... 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 🎉
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
@@mrjnutube , this video is not meant for a person at your mathematical level. It’s meant to help high school students who, most likely, are not yet that strong at substitution.
I know a good teacher when I see one. You are a good teacher with a patient heart. You're humble too. I am like that too. That's why I was able to recognize you, instantly. I will start following you, asap
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? 😅
@@gondaimoyana2691 What is your point? Take it for what it is please. Her back substitutions solved the equation. Your other factors as mentioned are inconsequential for this level!!!
*If m3^x - 2^x, for x>0 f'(x) = (3^x).(ln(3)) - (2^x).(ln(2)) >0 (as 3^x > 2^x > 0 and ln(3) > ln(2) > 0), so f strictly increases on R+* It means that if the equation f(x) = 65 has a solution then this solution is unique on R+* 4 is evident solution, so this is the only solution of the given equation.
Factors of 65 are 5x13, 1x65 and of course the negative integers which will not give solutions to m For factors of 1x65 a = 33, b = 32 which gives additional value of m
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
Actually, she posed the problem as a theorem to be demonstrated, and she went about using axioms, and previous known algebra integer properties of exponentials to demonstrate, and one has to have previous knowledge of those basics to understand her that's why a primary school children won't understand what her solution means because those materials are tought in 5,6,7 grade level depending of a country curriculum. What I mean mathematics are a building bloc science, you need to learn arithmetic to get to algebra, trigonometry, and geometry come in to help with spacial understanding, before moving on to more complex stuffs like integrals, derivatives and so on, if you skip one area you definitely going to have an acute problem understanding the next steps.
why do I feel this is pure luck/special case and wont' work in the general case... As a transcendental equation, you can't (in general) make algebraic manipulations to solve for 'm'. Good ideas though (finding the differences of squares and finding factors 3*15 that work for this case)
The solution in the video is arrived at via several unewarranted assumptions, such as 3^(m/2) and 2^(m/2) are integers; no such assumption is necessary. One can start with noticing that 3^m-2^m is an increasing function of the real variable m; hence the equation can have at most one solution. Then one can observe that we have 3
@@JJONLINEMATHSCLASSchannel It is trial and error, but I efficiently localize the solution. This is all you can do -- your approach is conceptually flawed.
Thank you so much my lady. Ladies of nowadays think that maths are only for men, buy u're just proving them that you ladies have capacity of knowing it. But I have one quetion, I love it ❤
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.
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.
@@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).
Good job. I like your approach and the step by step you showed to solve the problem. However, In the advanced algebra, there is a closed form solution without the long steps.
Lady: you are a good teacher and a bright mathematician. If I was a bit younger and if you are not married yet,... i would request a date. Anyway... power to you.
First of all 3 & 2 are not a common and breaking down 65 into powers of numbers will not have a coefficient of 3 or 2. Why don't you just take log of the same bases to reverse the equation.
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.
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 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 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?
You took one unknown and replaced it with two unknowns. Unfortunately, since you have one equation, you removed the certainty in the solution, and replaced it with speculation, unless you can prove that 5 and 13 are the only factors of 65. Unfortunately, 65 and 1 are also factors....
@PeaceWithYourNeighbour m =log 65/(log 3-log 2), implica que m=10.295... log 3^m - log 2^m, aplica si se tiene*log (3^m/2^m)* ya que es la propiedad de el *logaritmo de un cociente*, log (a/b) es igual a *log a - log b*, y en ningún renglón de la resolución, aparece que 2^m divida a 3^m, es decir (3^m/2^m) no es posible que sea parte de la resolución.
Wrong.you applied log term by term to the left handside instead of applying it once.it is suppose to be log(3^m-2^m) =log 65 and this will not give a solution
Excellent job. Enjoy it very much.
... 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 🎉
My pleasure!
Very good teaching
Thanks and welcome
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
Nice one
@@mrjnutube , this video is not meant for a person at your mathematical level. It’s meant to help high school students who, most likely, are not yet that strong at substitution.
Superb teaching technique, made so simple to explain and comprehend.
Beautiful teacher, Beautiful solution.
Thank you! Cheers!
Very clear and very helpful .
Glad it was helpful!
Waouh Excellente thanks you .
JJ nne, that is a good one. Keep it up. Lovely mode of deriving the solution.
Many thanks!
I like you and your short, logical and precise method !
