrt(2))*(3^n*sqrt(3) - 2^n*sqrt(2)) = 65. You must prove that this has no solution for any n. Although this is not the case, the factors don't have to necessarily be integers. For example, think how you would solve 3^m - 2^m = 211.

[@JJONLINEMATHSCLASSchannel]

Ok. Thanks

[@ifeanyianuka6835]

Why exactly should she prove that m is even? She was solving for me. Also, you solve as the occasion requires in mathematics, when you have 65, just consider the integer factors. You shouldn’t be worried about what happens when it’s a number like 211. Most times when these questions are presented, the numbers given have been selected to arrive at such junctions.

[@jansmotlacha1077]

@@ifeanyianuka6835 In the case 211, the exponent is 5 which is odd case, and as I showed, in such a case the factors are not integers - they include square roots. Regarding odd m in case 65 - you can say that you found one solution (for even m) and another solution does not exist, but, at least you must explain why only 1 solution exist (it is true but - WHY).

[@sirjuliusdeviscensus114]

gracias! buen desarrollo,

[@JJONLINEMATHSCLASSchannel]

U are welcome

[@abrahammelkonian9698]

Do you provide massages too ?

[@rammarocusa.1206]

I guess she will do better in massage then math hhhh

[@anasanasa645]

❤

[@ajokaefi]

JJ , ... dont pay attention to the HATERS (they only know how to hate) keep doing this for LOVE ...

[@JJONLINEMATHSCLASSchannel]

Thanks so much ❤️❤️

[@tomvitale3555]

Something didn't seem right when you said (a-b)(a+b)=5x13. From that you inferred (a-b)=5 and (a+b)=13. What if you said (a-b)(a+b)=2x32.5, thus (a-b)=2 and (a+b)=32.5? Or (a-b)(a+b)=sqrt(65)*sqrt(65), i.e., (a-b)=sqrt(65), (a+b)=sqrt(65)? Etc, etc, etc.

[@JJONLINEMATHSCLASSchannel]

Thanks so much for your contribution. Please kindly drop your solution as well.

[@tomvitale3555]

@@JJONLINEMATHSCLASSchannel If you considering all real solutions, there are an infinite amount of them. In this case, let p x q = 65. Then (a-b)(a+b)=pq, and using your method to solve for a and b, (a-b)=p and(a+b)=q you get a=(p+q)/2 and b=(q-p)/2. Of course, if you're just considering integer answers, for p/q you could have 5/13, 13/5, 1/65, 65/1, -1/-65, -13/-5, etc.

[@Ibrahimwodiali]

How can (a-b)(a+b) be sqrt(65)*sqrt(65)? According to your argument (a-b) gives me sqrt(65) and yet (a+b) would also give me the same sqrt(65)?

[@tomvitale3555]

@@Ibrahimwodiali Yes, I get your point, but what I'm saying is that the choice of 5 x 13 is arbitrary. Why not 13 x 5? Why not -5 x -13? There are an infinite amount of ways that the product of 2 factors can equal 65. And if you are just considering integer values, what about 1 x 65?

[@AbouDiawara-id6wk]

❤❤❤❤

[@JJONLINEMATHSCLASSchannel]

❤️❤️❤️

[@Divine-Education-1980]

Thankyou mam (from India)

[@JJONLINEMATHSCLASSchannel]

U are welcome

[@paulohgifoni5828]

What problem to use a method more extensive? The girl presented a method of resolving the issue. This method she finds interesting. If anyone finds an easier method, use it.

[@JJONLINEMATHSCLASSchannel]

Thanks so much. I appreciate this

[@jujumen]

what happens to your solution if (a-b)=13 and (a+b)=5 ?

[@JJONLINEMATHSCLASSchannel]

It's a simple logic

[@jujumen]

@@JJONLINEMATHSCLASSchannel ok but what happens to your solution if (a-b)=13 and (a+b)=5 ?

[@jujumen]

a=9 ; b=-4 then what happens to m in -4=2^(m/2) ? complex number?

[@troybingham6426]

You assumed that a and b are natural numbers. There are infinite real number factors of 65. On what basis are you excluding them?

[@JJONLINEMATHSCLASSchannel]

Thanks so much. The natural solutions were required

[@user-vy8nb6sd5f]

انت مدرسه جيده وجميله

[@rammarocusa.1206]

والله بعيدة كل البعد عن منطق الرياضيات انها تجخمط

[@donnyaruwata165]

Just guess since it's olympiad and pressed with time. The first biggest value of m=4 that is close to getting 65. 81-16 is 65. Next question please.

[@donnyaruwata165]

@@ae112 Olympiad was indeed a multiple choice bro.

[@JJONLINEMATHSCLASSchannel]

Thanks for your suggestions but we are here to show people how to solve it.

[@donnyaruwata165]

@@JJONLINEMATHSCLASSchannel You can still state clearly the method one used which is iteration. The method you used above still has that element of guessing or iteration because one cannot arbitrarily decide to square the elements!

[@doowadiwadi]

@@ae112 Nope

[@user-dq3uh6ee5w]

4.

[@kk12181]

These questions are for kids who doesn’t even get introduced to real number system. So chill!

[@pradyumnanayak9844]

🙏🇮🇳🇮🇳🇮🇳

[@JJONLINEMATHSCLASSchannel]

U are welcome

[@hermanzavaleta8879]

Sale al ojo 4

[@hermanzavaleta8879]

81.16 = 65

[@anuarsalomao9748]

COMICADO

[@harrymatabal8448]

Good. 65 can also be 1×65

[@JJONLINEMATHSCLASSchannel]

Thanks

[@jaggisaram4914]

4

[@kanyamagaraabdallah8300]

où pouvons-nous vous trouver

[@JJONLINEMATHSCLASSchannel]

Nigeria

[@GuyDedje]

😂😂😂😂

[@dwiyantowibisana4928]

m=4

[@moisesgustavodazcarbajal2741]

OK, it was easy to solve cause 65 is a number factorized by two numbers, it is not so easy when a number can be factorized with more than two numbers or there is a prime number equating the algebraical expresion. So you have to probe if there are more solutions too. Greetings from Mexico.

[@JJONLINEMATHSCLASSchannel]

Thanks

[@rubenfuentes205]

m=. 4

[@umarisiaku5500]

The method is deficult to understand

[@harrymatabal8448]

JJ did you know that you have a sexy smile❤

[@syedmdabid7191]

Id est m=4, responsi.

[@JJONLINEMATHSCLASSchannel]

🤝🤝

[@rammarocusa.1206]

That not we do math

[@hermanzavaleta8879]

Mucho chamul

[@omarchamiaa3216]

bad work

[@agrolactingenieriadealimentos]

M= 4

[@deepaksinghpaonam6753]

This concept is wrong if value of m is negative

[@success762]

Impossible. m must be positive only in this case

[@ifeanyianuka6835]

You’re wrong. Her concept is very correct. If m were a negative number, the values of the constants will change. For example if m were -4, the constant 65 should have been 1/65 instead.

[@akroy2422]

Mnnnnbbb0 , 0:06

[@deepaksinghpaonam6753]

Unbaised concept