Since you didn't say that m and n must be integers, technically there are infinitely many solutions for m and n in the form of n=log2(8064/(2^k-1)); m=log2(8064/(2^k-1)+8064) where k can be any number except 0. If you plug in k=6 you get the integer solution for m and n.
Iam very weak in math. I must be doing something wrong. Is n= log base 2(8064/(2^k-1)) ? Where if K=6 in both n and m, 2 will be raised the power of 5? I could not get n=7 and m=13. Help! Thanks
@@billl3936 Yes, n=log base 2 (8064/(2^k-1)). If you plug in k=6 you get n=log base2 (8064/(2^6-1)) 2^6-1=64-1=63 and 8064/63=128=2^7 finally log base 2 (2^7)=7 So n=7 and you get m with the same method.
I watch tons of maths videos, but I never comment. I liked the approach here a lot. I didn’t know how to solve it algebraically initially but your approach really gave me that “light bulb” moment that all maths people love. Great stuff!
Thanks a million and we are glad you gained some values from this video tutorial because, this is our utmost priority sir. We promise to give the best in all our teachings in order to serve you and other better by His grace. We love you sir...❤️❤️💖😍💕🙋🙋
In this case, the case of this problem, it's easy to solve mentally, because anyone knows that 2^10 = 1024 , so just double it until you reach 8,192 ( 2^13) . There it is simple, because 8192 - 128 = 8064 128 = 2^6 But for a higher base, then it starts to complicate, so let's use herr professor's method, which is simple and friendly !!!! Bingo, from Brazil !!!!!
My solution: In binary, 2ⁿ is represented as 1000... Witn n zeroes, so the diference 2^m - 2ⁿ, in binary, must be 11..11000..00, a number with m digits and n trailing zeroes. Since 8064 = 1111110000000_b, We have m=13 (13 digits) and n=7 (7 zeroes at the end)
Sorry am lost at your procedure sir. I will appreciate it more if you can put more light to this approach because I can see a wonderful and a unique method from it sir. Thanking you in advance sir.
@@onlineMathsTV I'll try to explain in simpler terms :) (When the number is in binary, I'll prefix it with b, that is, the binary 1101 will be written as b1101) To write a number in binary, we can only use 0 and 1, for example, 2 in binary would be b10; 3 would be b11 and 5 would be b101. As in base ten we have 1234 = 1*10³ + 2*10² + 3*10¹ + 4*10⁰ In binary we have b1010 = 1*2³ + 0*2² + 1*2¹ + 0*2⁰ What hapens if we have a power of two in binary? Well, the same way in base 10 we have 10³=1000, in binary, 2³ = b1000. What if, in base 10, we subtract two powers of 10? For example, 10⁶ - 10⁴ = 1,000,000 - 10,000 = 990,000. Note that the answer has 6 digits in total, and 4 trailing zeroes. This will hapen for any two powers of ten (you can see the why trying to subtract by the subtraction algorithm). And the inverse? How to find m and n such 10^m - 10^n = 999,900,000? Well, m=9 and n=7, we can do by just looking at the number! (Note: if the equation was 10^m - 10^n = 9876, we don't have integer solutions) The same thing happens in binary: 2⁴ - 2² = b10000 - b100 = b1100, 4 digits and 2 zeroes (the subtraction algorithm will work in a similar way) and, in a more general way, 2^m - 2^n will have m digits and n trailing zeroes. So, if we convert 8064 to binary, finding m and n would be trivial: just count the number of digits and the number of trailing zeroes. (Note that if instead of 8064 we had 5 = b101, there would be no integer solutions) To convert 8064 to binary, we could simply use a binary calculator (not so elegant uh?), or use an algorithm to do so. The algorithm consists of continuously divide the number by two, then read the remainders backwoods. Its kinda hard to explain in text, but the process is pretty simple, there are many videos explaining how to convert decimals to binary. After converting, we find 8064 = b1,1111,1000,0000 13 digits, 7 trailing zeroes
Although you did not specify m, n as integers, I assumed that was the intent. The first statement in my solution was 2^m > 8064 which leads to m >= 13 quickly. Different steps, same solution.
