If the negative root is used, then you are adding fractions since this is what a negative power means. If either root is negative, the sum won't be an integer.
The key in solving problem of this type is to manipulate the left hand side as the difference of 2 squares. (k^u)-(k^v)=n where k and n are constant while u and v are unknowns.
Before watching: Alright, so the first thing I did was find what powers of 2 (2^a and 2^b) can, when one is subtracted from the other, give you 60. The answer? 2^6 (64) - 2^2 (4) = 0. So, our first term must equal 64, and our second must equal 4. √2 = 2^(1/2), and (a^m)^n = a^(mn) So our initial terms can be rewritten as 2^((√x)/2) - 2^((√y)/2) = 60. From what I wrote above, this means that (√x)/2 = 6 -> √x = 12 -> x = 144 And (√y)/2 = 2 -> √y = 4 -> y=16 The solution is X=144, Y = 16.
Ici on trouve une solution mais quand on considère 2^n(2^k-1)=60 on considère que n et k sont des entiers mais on y apporte pas la preuve. Il faudrait prouver qu'en dehors des entiers on ne puisse pas avoir l'égalité 2^n(2^k-1)=60 ou alors prouver que la solution trouvée est unique soit un seul couple possible pour(x,y) ce qui n'est pas le cas. Pour l'instant je me casse les dents sur l'unicité et je ne suis pas davantage arrivé à prouver que n et k (utilisés dans la vidéo) ne peuvent être qu'entiers même si j'en ai la conviction profonde.
good video
good solution
Glad you liked it!
64-4=60
√2^(√x)=64=√2^(6×2)=√2^12
√2^(√y)=4=2^2=√2^4
√x=12 則x=144
√y=4 則y=16
Very Nice ❤
@ 2:40 / 11:08
There's no need to introduce an additional variable as k
2^(m) - 2^(n) = 60
2^(m + n - n) - 2^(n) = 60
2^(n + m - n) - 2^(n) = 60
[2^(n) * 2^(m - n)] - 2^(n) = 60
2^(n) * [2^(m - n) - 1] = 60
2^(n) * [2^(m - n) - 1] = 4 * 15
2^(n) * [2^(m - n) - 1] = 2^(2) * 15 → by identification
n = 2
2^(m - n) - 1 = 15
2^(m - n) = 16
2^(m - n) = 2^(4)
m - n = 4
m = 4 + n → recall: n = 2
m = 6
Very Nice ❤
It would have been helpful to mention that the square roots should be positive integers. For example, we can get solutions involving logarithms.
2:35 ( m>n) є N
2⁶ - 2² = 60
√x = 6 => x = 36
√y = 2 => y = 4
Thanks for your feedback ❤
exactly, the problem statement needs to specify the conditions on x and y.
If the negative root is used, then you are adding fractions since this is what a negative power means. If either root is negative, the sum won't be an integer.
The key in solving problem of this type is to manipulate the left hand side as the difference of 2 squares.
(k^u)-(k^v)=n
where k and n are constant while u and v are unknowns.
Yes, you are right! ❤
2⁶ - 2² = 60
√2¹² - √2⁴ = 60
√x = 12 => x = 144
√y = 4 => y = 16
Great! ❤
Ответ - 3.Обожаю африканских беженцев.
144 and 16 ❤
Before watching:
Alright, so the first thing I did was find what powers of 2 (2^a and 2^b) can, when one is subtracted from the other, give you 60.
The answer? 2^6 (64) - 2^2 (4) = 0.
So, our first term must equal 64, and our second must equal 4.
√2 = 2^(1/2), and (a^m)^n = a^(mn)
So our initial terms can be rewritten as 2^((√x)/2) - 2^((√y)/2) = 60.
From what I wrote above, this means that (√x)/2 = 6 -> √x = 12 -> x = 144
And (√y)/2 = 2 -> √y = 4 -> y=16
The solution is X=144, Y = 16.
Great! ❤
3a-2a=a => a=60
3a=3*60=180 => V2^Vx=180 => Vx = ln(180)/ ln(V2) => Vx= 5,1929.../ 0,34657....= 14,9837...
x= Vx^2 = 224,51....
2a=2*60=120 => V2^Vy=120 => Vy = ln(120)/ ln(V2) => Vy= 4,78749.../ 0,34657...= 13,81....
y= Vy^2 = 190,82....
Great! ❤
I feel like the solution to this problem you can always find it through trial and error.
Yes, you are right ❤
Do you need to require the condition that the exponents are natural numbers?
Thanks for your feedback ❤
X= 144, Y= 16.
Yes, you are right ❤
@@SALogics no entiendo.
Look at the original equation!!Thats all folks.
I can't understand what do you want to ask?? ❤
Ici on trouve une solution mais quand on considère 2^n(2^k-1)=60 on considère que n et k sont des entiers mais on y apporte pas la preuve.
Il faudrait prouver qu'en dehors des entiers on ne puisse pas avoir l'égalité 2^n(2^k-1)=60 ou alors prouver que la solution trouvée est unique soit un seul couple possible pour(x,y) ce qui n'est pas le cas. Pour l'instant je me casse les dents sur l'unicité et je ne suis pas davantage arrivé à prouver que n et k (utilisés dans la vidéo) ne peuvent être qu'entiers même si j'en ai la conviction profonde.
Merci pour vos commentaires. J'apprécie vos efforts! ❤