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.
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!!!
... 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 🎉
@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
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
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)
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.
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
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.
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 ❤
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
*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.
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).
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.
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.
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.
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....
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?
She did guess the factors of 65! What about 10*6.5, 100*.65, 8 * 65/8, … There are infinite number of factors if you involve rational and irrational numbers. Furthermore, using the method of quadratic equations can be used if one side of the equation is 0. One side is 65, therefore that’s not the proper way.
@@lagazofamily , the way the problem is structured, integer solutions are expected. Following your line of reasoning, of the infinite number of factors she just happened to guess the only non-one and itself pair of integers. The probability of that is 1/ infinity.
@@mawavoy How did you know that integer solution is expected? If I give you a very similar problem like 7^x - 5^x = 91, the way you’re solving the problem won’t hold for this. I’m not saying that your answer is incorrect but the way you’re doing it is non-standard.
@@lagazofamily , the problem didn’t say otherwise, and I know the integer factors for 65.. They come up all the time in high school and years 1 and two of college mathematics.
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.
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.
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.
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!!!
Very good teaching
Thanks and 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
... 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!
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
I like you and your short, logical and precise method !
Thanks
You did well by trying to make them appear as difference of two squares. Mathematics is all about manipulation. Thanks
U are welcome
Beautiful teacher, Beautiful solution.
Thank you! Cheers!
great vid
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
Beautiful math tutor
Very clear and very helpful .
Glad it was helpful!
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
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.
Good job my darling teacher
Thanks
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
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
Great job 🎉
Thank you
Elegant solution.
Thanks so much
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.
Very clear and logical method of solving this problem. Thank you
You are welcome
3*m- 2*m= 65 =81_16=3×3×3×3-2×2×2×2
= 3*4 - 2*4
m=4
Waouh Excellente thanks you .
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
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
Good😊
It would be easier by trial and error.
a=33, b=32, a-b=1, a+b=65
You are a superb master. Keep it up
.
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.
Thank you so much.
You're welcome!
Perfect.😀
Thanks! 😃
*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.
JJ nne, that is a good one. Keep it up. Lovely mode of deriving the solution.
Many thanks!
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).
Great job. Keep spreading it.
Thank you, I will
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 excellente explication grand merci madame ❤❤❤❤👍👋👋👋
U are welcome
The left side is an increasing function, so only one solution. You try m equals 1, 2, 3, 4
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.
Super solution MOM, (India)
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
3^m-2^m=81-16=3^4-2^4 -> m=4
How can you be sure that 3^^m/2 and 2^^m/2 are both integers ?
There are more than one case
Case 1
a-b = 1 and a+b = 65
Case 2
a-b = 5 and a+b = 13
These should be the cases
Congratulations
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.
That method only works when RHS is a prime number... Simply take logarithms to base 10 on both sides of the equation ASAT.
You should have mentioned that (m) belongs to natural numbers! Can you solve the same problem if you replace 65 by 60?
I appreciate your effort
JJ , ... dont pay attention to the HATERS (they only know how to hate) keep doing this for LOVE ...
Thanks so much ❤️❤️
I would get everything if you were my teacher....
Thanks
Very good
Thanks
Perfect! Thank you.
Good and well done...
Thank you!
It is a clever solution. Thank you!
U are welcome
Everyone with alternative solutions needs to show it.
Let your solutions do the talking... 🇯🇲🇯🇲🇯🇲🇯🇲
Thank you so much ❤️
You are great
Excellent!
Great job. Please keep it up!
Thanks, will do!
Beautiful 🥰, you are the best
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.
Thanks
Wao! I’m in love! ❤ pretty & Intellectuel woman!
Thanks
Impressive lesson
Thanks
You make the simplest thing looks difficult .That's why people are afraid of mathematics
Vous avez oublié le cas a=65 et b=1 et vice versa
Thank you.
Merci pour votre démonstrations mathématique logique.....
....
U are welcome
It should have been precised from the beginning that only the solution in integers is requested.
Amazing
Observen que 65 tiene sólo dos factores posibles 5 y 13 ... esa circunstancia da la posibilidad de aplicar el método propuesto.
GOOD
Thanks
Happened to see the video through out. However the beginning of the equation is the key.
🤝🤝
Many many thanks mam.
Most welcome 😊
Excellent ❤
Thanks 😊
let m=4, wow, m=4, end of solution
Easier to add equations 1 and 2 to eliminate b
Luckily 65 has just two factors. Suppose your number has several factors or it is prime?
Thanks
Why subtract equation 2 from equation 1 when you can just add them so that 2a =18 and a =9
Thanks for the suggestion
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.
to be honest (beyond the math):"love you".
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?
She didn’t guess the factors . Since 65 ends in 5 divide by 5 gives 13. Both 5 and 13 are prime numbers
Thanks so much
She did guess the factors of 65! What about 10*6.5, 100*.65, 8 * 65/8, … There are infinite number of factors if you involve rational and irrational numbers. Furthermore, using the method of quadratic equations can be used if one side of the equation is 0. One side is 65, therefore that’s not the proper way.
@@lagazofamily , the way the problem is structured, integer solutions are expected. Following your line of reasoning, of the infinite number of factors she just happened to guess the only non-one and itself pair of integers. The probability of that is 1/ infinity.
@@mawavoy How did you know that integer solution is expected? If I give you a very similar problem like 7^x - 5^x = 91, the way you’re solving the problem won’t hold for this. I’m not saying that your answer is incorrect but the way you’re doing it is non-standard.
@@lagazofamily , the problem didn’t say otherwise, and I know the integer factors for 65.. They come up all the time in high school and years 1 and two of college mathematics.
Brilliant
ممكن ان نكتب 65=65*1 وبذلك نحصل على
a+b=65
a_b=1
وعليه هل نستطيع اكمال الحل بنفس الطريقه ?….
Good afternoon my darling maths teacher
Good afternoon
Good one
Thanks for the visit
We are another solutions because 65=1.65
Congratulations. ❤
Interesting powerful 😜!
Thanks
Was really curious to see how you were going to do it. Well done!
Thanks!
65 is between 64 and 81=3^4
3^4_2^4=81_16=65 answer
Great indeed
Thank you
This is way too easy. Find m for 2^m + 3^m = 40…
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.
Thanks for your contribution. Nigeria
good
Thanks
You did not check for possible roots when m below 0. Possibly they are none, but you should have proved it.
It is getting more complex than expected.
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.
Thanks so much. I will appreciate it if you drop your solution as well
I just did
Mr Restanlex , you are also correct
Yes can be done by observing
You are guessing as well
m=4. i found it mentally in 5 seconds witout using any calculator or pen.
This exponential question you can find at least of 10 different videos, And where is Olympiad? 3< m= 65 3^5>>65 m=4