In the factored form (@3:10) you know that 2^n must be a power of 2, and therefore from the set {2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096}. You also know that 2^k - 1 is always an odd number. So, our factor pairs (2^n, 2^k - 1) are from the set {(2, 4032), (4, 2016), (8, 1008), (16, 504), (32, 252), (64, 126), (128, 63)} - the remaining factors are non-integer and are disqualified. The only factor pair that matchs the requirement for an integer pair (Even, Odd) is (128, 63) so it is the only factor pair that needs to be checked. J'espere que cela t'aider, et je te souhaite bonne chance mon ami.
You are most welcome sir. We are glad to have you here and at this time happy that you gained some values from our video tutorial. Much love from everyone here sir.💖💖💖
I don't understand the step where you come up with 2^n(2^k-1) =128 -63. A much better approach would be to think about the powers of 2: 1, 2, 4 ......... 1024, 2048, 4096, 8192. 1024 = 2^10. 8192 = 2 ^13, 8192 - 8064 = 128 = 2^7. Therefore m = 13, and n =7.
2^m - 2^n = 8064 2^m = 2^n + 8064 m = log(2^n + 8064)/log(2), for n real. If m and n are integers, then the solution requires more work: 2^m - 2^n = 8064 = 128*63 = 2^7*(2^6 - 2^0) = 2^13 - 2^7 m = 13, n = 7
8064 is close to 8196, the nearest integer power of 2. 8196-8064=132, not a power of 2 so try the next one. 16384-8064=8320, also not a power of 2 32768-8064 wait it’s never gonna be an integer because powers of 2 are separated by powers of 2. This must be solved algebraically. Edit: i subtracted wrong. 8192-8064=128, not 132
Don't voilate math's rules for short cuts, You can not cut the equal base, rather you can equate the powers in next step. Similarly in case of square of square root, don't cut their signs but remove root in next step. Cutting is only allowed in multiplications.
this is not adequately explained for students. how do they choose the two factors? if you tell students, here are two factors...you would need to tell them how to pick those 2 factors. 2^13-8064=128 and is the first positive integer. 8064/128 = 63. 2^12 and less yields negative integers, When students are stuck, find the answer and then work backwards to create the proof. By graphing, students can find m=13. 8064 - 2^13 = 128. the student can find 128 * 63. although graphing gives the answer, it is not a proof. they still need to do the proof.
Since you didn't say that m and n must be integers, technically there are infinitely many solutions for m and n in the form of n=log2(8064/(2^k-1)); m=log2(8064/(2^k-1)+8064) where k can be any number except 0. If you plug in k=6 you get the integer solution for m and n.
Thanks for this observation. Well received and it shall be implemented in subsequent videos.
Respect sir. ...👍👍👍
Iam very weak in math. I must be doing something wrong. Is n= log base 2(8064/(2^k-1)) ? Where if K=6 in both n and m, 2 will be raised the power of 5? I could not get n=7 and m=13. Help! Thanks
@@billl3936 Yes, n=log base 2 (8064/(2^k-1)).
If you plug in k=6 you get n=log base2 (8064/(2^6-1))
2^6-1=64-1=63
and 8064/63=128=2^7
finally log base 2 (2^7)=7
So n=7 and you get m with the same method.
Absolutely! In each of 3 videos, which I watched so far - such antimath "maybes". Shitty stuff. Blocking.
yes, the condition about integers is mandatory
I watch tons of maths videos, but I never comment. I liked the approach here a lot. I didn’t know how to solve it algebraically initially but your approach really gave me that “light bulb” moment that all maths people love. Great stuff!
Thanks a million and we are glad you gained some values from this video tutorial because, this is our utmost priority sir.
We promise to give the best in all our teachings in order to serve you and other better by His grace.
We love you sir...❤️❤️💖😍💕🙋🙋
In this case, the case of this problem, it's easy to solve mentally, because anyone knows that 2^10 = 1024 , so just double it until you reach 8,192 ( 2^13) .
There it is simple, because 8192 - 128 = 8064
128 = 2^6
But for a higher base, then it starts to complicate, so let's use herr professor's method, which is simple and friendly !!!!
Bingo, from Brazil !!!!!
The first explanation gives an insight into the solutions. Thanks for this solution Jakes.
Thanks a million ma.
My solution:
In binary, 2ⁿ is represented as 1000... Witn n zeroes, so the diference 2^m - 2ⁿ, in binary, must be
11..11000..00, a number with m digits and n trailing zeroes.
