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!!!
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.
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)
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.
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....
... 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 🎉
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.
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.
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.
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.
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
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 ❤
*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).
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.
@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
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
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
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?
@@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!
No. it means logically a- b must equal the smaller factor and a+b the larger factor. Remember. 3×4 = 4×3 and if you have a+b as factor and a-b as the second factor wrtten thus (a+b)(a-b).same as saying 4×3, a+b represents 4 , the larger part and a--b represents 3 the smaller part. And if the product is say 60 . We can write (a+b)(a-b) = 60 then we get factors of sixty ..There are several. But we chose those factors that align with the problem and solve the problem. Hopes this helps
@@merchantcharles1954. Great job. I think Maths/Algebra is universal so I do not feel very comftable in guessing instead of demonstrating. Never mind, I did Maths some years back so Maybe I may be out of date because I am doing totally different thing right now😊😊😊
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.
Thanks but in this platform, all I do is to show detailed solutions to questions. Am not in an exam hall. If I were to be, there are shortcuts to take but here, I am teaching, so all steps must be duly explained.
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.
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.
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.