Thanks
I know a good teacher when I see one. You are a good teacher with a patient heart. You're humble too. I am like that too. That's why I was able to recognize you, instantly. I will start following you, asap
You are a superb master. Keep it up
.
Good job my darling teacher
Thanks
Elegant solution.
Thanks so much
Great job. Keep spreading it.
Thank you, I will
Beautiful math tutor
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? 😅
Please remember that b=2^m/2 meaning b can't be -very number. Bcs the base is +ve.
@@gondaimoyana2691 What is your point? Take it for what it is please. Her back substitutions solved the equation. Your other factors as mentioned are inconsequential for this level!!!
Great job 🎉
Thank you
Great job. Please keep it up!
Thanks, will do!
Very clear and logical method of solving this problem. Thank you
You are welcome
Merci excellente explication grand merci madame ❤❤❤❤👍👋👋👋
U are welcome
Good and well done...
Thank you!
Well done
Thank you so much.
You're welcome!
great vid
Thanks
I appreciate your effort
Beautiful 🥰, you are the best
Many many thanks mam.
Most welcome 😊
Thank you for the strategy. My tactics: 3^m must be >65 let’s try m=4. 81-16=65. Stop
Thanks
Evidentemente el tanteo no siempre es posible, en este caso fue fácil, ahora bien se debería probar, que no existen otras soluciones
Strategic solutions
Impressive lesson
Thanks
Perfect.😀
Thanks! 😃
Super solution MOM, (India)
Good😊
3*m- 2*m= 65 =81_16=3×3×3×3-2×2×2×2
= 3*4 - 2*4
m=4
Excellent!
Good one
Thanks for the visit
Thank you.
*If m3^x - 2^x, for x>0
f'(x) = (3^x).(ln(3)) - (2^x).(ln(2)) >0 (as 3^x > 2^x > 0
and ln(3) > ln(2) > 0), so f strictly increases on R+*
It means that if the equation f(x) = 65 has a solution then this solution is unique on R+*
4 is evident solution, so this is the only solution of the given equation.
Factors of 65 are 5x13, 1x65 and of course the negative integers which will not give solutions to m
For factors of 1x65
a = 33, b = 32 which gives additional value of m
Perfect! Thank you.
The left side is an increasing function, so only one solution. You try m equals 1, 2, 3, 4
You are great
Excellent ❤
Thanks 😊
You did well by trying to make them appear as difference of two squares. Mathematics is all about manipulation. Thanks
U are welcome
Merci pour votre démonstrations mathématique logique.....
....
U are welcome
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
Thanks
Amazing
Actually, she posed the problem as a theorem to be demonstrated, and she went about using axioms, and previous known algebra integer properties of exponentials to demonstrate, and one has to have previous knowledge of those basics to understand her that's why a primary school children won't understand what her solution means because those materials are tought in 5,6,7 grade level depending of a country curriculum. What I mean mathematics are a building bloc science, you need to learn arithmetic to get to algebra, trigonometry, and geometry come in to help with spacial understanding, before moving on to more complex stuffs like integrals, derivatives and so on, if you skip one area you definitely going to have an acute problem understanding the next steps.
It is a clever solution. Thank you!
U are welcome
Great indeed
Thank you
Very good
Thanks
Brilliant
why do I feel this is pure luck/special case and wont' work in the general case...
As a transcendental equation, you can't (in general) make algebraic manipulations to solve for 'm'.
Good ideas though (finding the differences of squares and finding factors 3*15 that work for this case)
Thanks
Great job!!!
Thank you!!
Was really curious to see how you were going to do it. Well done!
Thanks!
How can you be sure that 3^^m/2 and 2^^m/2 are both integers ?
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.
Ok noted
Thank you ma . That will help us a lot.
Aren't there other solutions that you could get by factorising 65 as (65)(1) , (1)(65), (13)(5) and even 2 negative factors?
I like how you made it manageable by expressing it as the difference of 2 squares.
@@bentleybogle27 she made huge assumptions.
The solution in the video is arrived at via several unewarranted assumptions, such as 3^(m/2) and 2^(m/2) are integers; no such assumption is necessary. One can start with noticing that
3^m-2^m is an increasing function of the real variable m; hence the equation can have at most one solution. Then one can observe that we have 3
. Thanks for your suggestions but you are just doing trial and error. Show your workings
@@JJONLINEMATHSCLASSchannel It is trial and error, but I efficiently localize the solution. This is all you can do -- your approach is conceptually flawed.