Since
8064 = 1111110000000_b,
We have m=13 (13 digits) and n=7 (7 zeroes at the end)
Sorry am lost at your procedure sir. I will appreciate it more if you can put more light to this approach because I can see a wonderful and a unique method from it sir.
Thanking you in advance sir.
@@onlineMathsTV I'll try to explain in simpler terms :)
(When the number is in binary, I'll prefix it with b, that is, the binary 1101 will be written as b1101)
To write a number in binary, we can only use 0 and 1, for example, 2 in binary would be b10; 3 would be b11 and 5 would be b101.
As in base ten we have
1234 = 1*10³ + 2*10² + 3*10¹ + 4*10⁰
In binary we have
b1010 = 1*2³ + 0*2² + 1*2¹ + 0*2⁰
What hapens if we have a power of two in binary? Well, the same way in base 10 we have 10³=1000, in binary, 2³ = b1000.
What if, in base 10, we subtract two powers of 10? For example, 10⁶ - 10⁴ = 1,000,000 - 10,000 = 990,000. Note that the answer has 6 digits in total, and 4 trailing zeroes. This will hapen for any two powers of ten (you can see the why trying to subtract by the subtraction algorithm). And the inverse? How to find m and n such 10^m - 10^n = 999,900,000? Well, m=9 and n=7, we can do by just looking at the number! (Note: if the equation was 10^m - 10^n = 9876, we don't have integer solutions)
The same thing happens in binary: 2⁴ - 2² = b10000 - b100 = b1100, 4 digits and 2 zeroes (the subtraction algorithm will work in a similar way) and, in a more general way, 2^m - 2^n will have m digits and n trailing zeroes.
So, if we convert 8064 to binary, finding m and n would be trivial: just count the number of digits and the number of trailing zeroes. (Note that if instead of 8064 we had 5 = b101, there would be no integer solutions)
To convert 8064 to binary, we could simply use a binary calculator (not so elegant uh?), or use an algorithm to do so. The algorithm consists of continuously divide the number by two, then read the remainders backwoods. Its kinda hard to explain in text, but the process is pretty simple, there are many videos explaining how to convert decimals to binary.
After converting, we find
8064 = b1,1111,1000,0000
13 digits, 7 trailing zeroes
What a nice solution!
Although you did not specify m, n as integers, I assumed that was the intent. The first statement in my solution was 2^m > 8064 which leads to m >= 13 quickly. Different steps, same solution.
Respect sir. You are great at what you do sir.
Respect!!!
ok but how do you find that 8064 equal to 128x63 ?
In the factored form (@3:10) you know that 2^n must be a power of 2, and therefore from the set {2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096}. You also know that 2^k - 1 is always an odd number. So, our factor pairs (2^n, 2^k - 1) are from the set {(2, 4032), (4, 2016), (8, 1008), (16, 504), (32, 252), (64, 126), (128, 63)} - the remaining factors are non-integer and are disqualified. The only factor pair that matchs the requirement for an integer pair (Even, Odd) is (128, 63) so it is the only factor pair that needs to be checked. J'espere que cela t'aider, et je te souhaite bonne chance mon ami.
By expressing 8064 as a product of its prime factors and putting base two into consideration because both bases are in base two.
Thanks for this reply on behalf of onlinemathstv @Andrew Layton.
Thanks and much love sir...💕💕💖💖💖
@@andrewlayton9760 thanks a lot mon ami.
@@onlineMathsTV Thanks for reply
Very good and simple to understand.
Thanks a million sir/ma.
Merci professeur je vous suis très bien salut from morroco
You are most welcome sir. We are glad to have you here and at this time happy that you gained some values from our video tutorial.
Much love from everyone here sir.💖💖💖
Good explanation, thanks.
Glad you liked it and thanks for watching and leaving a comment behind sir/ma.
We love you dearly ❤️❤️💖💖
Exactly my solution and my steps . Thank u for ur fantastic content ❤🎉
Thanks for the complement sir.
Much love and more grace to carry on the good works in our times.
Love you sir....❤️❤️
@@onlineMathsTV Thank u for ur beautiful words.l hope u a lot of success and progress ❤️👍
How do we know or predict the breaking down of 8064 should be 128 and 63? There are so many combination. Thank you.
Simply consider the base numbers on the LHS
The answer is obvious. Too easy.
Sure, it shows you are the master here🙏🙏🙏
I don't understand the step where you come up with 2^n(2^k-1) =128 -63.
