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@ 2:46 / 10:41There's no need to introduce an additional variable as k2^(m) - 2^(n) = 282^(m + n - n) - 2^(n) = 282^(n + m - n) - 2^(n) = 28[2^(n) * 2^(m - n)] - 2^(n) = 282^(n) * [2^(m - n) - 1] = 282^(n) * [2^(m - n) - 1] = 4 * 72^(n) * [2^(m - n) - 1] = 2^(2) * 7 → by identificationn = 22^(m - n) - 1 = 72^(m - n) = 82^(m - n) = 2^(3)m - n = 3m = 3 + n → recall: n = 2m = 5
Nice trick! ❤
Thank you sir for sharing the good and challenging questions 🙃
You are most welcome! ❤
Go for x= 100 & y=4
It does,nt work! ❤
@@SALogics sorry overlooked one radical sign. then y=16 will do.
@ 2:46 / 10:41
There's no need to introduce an additional variable as k
2^(m) - 2^(n) = 28
2^(m + n - n) - 2^(n) = 28
2^(n + m - n) - 2^(n) = 28
[2^(n) * 2^(m - n)] - 2^(n) = 28
2^(n) * [2^(m - n) - 1] = 28
2^(n) * [2^(m - n) - 1] = 4 * 7
2^(n) * [2^(m - n) - 1] = 2^(2) * 7 → by identification
n = 2
2^(m - n) - 1 = 7
2^(m - n) = 8
2^(m - n) = 2^(3)
m - n = 3
m = 3 + n → recall: n = 2
m = 5
Nice trick! ❤
Thank you sir for sharing the good and challenging questions 🙃
You are most welcome! ❤
Go for x= 100 & y=4
It does,nt work! ❤
@@SALogics sorry overlooked one radical sign. then y=16 will do.