Congratulations
Thank you so much my lady. Ladies of nowadays think that maths are only for men, buy u're just proving them that you ladies have capacity of knowing it. But I have one quetion, I love it ❤
Thanks for the compliment
You should have mentioned that (m) belongs to natural numbers! Can you solve the same problem if you replace 65 by 60?
Do you provide massages too ?
I guess she will do better in massage then math hhhh
Congratulations. ❤
You did not check for possible roots when m below 0. Possibly they are none, but you should have proved it.
Good afternoon my darling maths teacher
Good afternoon
Interesting powerful 😜!
Thanks
Thanks so much
You're welcome!
That method only works when RHS is a prime number... Simply take logarithms to base 10 on both sides of the equation ASAT.
Happened to see the video through out. However the beginning of the equation is the key.
🤝🤝
Instead of subtracting equations 1 and 2, I added them. The answer is the same but easier to understand.
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.
Ok. Thanks
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.
@@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).
Good job. I like your approach and the step by step you showed to solve the problem. However, In the advanced algebra, there is a closed form solution without the long steps.
Thanks
a=33, b=32, a-b=1, a+b=65
Thank you
You're welcome
Lady: you are a good teacher and a bright mathematician. If I was a bit younger and if you are not married yet,... i would request a date. Anyway... power to you.
Y que paso con los factores 65 y 1 porqué no los consider
First of all 3 & 2 are not a common and breaking down 65 into powers of numbers will not have a coefficient of 3 or 2. Why don't you just take log of the same bases to reverse the equation.
I don't think 5 and 13 are the only factors that can give us 65. 1×65=?
Why do you skip this part?
By inspection, those two can't be the solution just like by inspection, you can bring out the roots of a quadratic equation out of so many factors
good
Thanks
gracias! buen desarrollo,
U are welcome
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.
3^m-2^m=81-16=3^4-2^4 -> m=4
m(=^)= 1 --> 3^ - 2^= 1
m= 2 --> 3^-2^= 5
m= 3 --> 3^-2^= 23
m= 4 --> 3^-2^= 65
m= 5 --> 3^-2^= 211
Merci ...
It is getting more complex than expected.
3^4=81 , 2^4=16 => 81-16 =65 => m=4
ممكن ان نكتب 65=65*1 وبذلك نحصل على
a+b=65
a_b=1
وعليه هل نستطيع اكمال الحل بنفس الطريقه ?….
Mathematical induction kudos
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.
Thanks so much for your contribution. Please kindly drop your solution as well.
@@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.
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)?
@@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?
Which mathematics is this?
How do you there aren't other solutions of m, also you did not classify m to be natural numbers.
Vous avez oublié le cas a=65 et b=1 et vice versa
انت مدرسه جيده وجميله
والله بعيدة كل البعد عن منطق الرياضيات
انها تجخمط
Observen que 65 tiene sólo dos factores posibles 5 y 13 ... esa circunstancia da la posibilidad de aplicar el método propuesto.
GOOD
Thanks
Thankyou mam (from India)
U are welcome
You took one unknown and replaced it with two unknowns. Unfortunately, since you have one equation, you removed the certainty in the solution, and replaced it with speculation, unless you can prove that 5 and 13 are the only factors of 65. Unfortunately, 65 and 1 are also factors....
Exactly, why not just use logarithm.
This looks assumpsive.
It should have been precised from the beginning that only the solution in integers is requested.
JJ , ... dont pay attention to the HATERS (they only know how to hate) keep doing this for LOVE ...
Thanks so much ❤️❤️
Use logarithm, take log of both sides of the equation.
log3^m - log2^m = log65
mlog3 - mlog2 = log65
m(log3 - log2) = log65
m = log65/(log3 - log2)
@PeaceWithYourNeighbour m =log 65/(log 3-log 2), implica que m=10.295...
log 3^m - log 2^m, aplica si se tiene*log (3^m/2^m)* ya que es la propiedad de el *logaritmo de un cociente*, log (a/b) es igual a *log a - log b*, y en ningún renglón de la resolución, aparece que 2^m divida a 3^m, es decir (3^m/2^m) no es posible que sea parte de la resolución.
Wrong.you applied log term by term to the left handside instead of applying it once.it is suppose to be log(3^m-2^m) =log 65 and this will not give a solution
Why subtract equation 2 from equation 1 when you can just add them so that 2a =18 and a =9
Thanks for the suggestion