A much better approach would be to think about the powers of 2:
1, 2, 4 ......... 1024, 2048, 4096, 8192. 1024 = 2^10. 8192 = 2 ^13, 8192 - 8064 = 128 = 2^7.
Therefore m = 13, and n =7.
great also
13 and 7 respectively
Muito bom! Muito bom!
Thanks for these words of encouragement sir.
Much love❤️❤️💕💕💖💖
@@onlineMathsTV a
A
Thanks good explanation
You are welcome sir, and thanks for watching and dropping a comment.
Much love...💖💖💖
2^m - 2^n = 8064
2^m = 2^n + 8064
m = log(2^n + 8064)/log(2), for n real.
If m and n are integers, then the solution requires more work:
2^m - 2^n = 8064 = 128*63 = 2^7*(2^6 - 2^0) = 2^13 - 2^7
m = 13, n = 7
Chia hai vế cho 2^n sẽ ra hai bên đều lẽ- suy ra n = 7
Congratulations
Thanks for the encouragement sir.
We love you ❤️❤️💕💕
What are the conditions for choosing the products of 8064, because there are may of the products like 32and252,64and126 etc
I see 8192-128=8064 from the beginning.
Bravo👍👍👍
2 سنيسيس m -2 sin n= enthing edpace
8064
=8192 - 128
=2^13 - 2^7
hence, m=13, n=7
يمكن بناء اي مفاعل نووي بنفس المساحة و الوزن كدالك .،
Facilis! m = 2, n=1 responsi.
Value of W- n(2) =???? ( valum numericus=????)
n^4_n^2=9900 correction please
Noted sir, thanks
Nice step 😅😅😅
Thanks sir. We are glad you gained some values.
Much love 💖💖💖
انشرها و عممها
Удивительно! 😯🔥🔥🔥
Thanks a million for watching and comment sir.
Much love💖💖💕💕
المتر و النيوتون متر بينهما مساحة و هي غير معيرة بالاسم و ما قبل متر و ما بعد نيومتر و كيف تسلمت علم المساحة بحساب العد الزمكوني ام بقياس نيتشه
كيف شكل المساحة .، اين اساس المساحة هو اتن من اين مساحة
هنا تصبح 2m-(enthing espace )
8064 is close to 8196, the nearest integer power of 2.
8196-8064=132, not a power of 2 so try the next one.
16384-8064=8320, also not a power of 2
32768-8064 wait it’s never gonna be an integer because powers of 2 are separated by powers of 2. This must be solved algebraically.
Edit: i subtracted wrong. 8192-8064=128, not 132
لا بد ان اخصل على معدل اسفل الارظ في اه من الاوزان و المساحة لكي احسب لك مثلا يا سيد رواتشلد تلك مفاعلة النووية الشينية مثلا مثلا
You alright bro?
8064 اجعلها اسطوانة مثلا
How many solutions does it have?
infinite
@@onlineMathsTV ok.
M=+4032 N=-4032
Not broblem it s good deal = قد تستلم اي رقم و المساحة الزمكونية بالمعدل الجدري
Thanks
How did you know to factor 8064 with 128 and 63?
Thanks for asking this hidden question sir.
We know that by putting into consideration the existing base on the LHS.
صعب ان اكتب في رياظبات دون ان استمنطق
Easy m=13 and n=7
Bravo 👍👍👍
Respect 🙏🙏🙏
هل لا زلت في كم ام ستنقلنا في سفينتك نحو المستقبل
m=13 n=7
Bravo👍👍👍
Don't voilate math's rules for short cuts,
You can not cut the equal base, rather you can equate the powers in next step. Similarly in case of square of square root, don't cut their signs but remove root in next step. Cutting is only allowed in multiplications.
Noted sir.
Thanks for this educative comment.
Respect 🙏🙏
this is not adequately explained for students. how do they choose the two factors?
if you tell students, here are two factors...you would need to tell them how to pick those 2 factors. 2^13-8064=128 and is the first positive integer. 8064/128 = 63.
2^12 and less yields negative integers,
When students are stuck, find the answer and then work backwards to create the proof.
By graphing, students can find m=13. 8064 - 2^13 = 128. the student can find 128 * 63. although graphing gives the answer, it is not a proof. they still need to do the proof.
👍🇧🇷
Many thanks to you sir
🎹
- 2 sinusus n = 8064 not
تتدكر